How to plot logistic regression decision boundary?Decision tree or logistic regression?Contrasting logistic regression vs decision tree performance in specific exampleSimple logistic regression wrong predictionsQuestion about Logistic RegressionLogistic Regression Independent Sampleslogistic regressionWhy is the logistic regression decision boundary linear in X?Why Decision trees performs better than logistic regressionLogistic regression in pythonlogistic regression : highly sensitive model

What are the steps to solving this definite integral?

Elements other than carbon that can form many different compounds by bonding to themselves?

What term is being referred to with "reflected-sound-of-underground-spirits"?

Can I criticise the more senior developers around me for not writing clean code?

What is the most expensive material in the world that could be used to create Pun-Pun's lute?

Like totally amazing interchangeable sister outfits II: The Revenge

Is there a way to generate a list of distinct numbers such that no two subsets ever have an equal sum?

Why did C use the -> operator instead of reusing the . operator?

What happens to Mjolnir (Thor's hammer) at the end of Endgame?

Can someone publish a story that happened to you?

A ​Note ​on ​N!

Alignment of various blocks in tikz

Apply MapThread to all but one variable

Was there a Viking Exchange as well as a Columbian one?

Philosophical question on logistic regression: why isn't the optimal threshold value trained?

What happened to Captain America in Endgame?

How do I deal with a coworker that keeps asking to make small superficial changes to a report, and it is seriously triggering my anxiety?

What makes accurate emulation of old systems a difficult task?

555 timer FM transmitter

Multiple options vs single option UI

How does Captain America channel this power?

Relationship between strut and baselineskip

Can I grease a crank spindle/bracket without disassembling the crank set?

Check if a string is entirely made of the same substring



How to plot logistic regression decision boundary?


Decision tree or logistic regression?Contrasting logistic regression vs decision tree performance in specific exampleSimple logistic regression wrong predictionsQuestion about Logistic RegressionLogistic Regression Independent Sampleslogistic regressionWhy is the logistic regression decision boundary linear in X?Why Decision trees performs better than logistic regressionLogistic regression in pythonlogistic regression : highly sensitive model













6












$begingroup$


I am running logistic regression on a small dataset which looks like this:



enter image description here



After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



enter image description here



Extracting data



clear all; close all; clc;

alpha = 0.01;
num_iters = 1000;

%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);

x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

x = [x1 x2]'; % X

subplot(2,2,1);
dat = [dat1 dat2]'; % Y

scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);


Computing Cost, Gradient and plotting



% Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);

% Add intercept term to x and X_test
x = [ones(m, 1) x];

% Initialize fitting parameters
theta = zeros(n + 1, 1);
%initial_theta = [0.2; 0.2];

J_history = zeros(num_iters, 1);

plot_x = [min(x(:,2))-2, max(x(:,2))+2]

for iter = 1:num_iters
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(theta, x, classdata);
theta = theta - alpha * grad;
J_history(iter) = cost;

fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


subplot(2,2,2);
hold on; grid on;
plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
xlabel('Iterations')
ylabel('MSE')
drawnow

subplot(2,2,3);
grid on;
plot3(theta(1), theta(2), J_history(iter),'o')
title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
xlabel('Tita0')
ylabel('Tita1')
zlabel('Cost')
hold on;
drawnow

subplot(2,2,1);
grid on;
% Calculate the decision boundary line
plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
% Plot, and adjust axes for better viewing
plot(plot_x, plot_y)
hold on;
drawnow

end

fprintf('Cost at initial theta (zeros): %fn', cost);
fprintf('Gradient at initial theta (zeros): n');
fprintf(' %f n', grad);


The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.




logistic_costFunction.m



function [J, grad] = logistic_costFunction(theta, X, y)

% Initialize some useful values
m = length(y); % number of training examples

grad = zeros(size(theta));

h = sigmoid(X * theta);
J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

for i = 1 : size(theta, 1)
grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
end

end


EDIT:



As per the below answer by @Esmailian, now I have something like this:



[m, n] = size(x);

x1_class = [ones(m, 1) x1' dat1'];
x2_class = [ones(m, 1) x2' dat2'];

x = [x1_class ; x2_class]









share|improve this question











$endgroup$
















    6












    $begingroup$


    I am running logistic regression on a small dataset which looks like this:



    enter image description here



    After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



    Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



    enter image description here



    Extracting data



    clear all; close all; clc;

    alpha = 0.01;
    num_iters = 1000;

    %% Plotting data
    x1 = linspace(0,3,50);
    mqtrue = 5;
    cqtrue = 30;
    dat1 = mqtrue*x1+5*randn(1,50);

    x2 = linspace(7,10,50);
    dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

    x = [x1 x2]'; % X

    subplot(2,2,1);
    dat = [dat1 dat2]'; % Y

    scatter(x1, dat1); hold on;
    scatter(x2, dat2, '*'); hold on;
    classdata = (dat>40);


    Computing Cost, Gradient and plotting



    % Setup the data matrix appropriately, and add ones for the intercept term
    [m, n] = size(x);

    % Add intercept term to x and X_test
    x = [ones(m, 1) x];

    % Initialize fitting parameters
    theta = zeros(n + 1, 1);
    %initial_theta = [0.2; 0.2];

    J_history = zeros(num_iters, 1);

    plot_x = [min(x(:,2))-2, max(x(:,2))+2]

    for iter = 1:num_iters
    % Compute and display initial cost and gradient
    [cost, grad] = logistic_costFunction(theta, x, classdata);
    theta = theta - alpha * grad;
    J_history(iter) = cost;

    fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


    subplot(2,2,2);
    hold on; grid on;
    plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
    xlabel('Iterations')
    ylabel('MSE')
    drawnow

    subplot(2,2,3);
    grid on;
    plot3(theta(1), theta(2), J_history(iter),'o')
    title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
    xlabel('Tita0')
    ylabel('Tita1')
    zlabel('Cost')
    hold on;
    drawnow

    subplot(2,2,1);
    grid on;
    % Calculate the decision boundary line
    plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
    % Plot, and adjust axes for better viewing
    plot(plot_x, plot_y)
    hold on;
    drawnow

    end

    fprintf('Cost at initial theta (zeros): %fn', cost);
    fprintf('Gradient at initial theta (zeros): n');
    fprintf(' %f n', grad);


    The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.




    logistic_costFunction.m



    function [J, grad] = logistic_costFunction(theta, X, y)

    % Initialize some useful values
    m = length(y); % number of training examples

    grad = zeros(size(theta));

    h = sigmoid(X * theta);
    J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

    for i = 1 : size(theta, 1)
    grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
    end

    end


    EDIT:



    As per the below answer by @Esmailian, now I have something like this:



    [m, n] = size(x);

    x1_class = [ones(m, 1) x1' dat1'];
    x2_class = [ones(m, 1) x2' dat2'];

    x = [x1_class ; x2_class]









    share|improve this question











    $endgroup$














      6












      6








      6


      1



      $begingroup$


      I am running logistic regression on a small dataset which looks like this:



      enter image description here



      After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



      Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



      enter image description here



      Extracting data



      clear all; close all; clc;

      alpha = 0.01;
      num_iters = 1000;

      %% Plotting data
      x1 = linspace(0,3,50);
      mqtrue = 5;
      cqtrue = 30;
      dat1 = mqtrue*x1+5*randn(1,50);

      x2 = linspace(7,10,50);
      dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

      x = [x1 x2]'; % X

      subplot(2,2,1);
      dat = [dat1 dat2]'; % Y

      scatter(x1, dat1); hold on;
      scatter(x2, dat2, '*'); hold on;
      classdata = (dat>40);


      Computing Cost, Gradient and plotting



      % Setup the data matrix appropriately, and add ones for the intercept term
      [m, n] = size(x);

      % Add intercept term to x and X_test
      x = [ones(m, 1) x];

      % Initialize fitting parameters
      theta = zeros(n + 1, 1);
      %initial_theta = [0.2; 0.2];

      J_history = zeros(num_iters, 1);

      plot_x = [min(x(:,2))-2, max(x(:,2))+2]

      for iter = 1:num_iters
      % Compute and display initial cost and gradient
      [cost, grad] = logistic_costFunction(theta, x, classdata);
      theta = theta - alpha * grad;
      J_history(iter) = cost;

      fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


      subplot(2,2,2);
      hold on; grid on;
      plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
      xlabel('Iterations')
      ylabel('MSE')
      drawnow

      subplot(2,2,3);
      grid on;
      plot3(theta(1), theta(2), J_history(iter),'o')
      title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
      xlabel('Tita0')
      ylabel('Tita1')
      zlabel('Cost')
      hold on;
      drawnow

      subplot(2,2,1);
      grid on;
      % Calculate the decision boundary line
      plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
      % Plot, and adjust axes for better viewing
      plot(plot_x, plot_y)
      hold on;
      drawnow

      end

      fprintf('Cost at initial theta (zeros): %fn', cost);
      fprintf('Gradient at initial theta (zeros): n');
      fprintf(' %f n', grad);


      The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.




      logistic_costFunction.m



      function [J, grad] = logistic_costFunction(theta, X, y)

      % Initialize some useful values
      m = length(y); % number of training examples

      grad = zeros(size(theta));

      h = sigmoid(X * theta);
      J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

      for i = 1 : size(theta, 1)
      grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
      end

      end


      EDIT:



      As per the below answer by @Esmailian, now I have something like this:



      [m, n] = size(x);

      x1_class = [ones(m, 1) x1' dat1'];
      x2_class = [ones(m, 1) x2' dat2'];

      x = [x1_class ; x2_class]









      share|improve this question











      $endgroup$




      I am running logistic regression on a small dataset which looks like this:



      enter image description here



      After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.



      Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.



      enter image description here



      Extracting data



      clear all; close all; clc;

      alpha = 0.01;
      num_iters = 1000;

      %% Plotting data
      x1 = linspace(0,3,50);
      mqtrue = 5;
      cqtrue = 30;
      dat1 = mqtrue*x1+5*randn(1,50);

      x2 = linspace(7,10,50);
      dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

      x = [x1 x2]'; % X

      subplot(2,2,1);
      dat = [dat1 dat2]'; % Y

      scatter(x1, dat1); hold on;
      scatter(x2, dat2, '*'); hold on;
      classdata = (dat>40);


      Computing Cost, Gradient and plotting



      % Setup the data matrix appropriately, and add ones for the intercept term
      [m, n] = size(x);

      % Add intercept term to x and X_test
      x = [ones(m, 1) x];

      % Initialize fitting parameters
      theta = zeros(n + 1, 1);
      %initial_theta = [0.2; 0.2];

      J_history = zeros(num_iters, 1);

      plot_x = [min(x(:,2))-2, max(x(:,2))+2]

      for iter = 1:num_iters
      % Compute and display initial cost and gradient
      [cost, grad] = logistic_costFunction(theta, x, classdata);
      theta = theta - alpha * grad;
      J_history(iter) = cost;

      fprintf('Iteration #%d - Cost = %d... rn',iter, cost);


      subplot(2,2,2);
      hold on; grid on;
      plot(iter, J_history(iter), '.r'); title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
      xlabel('Iterations')
      ylabel('MSE')
      drawnow

      subplot(2,2,3);
      grid on;
      plot3(theta(1), theta(2), J_history(iter),'o')
      title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
      xlabel('Tita0')
      ylabel('Tita1')
      zlabel('Cost')
      hold on;
      drawnow

      subplot(2,2,1);
      grid on;
      % Calculate the decision boundary line
      plot_y = theta(2).*plot_x + theta(1); % <--- Boundary line
      % Plot, and adjust axes for better viewing
      plot(plot_x, plot_y)
      hold on;
      drawnow

      end

      fprintf('Cost at initial theta (zeros): %fn', cost);
      fprintf('Gradient at initial theta (zeros): n');
      fprintf(' %f n', grad);


      The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.




      logistic_costFunction.m



      function [J, grad] = logistic_costFunction(theta, X, y)

      % Initialize some useful values
      m = length(y); % number of training examples

      grad = zeros(size(theta));

      h = sigmoid(X * theta);
      J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

      for i = 1 : size(theta, 1)
      grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
      end

      end


      EDIT:



      As per the below answer by @Esmailian, now I have something like this:



      [m, n] = size(x);

      x1_class = [ones(m, 1) x1' dat1'];
      x2_class = [ones(m, 1) x2' dat2'];

      x = [x1_class ; x2_class]






      machine-learning logistic-regression






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 20 at 9:46







      Rrz0

















      asked Apr 19 at 8:24









      Rrz0Rrz0

      1979




      1979




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Regarding the code



          1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.


          2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).


          Decision boundary



          Assuming that data is $boldsymbolx=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $boldsymboltheta=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2$$
          which can be drawn as a segment by connecting two points $(0, - fractheta_0theta_2)$ and $(- fractheta_0theta_1, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-fractheta_0theta_1$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbolx$ that satisfy
          $$Bbb P(y=1|boldsymbolx)=Bbb P(y=0|boldsymbolx) = frac12.$$
          Given
          $$Bbb P(y=1|boldsymbolx)=frac11+e^-boldsymboltheta^tboldsymbolx_+$$
          where $boldsymboltheta=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbolx$ is extended to $boldsymbolx_+=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymboltheta^tboldsymbolx_+=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac12 \
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          endalign*$$

          For two dimensional data $boldsymbolx=(x_1, x_2)$ we have
          $$beginalign*
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2
          endalign*$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$wBbb P(y=1|boldsymbolx) = Bbb P(y=0|boldsymbolx) = fracww+1$$



          For example, $w=2$ means point $boldsymbolx$ will be assigned to positive class if $Bbb P(y=1|boldsymbolx) > 0.33$ (or equivalently if $Bbb P(y=0|boldsymbolx) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac1w+1 \
          &Rightarrow e^-boldsymboltheta^tboldsymbolx_+ = w\
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = -textlnw\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -textlnw
          endalign*$$






          share|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
            $endgroup$
            – Rrz0
            Apr 20 at 8:53










          • $begingroup$
            Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
            $endgroup$
            – Rrz0
            Apr 20 at 9:06










          • $begingroup$
            @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
            $endgroup$
            – Esmailian
            Apr 20 at 9:11











          • $begingroup$
            Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
            $endgroup$
            – Rrz0
            Apr 20 at 9:15







          • 1




            $begingroup$
            @Rrz0 I spotted the problem. Check out the update.
            $endgroup$
            – Esmailian
            Apr 20 at 9:29


















          7












          $begingroup$

          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[x = Subdivide[0, 3, 50],
          dat1 = Transpose@x, mqtrue x + 5 RandomReal[1, Length@x];
          ];
          With[x = Subdivide[7, 10, 50],
          dat2 = Transpose@x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x];
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, #, Boole@Thread[#[[All, 2]] > 40]] & /@ dat1, dat2
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[dat = Join[dat1, dat2],
          model =
          LogitModelFit[
          MapThread[Append, dat, Boole@Thread[dat[[All, 2]] > 40]],
          x, y, x, y]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, MinMax /@ Transpose@Join[dat1, dat2], x, y]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          x, u, model["Function"][x, u], u, 0, 80, PlotStyle -> Orange],
          ParametricPlot3D[
          u, y, model["Function"][u, y], u, 0, 10, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ x, y, model["Function"][x, y]
          ],
          x, 6, Style["x", Orange, Bold], 0, 10, Appearance -> "Labeled",
          y, 40, Style["y", Purple, Bold], 0, 80, Appearance -> "Labeled"
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], x, 0, 10, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          y, 40, 0, yMax, Appearance -> "Labeled"
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, x, 0, 10, PlotStyle -> Red]
          ],
          p, .5, 0, 1, Appearance -> "Labeled"
          ]


          enter image description here



          Hope this helps.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            Apr 19 at 19:38






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            Apr 19 at 22:56










          • $begingroup$
            Thank you for your very helpful insight and excellent visualizations.
            $endgroup$
            – Rrz0
            Apr 20 at 8:54












          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "557"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f49573%2fhow-to-plot-logistic-regression-decision-boundary%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Regarding the code



          1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.


          2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).


          Decision boundary



          Assuming that data is $boldsymbolx=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $boldsymboltheta=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2$$
          which can be drawn as a segment by connecting two points $(0, - fractheta_0theta_2)$ and $(- fractheta_0theta_1, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-fractheta_0theta_1$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbolx$ that satisfy
          $$Bbb P(y=1|boldsymbolx)=Bbb P(y=0|boldsymbolx) = frac12.$$
          Given
          $$Bbb P(y=1|boldsymbolx)=frac11+e^-boldsymboltheta^tboldsymbolx_+$$
          where $boldsymboltheta=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbolx$ is extended to $boldsymbolx_+=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymboltheta^tboldsymbolx_+=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac12 \
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          endalign*$$

          For two dimensional data $boldsymbolx=(x_1, x_2)$ we have
          $$beginalign*
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2
          endalign*$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$wBbb P(y=1|boldsymbolx) = Bbb P(y=0|boldsymbolx) = fracww+1$$



          For example, $w=2$ means point $boldsymbolx$ will be assigned to positive class if $Bbb P(y=1|boldsymbolx) > 0.33$ (or equivalently if $Bbb P(y=0|boldsymbolx) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac1w+1 \
          &Rightarrow e^-boldsymboltheta^tboldsymbolx_+ = w\
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = -textlnw\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -textlnw
          endalign*$$






          share|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
            $endgroup$
            – Rrz0
            Apr 20 at 8:53










          • $begingroup$
            Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
            $endgroup$
            – Rrz0
            Apr 20 at 9:06










          • $begingroup$
            @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
            $endgroup$
            – Esmailian
            Apr 20 at 9:11











          • $begingroup$
            Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
            $endgroup$
            – Rrz0
            Apr 20 at 9:15







          • 1




            $begingroup$
            @Rrz0 I spotted the problem. Check out the update.
            $endgroup$
            – Esmailian
            Apr 20 at 9:29















          3












          $begingroup$

          Regarding the code



          1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.


          2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).


          Decision boundary



          Assuming that data is $boldsymbolx=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $boldsymboltheta=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2$$
          which can be drawn as a segment by connecting two points $(0, - fractheta_0theta_2)$ and $(- fractheta_0theta_1, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-fractheta_0theta_1$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbolx$ that satisfy
          $$Bbb P(y=1|boldsymbolx)=Bbb P(y=0|boldsymbolx) = frac12.$$
          Given
          $$Bbb P(y=1|boldsymbolx)=frac11+e^-boldsymboltheta^tboldsymbolx_+$$
          where $boldsymboltheta=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbolx$ is extended to $boldsymbolx_+=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymboltheta^tboldsymbolx_+=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac12 \
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          endalign*$$

          For two dimensional data $boldsymbolx=(x_1, x_2)$ we have
          $$beginalign*
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2
          endalign*$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$wBbb P(y=1|boldsymbolx) = Bbb P(y=0|boldsymbolx) = fracww+1$$



          For example, $w=2$ means point $boldsymbolx$ will be assigned to positive class if $Bbb P(y=1|boldsymbolx) > 0.33$ (or equivalently if $Bbb P(y=0|boldsymbolx) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac1w+1 \
          &Rightarrow e^-boldsymboltheta^tboldsymbolx_+ = w\
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = -textlnw\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -textlnw
          endalign*$$






          share|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
            $endgroup$
            – Rrz0
            Apr 20 at 8:53










          • $begingroup$
            Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
            $endgroup$
            – Rrz0
            Apr 20 at 9:06










          • $begingroup$
            @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
            $endgroup$
            – Esmailian
            Apr 20 at 9:11











          • $begingroup$
            Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
            $endgroup$
            – Rrz0
            Apr 20 at 9:15







          • 1




            $begingroup$
            @Rrz0 I spotted the problem. Check out the update.
            $endgroup$
            – Esmailian
            Apr 20 at 9:29













          3












          3








          3





          $begingroup$

          Regarding the code



          1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.


          2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).


          Decision boundary



          Assuming that data is $boldsymbolx=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $boldsymboltheta=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2$$
          which can be drawn as a segment by connecting two points $(0, - fractheta_0theta_2)$ and $(- fractheta_0theta_1, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-fractheta_0theta_1$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbolx$ that satisfy
          $$Bbb P(y=1|boldsymbolx)=Bbb P(y=0|boldsymbolx) = frac12.$$
          Given
          $$Bbb P(y=1|boldsymbolx)=frac11+e^-boldsymboltheta^tboldsymbolx_+$$
          where $boldsymboltheta=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbolx$ is extended to $boldsymbolx_+=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymboltheta^tboldsymbolx_+=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac12 \
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          endalign*$$

          For two dimensional data $boldsymbolx=(x_1, x_2)$ we have
          $$beginalign*
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2
          endalign*$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$wBbb P(y=1|boldsymbolx) = Bbb P(y=0|boldsymbolx) = fracww+1$$



          For example, $w=2$ means point $boldsymbolx$ will be assigned to positive class if $Bbb P(y=1|boldsymbolx) > 0.33$ (or equivalently if $Bbb P(y=0|boldsymbolx) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac1w+1 \
          &Rightarrow e^-boldsymboltheta^tboldsymbolx_+ = w\
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = -textlnw\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -textlnw
          endalign*$$






          share|improve this answer











          $endgroup$



          Regarding the code



          1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.


          2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).


          Decision boundary



          Assuming that data is $boldsymbolx=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $boldsymboltheta=(theta_0, theta_1,theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary:
          $$x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2$$
          which can be drawn as a segment by connecting two points $(0, - fractheta_0theta_2)$ and $(- fractheta_0theta_1, 0)$.
          However, if $theta_2=0$, the line would be $x_1=-fractheta_0theta_1$.



          Where this comes from?



          Decision boundary of Logistic regression is the set of all points $boldsymbolx$ that satisfy
          $$Bbb P(y=1|boldsymbolx)=Bbb P(y=0|boldsymbolx) = frac12.$$
          Given
          $$Bbb P(y=1|boldsymbolx)=frac11+e^-boldsymboltheta^tboldsymbolx_+$$
          where $boldsymboltheta=(theta_0, theta_1,cdots,theta_d)$, and $boldsymbolx$ is extended to $boldsymbolx_+=(1, x_1, cdots, x_d)$ for the sake of readability to have$$boldsymboltheta^tboldsymbolx_+=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
          decision boundary can be derived as follows
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac12 \
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = 0\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
          endalign*$$

          For two dimensional data $boldsymbolx=(x_1, x_2)$ we have
          $$beginalign*
          & theta_0 + theta_1 x_1+theta_2 x_2 = 0 \
          & Rightarrow x_2 = -fractheta_1theta_2 x_1 - fractheta_0theta_2
          endalign*$$

          which is the separation line that should be drawn in $(x_1, x_2)$ plane.



          Weighted decision boundary



          If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary:
          $$wBbb P(y=1|boldsymbolx) = Bbb P(y=0|boldsymbolx) = fracww+1$$



          For example, $w=2$ means point $boldsymbolx$ will be assigned to positive class if $Bbb P(y=1|boldsymbolx) > 0.33$ (or equivalently if $Bbb P(y=0|boldsymbolx) < 0.66$), which implies favoring the positive class (increasing the true positive rate).



          Here is the line for this general case:
          $$beginalign*
          &frac11+e^-boldsymboltheta^tboldsymbolx_+ = frac1w+1 \
          &Rightarrow e^-boldsymboltheta^tboldsymbolx_+ = w\
          &Rightarrow boldsymboltheta^tboldsymbolx_+ = -textlnw\
          &Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -textlnw
          endalign*$$







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 20 at 13:17

























          answered Apr 19 at 11:10









          EsmailianEsmailian

          3,896421




          3,896421











          • $begingroup$
            Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
            $endgroup$
            – Rrz0
            Apr 20 at 8:53










          • $begingroup$
            Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
            $endgroup$
            – Rrz0
            Apr 20 at 9:06










          • $begingroup$
            @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
            $endgroup$
            – Esmailian
            Apr 20 at 9:11











          • $begingroup$
            Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
            $endgroup$
            – Rrz0
            Apr 20 at 9:15







          • 1




            $begingroup$
            @Rrz0 I spotted the problem. Check out the update.
            $endgroup$
            – Esmailian
            Apr 20 at 9:29
















          • $begingroup$
            Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
            $endgroup$
            – Rrz0
            Apr 20 at 8:53










          • $begingroup$
            Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
            $endgroup$
            – Rrz0
            Apr 20 at 9:06










          • $begingroup$
            @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
            $endgroup$
            – Esmailian
            Apr 20 at 9:11











          • $begingroup$
            Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
            $endgroup$
            – Rrz0
            Apr 20 at 9:15







          • 1




            $begingroup$
            @Rrz0 I spotted the problem. Check out the update.
            $endgroup$
            – Esmailian
            Apr 20 at 9:29















          $begingroup$
          Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
          $endgroup$
          – Rrz0
          Apr 20 at 8:53




          $begingroup$
          Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above?
          $endgroup$
          – Rrz0
          Apr 20 at 8:53












          $begingroup$
          Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
          $endgroup$
          – Rrz0
          Apr 20 at 9:06




          $begingroup$
          Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1]
          $endgroup$
          – Rrz0
          Apr 20 at 9:06












          $begingroup$
          @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
          $endgroup$
          – Esmailian
          Apr 20 at 9:11





          $begingroup$
          @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'.
          $endgroup$
          – Esmailian
          Apr 20 at 9:11













          $begingroup$
          Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
          $endgroup$
          – Rrz0
          Apr 20 at 9:15





          $begingroup$
          Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3]
          $endgroup$
          – Rrz0
          Apr 20 at 9:15





          1




          1




          $begingroup$
          @Rrz0 I spotted the problem. Check out the update.
          $endgroup$
          – Esmailian
          Apr 20 at 9:29




          $begingroup$
          @Rrz0 I spotted the problem. Check out the update.
          $endgroup$
          – Esmailian
          Apr 20 at 9:29











          7












          $begingroup$

          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[x = Subdivide[0, 3, 50],
          dat1 = Transpose@x, mqtrue x + 5 RandomReal[1, Length@x];
          ];
          With[x = Subdivide[7, 10, 50],
          dat2 = Transpose@x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x];
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, #, Boole@Thread[#[[All, 2]] > 40]] & /@ dat1, dat2
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[dat = Join[dat1, dat2],
          model =
          LogitModelFit[
          MapThread[Append, dat, Boole@Thread[dat[[All, 2]] > 40]],
          x, y, x, y]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, MinMax /@ Transpose@Join[dat1, dat2], x, y]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          x, u, model["Function"][x, u], u, 0, 80, PlotStyle -> Orange],
          ParametricPlot3D[
          u, y, model["Function"][u, y], u, 0, 10, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ x, y, model["Function"][x, y]
          ],
          x, 6, Style["x", Orange, Bold], 0, 10, Appearance -> "Labeled",
          y, 40, Style["y", Purple, Bold], 0, 80, Appearance -> "Labeled"
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], x, 0, 10, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          y, 40, 0, yMax, Appearance -> "Labeled"
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, x, 0, 10, PlotStyle -> Red]
          ],
          p, .5, 0, 1, Appearance -> "Labeled"
          ]


          enter image description here



          Hope this helps.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            Apr 19 at 19:38






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            Apr 19 at 22:56










          • $begingroup$
            Thank you for your very helpful insight and excellent visualizations.
            $endgroup$
            – Rrz0
            Apr 20 at 8:54
















          7












          $begingroup$

          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[x = Subdivide[0, 3, 50],
          dat1 = Transpose@x, mqtrue x + 5 RandomReal[1, Length@x];
          ];
          With[x = Subdivide[7, 10, 50],
          dat2 = Transpose@x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x];
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, #, Boole@Thread[#[[All, 2]] > 40]] & /@ dat1, dat2
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[dat = Join[dat1, dat2],
          model =
          LogitModelFit[
          MapThread[Append, dat, Boole@Thread[dat[[All, 2]] > 40]],
          x, y, x, y]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, MinMax /@ Transpose@Join[dat1, dat2], x, y]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          x, u, model["Function"][x, u], u, 0, 80, PlotStyle -> Orange],
          ParametricPlot3D[
          u, y, model["Function"][u, y], u, 0, 10, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ x, y, model["Function"][x, y]
          ],
          x, 6, Style["x", Orange, Bold], 0, 10, Appearance -> "Labeled",
          y, 40, Style["y", Purple, Bold], 0, 80, Appearance -> "Labeled"
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], x, 0, 10, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          y, 40, 0, yMax, Appearance -> "Labeled"
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, x, 0, 10, PlotStyle -> Red]
          ],
          p, .5, 0, 1, Appearance -> "Labeled"
          ]


          enter image description here



          Hope this helps.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            Apr 19 at 19:38






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            Apr 19 at 22:56










          • $begingroup$
            Thank you for your very helpful insight and excellent visualizations.
            $endgroup$
            – Rrz0
            Apr 20 at 8:54














          7












          7








          7





          $begingroup$

          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[x = Subdivide[0, 3, 50],
          dat1 = Transpose@x, mqtrue x + 5 RandomReal[1, Length@x];
          ];
          With[x = Subdivide[7, 10, 50],
          dat2 = Transpose@x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x];
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, #, Boole@Thread[#[[All, 2]] > 40]] & /@ dat1, dat2
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[dat = Join[dat1, dat2],
          model =
          LogitModelFit[
          MapThread[Append, dat, Boole@Thread[dat[[All, 2]] > 40]],
          x, y, x, y]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, MinMax /@ Transpose@Join[dat1, dat2], x, y]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          x, u, model["Function"][x, u], u, 0, 80, PlotStyle -> Orange],
          ParametricPlot3D[
          u, y, model["Function"][u, y], u, 0, 10, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ x, y, model["Function"][x, y]
          ],
          x, 6, Style["x", Orange, Bold], 0, 10, Appearance -> "Labeled",
          y, 40, Style["y", Purple, Bold], 0, 80, Appearance -> "Labeled"
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], x, 0, 10, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          y, 40, 0, yMax, Appearance -> "Labeled"
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, x, 0, 10, PlotStyle -> Red]
          ],
          p, .5, 0, 1, Appearance -> "Labeled"
          ]


          enter image description here



          Hope this helps.






          share|improve this answer











          $endgroup$



          Your decision boundary is a surface in 3D as your points are in 2D.



          With Wolfram Language



          Create the data sets.



          mqtrue = 5;
          cqtrue = 30;
          With[x = Subdivide[0, 3, 50],
          dat1 = Transpose@x, mqtrue x + 5 RandomReal[1, Length@x];
          ];
          With[x = Subdivide[7, 10, 50],
          dat2 = Transpose@x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x];
          ];


          View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.



          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]



          Mathematica graphics




          I Append the response variable to the data.



          datPlot =
          ListPointPlot3D[
          MapThread[Append, #, Boole@Thread[#[[All, 2]] > 40]] & /@ dat1, dat2
          ]



          enter image description here




          Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.



          With[dat = Join[dat1, dat2],
          model =
          LogitModelFit[
          MapThread[Append, dat, Boole@Thread[dat[[All, 2]] > 40]],
          x, y, x, y]
          ]



          Mathematica graphics




          From the FittedModel "Properties" we need "Function".



          model["Properties"]



          AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries,
          AIC, DevianceTableEntries, ParameterConfidenceRegion,
          AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors,
          BasisFunctions, DevianceTableResidualDeviances, ParameterPValues,
          BestFit, EfronPseudoRSquared, ParameterTable,
          BestFitParameters, EstimatedDispersion, ParameterTableEntries,
          BIC, FitResiduals, ParameterZStatistics,
          CookDistances, Function, PearsonChiSquare,
          CorrelationMatrix, HatDiagonal, PearsonResiduals,
          CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse,
          CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties,
          CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance,
          Data, LinearPredictor, ResidualDegreesOfFreedom,
          DesignMatrix, LogLikelihood, Response,
          DevianceResiduals, NullDeviance, StandardizedDevianceResiduals,
          Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals,
          DevianceTable, ParameterConfidenceIntervals, WorkingResiduals,
          DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable




          model["Function"]



          Mathematica graphics




          Use this for prediction



          model["Function"][8, 54]



          0.0196842



          and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D



          modelPlot =
          Show[
          datPlot,
          Plot3D[
          model["Function"][x, y],
          Evaluate[
          Sequence @@
          MapThread[Prepend, MinMax /@ Transpose@Join[dat1, dat2], x, y]],
          Mesh -> None,
          PlotStyle -> Opacity[.25, Green],
          PlotPoints -> 30
          ]
          ]


          enter image description here



          With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.



          Manipulate[
          Show[
          modelPlot,
          ParametricPlot3D[
          x, u, model["Function"][x, u], u, 0, 80, PlotStyle -> Orange],
          ParametricPlot3D[
          u, y, model["Function"][u, y], u, 0, 10, PlotStyle -> Purple],
          PlotLabel ->
          StringTemplate["model[`1`, `2`] = `3`"] @@ x, y, model["Function"][x, y]
          ],
          x, 6, Style["x", Orange, Bold], 0, 10, Appearance -> "Labeled",
          y, 40, Style["y", Purple, Bold], 0, 80, Appearance -> "Labeled"
          ]


          enter image description here



          You can also project contours of this surface into 2D (Plot). For example, keeping y fixed and letting x vary.



          yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
          Manipulate[
          Show[
          ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers",
          PlotTheme -> "Detailed"],
          Plot[yMax model["Function"][x, y], x, 0, 10, PlotStyle -> Purple,
          Exclusions -> None]
          ],
          y, 40, 0, yMax, Appearance -> "Labeled"
          ]


          enter image description here



          Update



          You can also plot contours of the probability in 2D.



          plot = ListPlot[dat1, dat2, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

          Manipulate[
          db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
          Show[
          plot,
          Plot[db, x, 0, 10, PlotStyle -> Red]
          ],
          p, .5, 0, 1, Appearance -> "Labeled"
          ]


          enter image description here



          Hope this helps.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 19 at 22:56

























          answered Apr 19 at 13:55









          EdmundEdmund

          280311




          280311











          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            Apr 19 at 19:38






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            Apr 19 at 22:56










          • $begingroup$
            Thank you for your very helpful insight and excellent visualizations.
            $endgroup$
            – Rrz0
            Apr 20 at 8:54

















          • $begingroup$
            Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
            $endgroup$
            – Esmailian
            Apr 19 at 19:38






          • 1




            $begingroup$
            @Esmailian See update.
            $endgroup$
            – Edmund
            Apr 19 at 22:56










          • $begingroup$
            Thank you for your very helpful insight and excellent visualizations.
            $endgroup$
            – Rrz0
            Apr 20 at 8:54
















          $begingroup$
          Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
          $endgroup$
          – Esmailian
          Apr 19 at 19:38




          $begingroup$
          Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(boldsymbolx,P(y=1|boldsymbolx))$, as you have beautifully illustrated, needs $d+1$ for visualization.
          $endgroup$
          – Esmailian
          Apr 19 at 19:38




          1




          1




          $begingroup$
          @Esmailian See update.
          $endgroup$
          – Edmund
          Apr 19 at 22:56




          $begingroup$
          @Esmailian See update.
          $endgroup$
          – Edmund
          Apr 19 at 22:56












          $begingroup$
          Thank you for your very helpful insight and excellent visualizations.
          $endgroup$
          – Rrz0
          Apr 20 at 8:54





          $begingroup$
          Thank you for your very helpful insight and excellent visualizations.
          $endgroup$
          – Rrz0
          Apr 20 at 8:54


















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Data Science Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f49573%2fhow-to-plot-logistic-regression-decision-boundary%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Wikipedia:Vital articles Мазмуну Biography - Өмүр баян Philosophy and psychology - Философия жана психология Religion - Дин Social sciences - Коомдук илимдер Language and literature - Тил жана адабият Science - Илим Technology - Технология Arts and recreation - Искусство жана эс алуу History and geography - Тарых жана география Навигация менюсу

          Bruxelas-Capital Índice Historia | Composición | Situación lingüística | Clima | Cidades irmandadas | Notas | Véxase tamén | Menú de navegacióneO uso das linguas en Bruxelas e a situación do neerlandés"Rexión de Bruxelas Capital"o orixinalSitio da rexiónPáxina de Bruselas no sitio da Oficina de Promoción Turística de Valonia e BruxelasMapa Interactivo da Rexión de Bruxelas-CapitaleeWorldCat332144929079854441105155190212ID28008674080552-90000 0001 0666 3698n94104302ID540940339365017018237

          What should I write in an apology letter, since I have decided not to join a company after accepting an offer letterShould I keep looking after accepting a job offer?What should I do when I've been verbally told I would get an offer letter, but still haven't gotten one after 4 weeks?Do I accept an offer from a company that I am not likely to join?New job hasn't confirmed starting date and I want to give current employer as much notice as possibleHow should I address my manager in my resignation letter?HR delayed background verification, now jobless as resignedNo email communication after accepting a formal written offer. How should I phrase the call?What should I do if after receiving a verbal offer letter I am informed that my written job offer is put on hold due to some internal issues?Should I inform the current employer that I am about to resign within 1-2 weeks since I have signed the offer letter and waiting for visa?What company will do, if I send their offer letter to another company