Select row of data if next row contains zeroData Table Manipulation in MathematicaData Table Manipulation in Mathematica: Step 2Using levels in MapIndexedCompute the evolution of percentages from a data fileGenerate a new listSort columns in a TableFormmaking two columns from 1D-horizontal dataLoading and formatting data for plotting standard deviation bars on top of dataSelect rows of data that satisfies both conditions simultaneouslySelecting matrices based on existence of unique column and row of all equal terms

Is it advisable to add a location heads-up when a scene changes in a novel?

How can powerful telekinesis avoid violating Newton's 3rd Law?

Why would a home insurer offer a discount based on credit score?

What did the 8086 (and 8088) do upon encountering an illegal instruction?

Is fission/fusion to iron the most efficient way to convert mass to energy?

How to import .txt file with missing data?

Keeping track of theme when improvising

Approach sick days in feedback meeting

Is the first of the 10 Commandments considered a mitzvah?

As easy as Three, Two, One... How fast can you go from Five to Four?

David slept with Bathsheba because she was pure?? What does that mean?

If absolute velocity does not exist, how can we say a rocket accelerates in empty space?

What's the difference between DHCP and NAT? Are they mutually exclusive?

Am I allowed to determine tenets of my contract as a warlock?

Was the Lonely Mountain, where Smaug lived, a volcano?

Undocumented incompatibility between changes and siunitx?

Part of my house is inexplicably gone

Do they make "karaoke" versions of concertos for solo practice?

How do I properly use a function under a class?

Idiom for 'person who gets violent when drunk"

What class is best to play when a level behind the rest of the party?

Why did the Death Eaters wait to reopen the Chamber of Secrets?

Can I attach a DC blower to intake manifold of my 150CC Yamaha FZS FI engine?

Does the UK delegate some immigration control to the Republic of Ireland?



Select row of data if next row contains zero


Data Table Manipulation in MathematicaData Table Manipulation in Mathematica: Step 2Using levels in MapIndexedCompute the evolution of percentages from a data fileGenerate a new listSort columns in a TableFormmaking two columns from 1D-horizontal dataLoading and formatting data for plotting standard deviation bars on top of dataSelect rows of data that satisfies both conditions simultaneouslySelecting matrices based on existence of unique column and row of all equal terms













9












$begingroup$


I am a new user of Mathematica, and just faced a problem.
I have a data set of three columns, and want to select only those rows that go before the rows, which contain 0 in the third column.
Eventually I want to create a new data set containing only these selected rows.



Can it be performed in Mathematica?
Thank you in advance!



(edited)










share|improve this question











$endgroup$
















    9












    $begingroup$


    I am a new user of Mathematica, and just faced a problem.
    I have a data set of three columns, and want to select only those rows that go before the rows, which contain 0 in the third column.
    Eventually I want to create a new data set containing only these selected rows.



    Can it be performed in Mathematica?
    Thank you in advance!



    (edited)










    share|improve this question











    $endgroup$














      9












      9








      9





      $begingroup$


      I am a new user of Mathematica, and just faced a problem.
      I have a data set of three columns, and want to select only those rows that go before the rows, which contain 0 in the third column.
      Eventually I want to create a new data set containing only these selected rows.



      Can it be performed in Mathematica?
      Thank you in advance!



      (edited)










      share|improve this question











      $endgroup$




      I am a new user of Mathematica, and just faced a problem.
      I have a data set of three columns, and want to select only those rows that go before the rows, which contain 0 in the third column.
      Eventually I want to create a new data set containing only these selected rows.



      Can it be performed in Mathematica?
      Thank you in advance!



      (edited)







      list-manipulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 29 at 6:39







      Anna

















      asked May 28 at 14:44









      AnnaAnna

      463




      463




















          4 Answers
          4






          active

          oldest

          votes


















          15












          $begingroup$

          You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.



          data = 1, 2, 8, 4, 5, 0, 3, 2, 8, 9, 7, 0, 2, 3;

          SequenceCases[data, a_, ___, 0 :> a]
          (* 8, 4, 8, 9 *)


          Here a_, ___, 0 is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
            $endgroup$
            – Roman
            May 28 at 16:57











          • $begingroup$
            Thanks for the suggestions!
            $endgroup$
            – Jason B.
            May 28 at 17:10


















          13












          $begingroup$

          You can also use ReplaceList:



          data = 1, 1, 2, 2, 8, 4, 3, 5, 0, 4, 3, 2, 5, 8, 9, 6, 7, 0, 7, 2, 3;
          ReplaceList[data, ___, a_, __, 0, ___ :> a]



          2, 8, 4, 5, 8, 9




          Or combinations of



          Extract and Position:



          Extract[data, Position[data[[2 ;;, -1]], 0]]



          2, 8, 4, 5, 8, 9




          PositionIndex and Part:



          data[[PositionIndex[data[[2 ;;, -1]]][0]]]



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
            $endgroup$
            – Anna
            May 28 at 15:25











          • $begingroup$
            @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
            $endgroup$
            – kglr
            May 28 at 15:50










          • $begingroup$
            @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
            $endgroup$
            – Christopher Lamb
            May 28 at 18:49










          • $begingroup$
            @ChristopherLamb, thank you; it is fixed now.
            $endgroup$
            – kglr
            May 28 at 18:58


















          7












          $begingroup$

          data //Pick[Most@#, #[[2;;,3]],0]&



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This is much faster than other methods posted for large input. (+1)
            $endgroup$
            – kglr
            May 29 at 6:55


















          3












          $begingroup$

          Assuming we are interested in every row exactly preceding a row whose last entry (column) is equal to zero then it is possible to use Partition and its undocumented last argument.



          In short, the last argument of Partition allows to apply a user defined function to every n-tuple of elements generated; eg. evaluating Partition[1,2,3,2,1,1,-1,Null] produces




          1, 2, 2, 3




          while using the last argument and evaluating , Partition[1,2,3,2,1,1,-1,Null,f] produces




          f[1, 2], f[2, 3]




          where f can be any user-defined (or built-in, for that matter) function of an appropriate number of arguments.



          Returning to the question, assuming data is a list of lists of three elements, then evaluating



          Partition[data, 2, 1, 1, -1, Null, If[#2[[-1]] == 0&&#1[[-1]] != 0, #1, Nothing] &]


          should return the desired result.




          -3,2,2,1,2,2,3,0,-3,-2,-3,-2,-2,1,-2,2,3,2,
          2,-1,2,2,3,1,2,3,-3,0,1,2,2,0,-3,2,-3,-3



          Please note how the second and third arguments to Partition instruct it to create pairs (2) of elements from the initial list that differ by one entry (1) while the fourth and fifth arguments are used to instruct Partition where in the new lists it should place the first and last element of the list and what to use for padding new lists, if needed (More details and examples are available at the documentation page for Partition).




          Here are the example data used



          BlockRandom[
          data = RandomVariate[DiscreteUniformDistribution[-3, 3], 100, 3];
          RandomSeeding -> 123456789]





          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
            $endgroup$
            – kglr
            May 29 at 9:32






          • 1




            $begingroup$
            @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
            $endgroup$
            – user42582
            May 29 at 11:50












          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "387"
          ;
          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%2fmathematica.stackexchange.com%2fquestions%2f199260%2fselect-row-of-data-if-next-row-contains-zero%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          15












          $begingroup$

          You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.



          data = 1, 2, 8, 4, 5, 0, 3, 2, 8, 9, 7, 0, 2, 3;

          SequenceCases[data, a_, ___, 0 :> a]
          (* 8, 4, 8, 9 *)


          Here a_, ___, 0 is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
            $endgroup$
            – Roman
            May 28 at 16:57











          • $begingroup$
            Thanks for the suggestions!
            $endgroup$
            – Jason B.
            May 28 at 17:10















          15












          $begingroup$

          You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.



          data = 1, 2, 8, 4, 5, 0, 3, 2, 8, 9, 7, 0, 2, 3;

          SequenceCases[data, a_, ___, 0 :> a]
          (* 8, 4, 8, 9 *)


          Here a_, ___, 0 is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
            $endgroup$
            – Roman
            May 28 at 16:57











          • $begingroup$
            Thanks for the suggestions!
            $endgroup$
            – Jason B.
            May 28 at 17:10













          15












          15








          15





          $begingroup$

          You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.



          data = 1, 2, 8, 4, 5, 0, 3, 2, 8, 9, 7, 0, 2, 3;

          SequenceCases[data, a_, ___, 0 :> a]
          (* 8, 4, 8, 9 *)


          Here a_, ___, 0 is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.






          share|improve this answer











          $endgroup$



          You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.



          data = 1, 2, 8, 4, 5, 0, 3, 2, 8, 9, 7, 0, 2, 3;

          SequenceCases[data, a_, ___, 0 :> a]
          (* 8, 4, 8, 9 *)


          Here a_, ___, 0 is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited May 28 at 17:10

























          answered May 28 at 14:56









          Jason B.Jason B.

          49.8k392199




          49.8k392199











          • $begingroup$
            Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
            $endgroup$
            – Roman
            May 28 at 16:57











          • $begingroup$
            Thanks for the suggestions!
            $endgroup$
            – Jason B.
            May 28 at 17:10
















          • $begingroup$
            Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
            $endgroup$
            – Roman
            May 28 at 16:57











          • $begingroup$
            Thanks for the suggestions!
            $endgroup$
            – Jason B.
            May 28 at 17:10















          $begingroup$
          Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
          $endgroup$
          – Roman
          May 28 at 16:57





          $begingroup$
          Great solution! There's no need to name b, you can use the pattern a_, _, 0 or even a_, ___, 0 (to be more flexible, see @kglr's comment below other solution).
          $endgroup$
          – Roman
          May 28 at 16:57













          $begingroup$
          Thanks for the suggestions!
          $endgroup$
          – Jason B.
          May 28 at 17:10




          $begingroup$
          Thanks for the suggestions!
          $endgroup$
          – Jason B.
          May 28 at 17:10











          13












          $begingroup$

          You can also use ReplaceList:



          data = 1, 1, 2, 2, 8, 4, 3, 5, 0, 4, 3, 2, 5, 8, 9, 6, 7, 0, 7, 2, 3;
          ReplaceList[data, ___, a_, __, 0, ___ :> a]



          2, 8, 4, 5, 8, 9




          Or combinations of



          Extract and Position:



          Extract[data, Position[data[[2 ;;, -1]], 0]]



          2, 8, 4, 5, 8, 9




          PositionIndex and Part:



          data[[PositionIndex[data[[2 ;;, -1]]][0]]]



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
            $endgroup$
            – Anna
            May 28 at 15:25











          • $begingroup$
            @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
            $endgroup$
            – kglr
            May 28 at 15:50










          • $begingroup$
            @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
            $endgroup$
            – Christopher Lamb
            May 28 at 18:49










          • $begingroup$
            @ChristopherLamb, thank you; it is fixed now.
            $endgroup$
            – kglr
            May 28 at 18:58















          13












          $begingroup$

          You can also use ReplaceList:



          data = 1, 1, 2, 2, 8, 4, 3, 5, 0, 4, 3, 2, 5, 8, 9, 6, 7, 0, 7, 2, 3;
          ReplaceList[data, ___, a_, __, 0, ___ :> a]



          2, 8, 4, 5, 8, 9




          Or combinations of



          Extract and Position:



          Extract[data, Position[data[[2 ;;, -1]], 0]]



          2, 8, 4, 5, 8, 9




          PositionIndex and Part:



          data[[PositionIndex[data[[2 ;;, -1]]][0]]]



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
            $endgroup$
            – Anna
            May 28 at 15:25











          • $begingroup$
            @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
            $endgroup$
            – kglr
            May 28 at 15:50










          • $begingroup$
            @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
            $endgroup$
            – Christopher Lamb
            May 28 at 18:49










          • $begingroup$
            @ChristopherLamb, thank you; it is fixed now.
            $endgroup$
            – kglr
            May 28 at 18:58













          13












          13








          13





          $begingroup$

          You can also use ReplaceList:



          data = 1, 1, 2, 2, 8, 4, 3, 5, 0, 4, 3, 2, 5, 8, 9, 6, 7, 0, 7, 2, 3;
          ReplaceList[data, ___, a_, __, 0, ___ :> a]



          2, 8, 4, 5, 8, 9




          Or combinations of



          Extract and Position:



          Extract[data, Position[data[[2 ;;, -1]], 0]]



          2, 8, 4, 5, 8, 9




          PositionIndex and Part:



          data[[PositionIndex[data[[2 ;;, -1]]][0]]]



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$



          You can also use ReplaceList:



          data = 1, 1, 2, 2, 8, 4, 3, 5, 0, 4, 3, 2, 5, 8, 9, 6, 7, 0, 7, 2, 3;
          ReplaceList[data, ___, a_, __, 0, ___ :> a]



          2, 8, 4, 5, 8, 9




          Or combinations of



          Extract and Position:



          Extract[data, Position[data[[2 ;;, -1]], 0]]



          2, 8, 4, 5, 8, 9




          PositionIndex and Part:



          data[[PositionIndex[data[[2 ;;, -1]]][0]]]



          2, 8, 4, 5, 8, 9








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited May 28 at 23:38

























          answered May 28 at 15:07









          kglrkglr

          197k10222445




          197k10222445











          • $begingroup$
            Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
            $endgroup$
            – Anna
            May 28 at 15:25











          • $begingroup$
            @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
            $endgroup$
            – kglr
            May 28 at 15:50










          • $begingroup$
            @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
            $endgroup$
            – Christopher Lamb
            May 28 at 18:49










          • $begingroup$
            @ChristopherLamb, thank you; it is fixed now.
            $endgroup$
            – kglr
            May 28 at 18:58
















          • $begingroup$
            Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
            $endgroup$
            – Anna
            May 28 at 15:25











          • $begingroup$
            @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
            $endgroup$
            – kglr
            May 28 at 15:50










          • $begingroup$
            @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
            $endgroup$
            – Christopher Lamb
            May 28 at 18:49










          • $begingroup$
            @ChristopherLamb, thank you; it is fixed now.
            $endgroup$
            – kglr
            May 28 at 18:58















          $begingroup$
          Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
          $endgroup$
          – Anna
          May 28 at 15:25





          $begingroup$
          Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
          $endgroup$
          – Anna
          May 28 at 15:25













          $begingroup$
          @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
          $endgroup$
          – kglr
          May 28 at 15:50




          $begingroup$
          @Anna, made a small change ( replaced _,0 with __,0) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
          $endgroup$
          – kglr
          May 28 at 15:50












          $begingroup$
          @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
          $endgroup$
          – Christopher Lamb
          May 28 at 18:49




          $begingroup$
          @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], , a_, , 0 :> a] returns for me.
          $endgroup$
          – Christopher Lamb
          May 28 at 18:49












          $begingroup$
          @ChristopherLamb, thank you; it is fixed now.
          $endgroup$
          – kglr
          May 28 at 18:58




          $begingroup$
          @ChristopherLamb, thank you; it is fixed now.
          $endgroup$
          – kglr
          May 28 at 18:58











          7












          $begingroup$

          data //Pick[Most@#, #[[2;;,3]],0]&



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This is much faster than other methods posted for large input. (+1)
            $endgroup$
            – kglr
            May 29 at 6:55















          7












          $begingroup$

          data //Pick[Most@#, #[[2;;,3]],0]&



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This is much faster than other methods posted for large input. (+1)
            $endgroup$
            – kglr
            May 29 at 6:55













          7












          7








          7





          $begingroup$

          data //Pick[Most@#, #[[2;;,3]],0]&



          2, 8, 4, 5, 8, 9







          share|improve this answer











          $endgroup$



          data //Pick[Most@#, #[[2;;,3]],0]&



          2, 8, 4, 5, 8, 9








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited May 28 at 23:46

























          answered May 28 at 19:03









          user1066user1066

          6,27822033




          6,27822033







          • 1




            $begingroup$
            This is much faster than other methods posted for large input. (+1)
            $endgroup$
            – kglr
            May 29 at 6:55












          • 1




            $begingroup$
            This is much faster than other methods posted for large input. (+1)
            $endgroup$
            – kglr
            May 29 at 6:55







          1




          1




          $begingroup$
          This is much faster than other methods posted for large input. (+1)
          $endgroup$
          – kglr
          May 29 at 6:55




          $begingroup$
          This is much faster than other methods posted for large input. (+1)
          $endgroup$
          – kglr
          May 29 at 6:55











          3












          $begingroup$

          Assuming we are interested in every row exactly preceding a row whose last entry (column) is equal to zero then it is possible to use Partition and its undocumented last argument.



          In short, the last argument of Partition allows to apply a user defined function to every n-tuple of elements generated; eg. evaluating Partition[1,2,3,2,1,1,-1,Null] produces




          1, 2, 2, 3




          while using the last argument and evaluating , Partition[1,2,3,2,1,1,-1,Null,f] produces




          f[1, 2], f[2, 3]




          where f can be any user-defined (or built-in, for that matter) function of an appropriate number of arguments.



          Returning to the question, assuming data is a list of lists of three elements, then evaluating



          Partition[data, 2, 1, 1, -1, Null, If[#2[[-1]] == 0&&#1[[-1]] != 0, #1, Nothing] &]


          should return the desired result.




          -3,2,2,1,2,2,3,0,-3,-2,-3,-2,-2,1,-2,2,3,2,
          2,-1,2,2,3,1,2,3,-3,0,1,2,2,0,-3,2,-3,-3



          Please note how the second and third arguments to Partition instruct it to create pairs (2) of elements from the initial list that differ by one entry (1) while the fourth and fifth arguments are used to instruct Partition where in the new lists it should place the first and last element of the list and what to use for padding new lists, if needed (More details and examples are available at the documentation page for Partition).




          Here are the example data used



          BlockRandom[
          data = RandomVariate[DiscreteUniformDistribution[-3, 3], 100, 3];
          RandomSeeding -> 123456789]





          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
            $endgroup$
            – kglr
            May 29 at 9:32






          • 1




            $begingroup$
            @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
            $endgroup$
            – user42582
            May 29 at 11:50
















          3












          $begingroup$

          Assuming we are interested in every row exactly preceding a row whose last entry (column) is equal to zero then it is possible to use Partition and its undocumented last argument.



          In short, the last argument of Partition allows to apply a user defined function to every n-tuple of elements generated; eg. evaluating Partition[1,2,3,2,1,1,-1,Null] produces




          1, 2, 2, 3




          while using the last argument and evaluating , Partition[1,2,3,2,1,1,-1,Null,f] produces




          f[1, 2], f[2, 3]




          where f can be any user-defined (or built-in, for that matter) function of an appropriate number of arguments.



          Returning to the question, assuming data is a list of lists of three elements, then evaluating



          Partition[data, 2, 1, 1, -1, Null, If[#2[[-1]] == 0&&#1[[-1]] != 0, #1, Nothing] &]


          should return the desired result.




          -3,2,2,1,2,2,3,0,-3,-2,-3,-2,-2,1,-2,2,3,2,
          2,-1,2,2,3,1,2,3,-3,0,1,2,2,0,-3,2,-3,-3



          Please note how the second and third arguments to Partition instruct it to create pairs (2) of elements from the initial list that differ by one entry (1) while the fourth and fifth arguments are used to instruct Partition where in the new lists it should place the first and last element of the list and what to use for padding new lists, if needed (More details and examples are available at the documentation page for Partition).




          Here are the example data used



          BlockRandom[
          data = RandomVariate[DiscreteUniformDistribution[-3, 3], 100, 3];
          RandomSeeding -> 123456789]





          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
            $endgroup$
            – kglr
            May 29 at 9:32






          • 1




            $begingroup$
            @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
            $endgroup$
            – user42582
            May 29 at 11:50














          3












          3








          3





          $begingroup$

          Assuming we are interested in every row exactly preceding a row whose last entry (column) is equal to zero then it is possible to use Partition and its undocumented last argument.



          In short, the last argument of Partition allows to apply a user defined function to every n-tuple of elements generated; eg. evaluating Partition[1,2,3,2,1,1,-1,Null] produces




          1, 2, 2, 3




          while using the last argument and evaluating , Partition[1,2,3,2,1,1,-1,Null,f] produces




          f[1, 2], f[2, 3]




          where f can be any user-defined (or built-in, for that matter) function of an appropriate number of arguments.



          Returning to the question, assuming data is a list of lists of three elements, then evaluating



          Partition[data, 2, 1, 1, -1, Null, If[#2[[-1]] == 0&&#1[[-1]] != 0, #1, Nothing] &]


          should return the desired result.




          -3,2,2,1,2,2,3,0,-3,-2,-3,-2,-2,1,-2,2,3,2,
          2,-1,2,2,3,1,2,3,-3,0,1,2,2,0,-3,2,-3,-3



          Please note how the second and third arguments to Partition instruct it to create pairs (2) of elements from the initial list that differ by one entry (1) while the fourth and fifth arguments are used to instruct Partition where in the new lists it should place the first and last element of the list and what to use for padding new lists, if needed (More details and examples are available at the documentation page for Partition).




          Here are the example data used



          BlockRandom[
          data = RandomVariate[DiscreteUniformDistribution[-3, 3], 100, 3];
          RandomSeeding -> 123456789]





          share|improve this answer









          $endgroup$



          Assuming we are interested in every row exactly preceding a row whose last entry (column) is equal to zero then it is possible to use Partition and its undocumented last argument.



          In short, the last argument of Partition allows to apply a user defined function to every n-tuple of elements generated; eg. evaluating Partition[1,2,3,2,1,1,-1,Null] produces




          1, 2, 2, 3




          while using the last argument and evaluating , Partition[1,2,3,2,1,1,-1,Null,f] produces




          f[1, 2], f[2, 3]




          where f can be any user-defined (or built-in, for that matter) function of an appropriate number of arguments.



          Returning to the question, assuming data is a list of lists of three elements, then evaluating



          Partition[data, 2, 1, 1, -1, Null, If[#2[[-1]] == 0&&#1[[-1]] != 0, #1, Nothing] &]


          should return the desired result.




          -3,2,2,1,2,2,3,0,-3,-2,-3,-2,-2,1,-2,2,3,2,
          2,-1,2,2,3,1,2,3,-3,0,1,2,2,0,-3,2,-3,-3



          Please note how the second and third arguments to Partition instruct it to create pairs (2) of elements from the initial list that differ by one entry (1) while the fourth and fifth arguments are used to instruct Partition where in the new lists it should place the first and last element of the list and what to use for padding new lists, if needed (More details and examples are available at the documentation page for Partition).




          Here are the example data used



          BlockRandom[
          data = RandomVariate[DiscreteUniformDistribution[-3, 3], 100, 3];
          RandomSeeding -> 123456789]






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered May 29 at 8:25









          user42582user42582

          3,1101526




          3,1101526







          • 1




            $begingroup$
            I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
            $endgroup$
            – kglr
            May 29 at 9:32






          • 1




            $begingroup$
            @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
            $endgroup$
            – user42582
            May 29 at 11:50













          • 1




            $begingroup$
            I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
            $endgroup$
            – kglr
            May 29 at 9:32






          • 1




            $begingroup$
            @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
            $endgroup$
            – user42582
            May 29 at 11:50








          1




          1




          $begingroup$
          I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
          $endgroup$
          – kglr
          May 29 at 9:32




          $begingroup$
          I used this feature in these q/a s. Btw, you can also use BlockMap[ If[#[[2,-1]] == 0&&#[[1,-1]] != 0, #[[1]], Nothing] &,data,2,1]; (+1)
          $endgroup$
          – kglr
          May 29 at 9:32




          1




          1




          $begingroup$
          @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
          $endgroup$
          – user42582
          May 29 at 11:50





          $begingroup$
          @kglr indeed it is the case that I learned about this feature from one of your answers; thanks for pointing out BlockMap
          $endgroup$
          – user42582
          May 29 at 11:50


















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematica 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%2fmathematica.stackexchange.com%2fquestions%2f199260%2fselect-row-of-data-if-next-row-contains-zero%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

          Club Baloncesto Breogán Índice Historia | Pavillón | Nome | O Breogán na cultura popular | Xogadores | Adestradores | Presidentes | Palmarés | Historial | Líderes | Notas | Véxase tamén | Menú de navegacióncbbreogan.galCadroGuía oficial da ACB 2009-10, páxina 201Guía oficial ACB 1992, páxina 183. Editorial DB.É de 6.500 espectadores sentados axeitándose á última normativa"Estudiantes Junior, entre as mellores canteiras"o orixinalHemeroteca El Mundo Deportivo, 16 setembro de 1970, páxina 12Historia do BreogánAlfredo Pérez, o último canoneiroHistoria C.B. BreogánHemeroteca de El Mundo DeportivoJimmy Wright, norteamericano do Breogán deixará Lugo por ameazas de morteResultados de Breogán en 1986-87Resultados de Breogán en 1990-91Ficha de Velimir Perasović en acb.comResultados de Breogán en 1994-95Breogán arrasa al Barça. "El Mundo Deportivo", 27 de setembro de 1999, páxina 58CB Breogán - FC BarcelonaA FEB invita a participar nunha nova Liga EuropeaCharlie Bell na prensa estatalMáximos anotadores 2005Tempada 2005-06 : Tódolos Xogadores da Xornada""Non quero pensar nunha man negra, mais pregúntome que está a pasar""o orixinalRaúl López, orgulloso dos xogadores, presume da boa saúde económica do BreogánJulio González confirma que cesa como presidente del BreogánHomenaxe a Lisardo GómezA tempada do rexurdimento celesteEntrevista a Lisardo GómezEl COB dinamita el Pazo para forzar el quinto (69-73)Cafés Candelas, patrocinador del CB Breogán"Suso Lázare, novo presidente do Breogán"o orixinalCafés Candelas Breogán firma el mayor triunfo de la historiaEl Breogán realizará 17 homenajes por su cincuenta aniversario"O Breogán honra ao seu fundador e primeiro presidente"o orixinalMiguel Giao recibiu a homenaxe do PazoHomenaxe aos primeiros gladiadores celestesO home que nos amosa como ver o Breo co corazónTita Franco será homenaxeada polos #50anosdeBreoJulio Vila recibirá unha homenaxe in memoriam polos #50anosdeBreo"O Breogán homenaxeará aos seus aboados máis veteráns"Pechada ovación a «Capi» Sanmartín e Ricardo «Corazón de González»Homenaxe por décadas de informaciónPaco García volve ao Pazo con motivo do 50 aniversario"Resultados y clasificaciones""O Cafés Candelas Breogán, campión da Copa Princesa""O Cafés Candelas Breogán, equipo ACB"C.B. Breogán"Proxecto social"o orixinal"Centros asociados"o orixinalFicha en imdb.comMario Camus trata la recuperación del amor en 'La vieja música', su última película"Páxina web oficial""Club Baloncesto Breogán""C. B. Breogán S.A.D."eehttp://www.fegaba.com

          Vilaño, A Laracha Índice Patrimonio | Lugares e parroquias | Véxase tamén | Menú de navegación43°14′52″N 8°36′03″O / 43.24775, -8.60070

          Cegueira Índice Epidemioloxía | Deficiencia visual | Tipos de cegueira | Principais causas de cegueira | Tratamento | Técnicas de adaptación e axudas | Vida dos cegos | Primeiros auxilios | Crenzas respecto das persoas cegas | Crenzas das persoas cegas | O neno deficiente visual | Aspectos psicolóxicos da cegueira | Notas | Véxase tamén | Menú de navegación54.054.154.436928256blindnessDicionario da Real Academia GalegaPortal das Palabras"International Standards: Visual Standards — Aspects and Ranges of Vision Loss with Emphasis on Population Surveys.""Visual impairment and blindness""Presentan un plan para previr a cegueira"o orixinalACCDV Associació Catalana de Cecs i Disminuïts Visuals - PMFTrachoma"Effect of gene therapy on visual function in Leber's congenital amaurosis"1844137110.1056/NEJMoa0802268Cans guía - os mellores amigos dos cegosArquivadoEscola de cans guía para cegos en Mortágua, PortugalArquivado"Tecnología para ciegos y deficientes visuales. Recopilación de recursos gratuitos en la Red""Colorino""‘COL.diesis’, escuchar los sonidos del color""COL.diesis: Transforming Colour into Melody and Implementing the Result in a Colour Sensor Device"o orixinal"Sistema de desarrollo de sinestesia color-sonido para invidentes utilizando un protocolo de audio""Enseñanza táctil - geometría y color. Juegos didácticos para niños ciegos y videntes""Sistema Constanz"L'ocupació laboral dels cecs a l'Estat espanyol està pràcticament equiparada a la de les persones amb visió, entrevista amb Pedro ZuritaONCE (Organización Nacional de Cegos de España)Prevención da cegueiraDescrición de deficiencias visuais (Disc@pnet)Braillín, un boneco atractivo para calquera neno, con ou sen discapacidade, que permite familiarizarse co sistema de escritura e lectura brailleAxudas Técnicas36838ID00897494007150-90057129528256DOID:1432HP:0000618D001766C10.597.751.941.162C97109C0155020