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How to obtain original coefficients after performing linear regression on normalized data?
Stock Price Data Manipulation for LSTMHow to use the output of GridSearch?Summary statistics by category using PythonOutlier detection with sklearnObtaining a confidence interval for the prediction of a linear regressionNeural Network Data Normalization Setupsklearn preprocessing MinMaxScalerHow to plot logistic regression decision boundary?How do I implement stochastic gradient descent correctly?How to visualize multivariate regression results by plotting plane of best fit?
$begingroup$
I am reading data from a file using pandas which looks like this:
data.head()
ldr1 ldr2 servo
0 971 956 -2
1 691 825 -105
2 841 963 -26
3 970 731 44
4 755 939 -69
I proceed to normalize this data to perform gradient descent:
my_data = (my_data - my_data.mean())/my_data.std()
my_data.head()
ldr1 ldr2 servo
0 1.419949 1.289668 0.366482
1 -0.242834 0.591311 -1.580420
2 0.647943 1.326984 -0.087165
3 1.414011 0.090200 1.235972
4 0.137231 1.199041 -0.899949
I perform multivariate regression and end up with fitted parameters on the normalized data:
Thetas: [[-3.86865143e-17, 8.47885685e-01, -5.39083511e-01]]
I would like to plot the plane of best fit on the original data and not the normalized data using the normalized thetas.
I used scipy.optimize.curve_fit to perform multivariate linear regression and come up with the optimal fitted parameters. I know that the original thetas should be close to the following:
[ 0.26654135 -0.15218007 -107.79915373]
How can I get the 'original' thetas for the original data-set in order to plot, without using Scikit-Learn?
Any suggestions will be appreciated.
machine-learning python
edited Apr 25 at 11:03
Esmailian
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
$endgroup$
add a comment |
$begingroup$
I am reading data from a file using pandas which looks like this:
data.head()
ldr1 ldr2 servo
0 971 956 -2
1 691 825 -105
2 841 963 -26
3 970 731 44
4 755 939 -69
I proceed to normalize this data to perform gradient descent:
my_data = (my_data - my_data.mean())/my_data.std()
my_data.head()
ldr1 ldr2 servo
0 1.419949 1.289668 0.366482
1 -0.242834 0.591311 -1.580420
2 0.647943 1.326984 -0.087165
3 1.414011 0.090200 1.235972
4 0.137231 1.199041 -0.899949
I perform multivariate regression and end up with fitted parameters on the normalized data:
Thetas: [[-3.86865143e-17, 8.47885685e-01, -5.39083511e-01]]
I would like to plot the plane of best fit on the original data and not the normalized data using the normalized thetas.
I used scipy.optimize.curve_fit to perform multivariate linear regression and come up with the optimal fitted parameters. I know that the original thetas should be close to the following:
[ 0.26654135 -0.15218007 -107.79915373]
How can I get the 'original' thetas for the original data-set in order to plot, without using Scikit-Learn?
Any suggestions will be appreciated.
machine-learning python
edited Apr 25 at 11:03
Esmailian
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
$endgroup$
add a comment |
$begingroup$
I am reading data from a file using pandas which looks like this:
data.head()
ldr1 ldr2 servo
0 971 956 -2
1 691 825 -105
2 841 963 -26
3 970 731 44
4 755 939 -69
I proceed to normalize this data to perform gradient descent:
my_data = (my_data - my_data.mean())/my_data.std()
my_data.head()
ldr1 ldr2 servo
0 1.419949 1.289668 0.366482
1 -0.242834 0.591311 -1.580420
2 0.647943 1.326984 -0.087165
3 1.414011 0.090200 1.235972
4 0.137231 1.199041 -0.899949
I perform multivariate regression and end up with fitted parameters on the normalized data:
Thetas: [[-3.86865143e-17, 8.47885685e-01, -5.39083511e-01]]
I would like to plot the plane of best fit on the original data and not the normalized data using the normalized thetas.
I used scipy.optimize.curve_fit to perform multivariate linear regression and come up with the optimal fitted parameters. I know that the original thetas should be close to the following:
[ 0.26654135 -0.15218007 -107.79915373]
How can I get the 'original' thetas for the original data-set in order to plot, without using Scikit-Learn?
Any suggestions will be appreciated.
machine-learning python
edited Apr 25 at 11:03
Esmailian
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
$endgroup$
I am reading data from a file using pandas which looks like this:
data.head()
ldr1 ldr2 servo
0 971 956 -2
1 691 825 -105
2 841 963 -26
3 970 731 44
4 755 939 -69
I proceed to normalize this data to perform gradient descent:
my_data = (my_data - my_data.mean())/my_data.std()
my_data.head()
ldr1 ldr2 servo
0 1.419949 1.289668 0.366482
1 -0.242834 0.591311 -1.580420
2 0.647943 1.326984 -0.087165
3 1.414011 0.090200 1.235972
4 0.137231 1.199041 -0.899949
I perform multivariate regression and end up with fitted parameters on the normalized data:
Thetas: [[-3.86865143e-17, 8.47885685e-01, -5.39083511e-01]]
I would like to plot the plane of best fit on the original data and not the normalized data using the normalized thetas.
I used scipy.optimize.curve_fit to perform multivariate linear regression and come up with the optimal fitted parameters. I know that the original thetas should be close to the following:
[ 0.26654135 -0.15218007 -107.79915373]
How can I get the 'original' thetas for the original data-set in order to plot, without using Scikit-Learn?
Any suggestions will be appreciated.
machine-learning python
machine-learning python
edited Apr 25 at 11:03
Esmailian
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
edited Apr 25 at 11:03
Esmailian
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
edited Apr 25 at 11:03
Esmailian
4,114422
edited Apr 25 at 11:03
Esmailian
4,114422
edited Apr 25 at 11:03
Esmailian
4,114422
4,114422
asked Apr 23 at 9:08
Rrz0Rrz0
1979
asked Apr 23 at 9:08
Rrz0Rrz0
1979
asked Apr 23 at 9:08
Rrz0Rrz0
1979
1979
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Coefficients of the linear regression for unnormalized features
If parameters in the normalized space are denoted as $(theta_0', theta_1', theta_2')$, parameters in the original space $(theta_0, theta_1, theta_2)$ can be derived as follows
$$beginalign*
y' &= theta_2'x_2'+theta_1'x_1'+theta_0'\
fracy-mu_Ysigma_Y &= theta_2'fracx_2 - mu_X_2sigma_X_2 + theta_1'fracx_1 - mu_X_1sigma_X_1 +theta_0' \
y &= overbraceleft(fracsigma_Ysigma_X_2theta_2'right)^theta_2x_2+ overbraceleft(fracsigma_Ysigma_X_1theta_1'right)^theta_1x_1
+ overbracesigma_Yleft(-theta_2'fracmu_X_2sigma_X_2-theta_1'fracmu_X_1sigma_X_1 + theta_0'right) + mu_Y^theta_0
endalign*$$
Generalization to D features
$$beginalign*
theta_d = left{beginmatrix
sigma_Y left(theta_0' - sum_i=1^Dtheta_i'fracmu_X_isigma_X_i right) + mu_Y& d=0\
fracsigma_Ysigma_X_dtheta_d' & d > 0
endmatrixright.
endalign*$$
A trick
For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.
For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(sigma_X_1x_1'+mu_X_1, sigma_X_2x_2'+mu_X_2, sigma_Yy'+mu_Y)$$
in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.
Note that $y'=theta_2'x_2'+theta_1'x_1'+theta_0'$ is still calculated using normalized features.
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
$endgroup$
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual onplot_surfacemaybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.
$endgroup$
– Esmailian
Apr 23 at 14:02
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
|
show 3 more comments
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
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oldest
votes
$begingroup$
Coefficients of the linear regression for unnormalized features
If parameters in the normalized space are denoted as $(theta_0', theta_1', theta_2')$, parameters in the original space $(theta_0, theta_1, theta_2)$ can be derived as follows
$$beginalign*
y' &= theta_2'x_2'+theta_1'x_1'+theta_0'\
fracy-mu_Ysigma_Y &= theta_2'fracx_2 - mu_X_2sigma_X_2 + theta_1'fracx_1 - mu_X_1sigma_X_1 +theta_0' \
y &= overbraceleft(fracsigma_Ysigma_X_2theta_2'right)^theta_2x_2+ overbraceleft(fracsigma_Ysigma_X_1theta_1'right)^theta_1x_1
+ overbracesigma_Yleft(-theta_2'fracmu_X_2sigma_X_2-theta_1'fracmu_X_1sigma_X_1 + theta_0'right) + mu_Y^theta_0
endalign*$$
Generalization to D features
$$beginalign*
theta_d = left{beginmatrix
sigma_Y left(theta_0' - sum_i=1^Dtheta_i'fracmu_X_isigma_X_i right) + mu_Y& d=0\
fracsigma_Ysigma_X_dtheta_d' & d > 0
endmatrixright.
endalign*$$
A trick
For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.
For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(sigma_X_1x_1'+mu_X_1, sigma_X_2x_2'+mu_X_2, sigma_Yy'+mu_Y)$$
in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.
Note that $y'=theta_2'x_2'+theta_1'x_1'+theta_0'$ is still calculated using normalized features.
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
$endgroup$
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual onplot_surfacemaybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.
$endgroup$
– Esmailian
Apr 23 at 14:02
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
|
show 3 more comments
$begingroup$
Coefficients of the linear regression for unnormalized features
If parameters in the normalized space are denoted as $(theta_0', theta_1', theta_2')$, parameters in the original space $(theta_0, theta_1, theta_2)$ can be derived as follows
$$beginalign*
y' &= theta_2'x_2'+theta_1'x_1'+theta_0'\
fracy-mu_Ysigma_Y &= theta_2'fracx_2 - mu_X_2sigma_X_2 + theta_1'fracx_1 - mu_X_1sigma_X_1 +theta_0' \
y &= overbraceleft(fracsigma_Ysigma_X_2theta_2'right)^theta_2x_2+ overbraceleft(fracsigma_Ysigma_X_1theta_1'right)^theta_1x_1
+ overbracesigma_Yleft(-theta_2'fracmu_X_2sigma_X_2-theta_1'fracmu_X_1sigma_X_1 + theta_0'right) + mu_Y^theta_0
endalign*$$
Generalization to D features
$$beginalign*
theta_d = left{beginmatrix
sigma_Y left(theta_0' - sum_i=1^Dtheta_i'fracmu_X_isigma_X_i right) + mu_Y& d=0\
fracsigma_Ysigma_X_dtheta_d' & d > 0
endmatrixright.
endalign*$$
A trick
For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.
For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(sigma_X_1x_1'+mu_X_1, sigma_X_2x_2'+mu_X_2, sigma_Yy'+mu_Y)$$
in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.
Note that $y'=theta_2'x_2'+theta_1'x_1'+theta_0'$ is still calculated using normalized features.
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
$endgroup$
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual onplot_surfacemaybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.
$endgroup$
– Esmailian
Apr 23 at 14:02
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
|
show 3 more comments
$begingroup$
Coefficients of the linear regression for unnormalized features
If parameters in the normalized space are denoted as $(theta_0', theta_1', theta_2')$, parameters in the original space $(theta_0, theta_1, theta_2)$ can be derived as follows
$$beginalign*
y' &= theta_2'x_2'+theta_1'x_1'+theta_0'\
fracy-mu_Ysigma_Y &= theta_2'fracx_2 - mu_X_2sigma_X_2 + theta_1'fracx_1 - mu_X_1sigma_X_1 +theta_0' \
y &= overbraceleft(fracsigma_Ysigma_X_2theta_2'right)^theta_2x_2+ overbraceleft(fracsigma_Ysigma_X_1theta_1'right)^theta_1x_1
+ overbracesigma_Yleft(-theta_2'fracmu_X_2sigma_X_2-theta_1'fracmu_X_1sigma_X_1 + theta_0'right) + mu_Y^theta_0
endalign*$$
Generalization to D features
$$beginalign*
theta_d = left{beginmatrix
sigma_Y left(theta_0' - sum_i=1^Dtheta_i'fracmu_X_isigma_X_i right) + mu_Y& d=0\
fracsigma_Ysigma_X_dtheta_d' & d > 0
endmatrixright.
endalign*$$
A trick
For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.
For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(sigma_X_1x_1'+mu_X_1, sigma_X_2x_2'+mu_X_2, sigma_Yy'+mu_Y)$$
in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.
Note that $y'=theta_2'x_2'+theta_1'x_1'+theta_0'$ is still calculated using normalized features.
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
$endgroup$
Coefficients of the linear regression for unnormalized features
If parameters in the normalized space are denoted as $(theta_0', theta_1', theta_2')$, parameters in the original space $(theta_0, theta_1, theta_2)$ can be derived as follows
$$beginalign*
y' &= theta_2'x_2'+theta_1'x_1'+theta_0'\
fracy-mu_Ysigma_Y &= theta_2'fracx_2 - mu_X_2sigma_X_2 + theta_1'fracx_1 - mu_X_1sigma_X_1 +theta_0' \
y &= overbraceleft(fracsigma_Ysigma_X_2theta_2'right)^theta_2x_2+ overbraceleft(fracsigma_Ysigma_X_1theta_1'right)^theta_1x_1
+ overbracesigma_Yleft(-theta_2'fracmu_X_2sigma_X_2-theta_1'fracmu_X_1sigma_X_1 + theta_0'right) + mu_Y^theta_0
endalign*$$
Generalization to D features
$$beginalign*
theta_d = left{beginmatrix
sigma_Y left(theta_0' - sum_i=1^Dtheta_i'fracmu_X_isigma_X_i right) + mu_Y& d=0\
fracsigma_Ysigma_X_dtheta_d' & d > 0
endmatrixright.
endalign*$$
A trick
For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.
For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(sigma_X_1x_1'+mu_X_1, sigma_X_2x_2'+mu_X_2, sigma_Yy'+mu_Y)$$
in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.
Note that $y'=theta_2'x_2'+theta_1'x_1'+theta_0'$ is still calculated using normalized features.
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
edited Apr 23 at 16:36
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
answered Apr 23 at 10:52
EsmailianEsmailian
4,114422
4,114422
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual onplot_surfacemaybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.
$endgroup$
– Esmailian
Apr 23 at 14:02
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
|
show 3 more comments
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual onplot_surfacemaybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.
$endgroup$
– Esmailian
Apr 23 at 14:02
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
$begingroup$
Thank you, this helps! I will try your suggestions and get back to you. Confirming that I understood correctly, in reference to your last equation, should I multiply the standard deviation of both normalized and original parameters and add to it the variance of the original parameter?
$endgroup$
– Rrz0
Apr 23 at 12:52
1
1
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
@Rrz0 There is no variance or std for parameters $theta$. $sigma$ and $mu$ are calculated from original data.
$endgroup$
– Esmailian
Apr 23 at 12:59
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
$begingroup$
Right, thanks for confirming. I am still not able to visualize the proper plane. I managed to convert the normalized data into the original space, however the plane appears incorrectly. I edited the question.
$endgroup$
– Rrz0
Apr 23 at 13:06
1
1
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual on
plot_surface maybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.$endgroup$
– Esmailian
Apr 23 at 14:02
$begingroup$
@Rrz0 you should plot (x_scaled, y_scaled, z_scaled), also check out the manual on
plot_surface maybe that is the problem, see whether you can plot an arbitrary surface. I suggest also implementing the first approach to better find the problem.$endgroup$
– Esmailian
Apr 23 at 14:02
1
1
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
$begingroup$
@Rrz0 remove $x_2$, it should not be included
$endgroup$
– Esmailian
Apr 23 at 14:48
|
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