Select row of data if next row contains zero





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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)






list-manipulation







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)






    list-manipulation







    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)






      list-manipulation







      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




      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 bronze badges




      463 bronze badges

























          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















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          4 Answers
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          active

          oldest

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          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.

          50.9k3 gold badges96 silver badges203 bronze badges




          50.9k3 gold badges96 silver badges203 bronze badges















          • $begingroup$
            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

          221k10 gold badges251 silver badges509 bronze badges




          221k10 gold badges251 silver badges509 bronze badges















          • $begingroup$
            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

          6,5782 gold badges20 silver badges33 bronze badges




          6,5782 gold badges20 silver badges33 bronze badges











          • 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

          3,2501 gold badge6 silver badges26 bronze badges




          3,2501 gold badge6 silver badges26 bronze badges











          • 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





















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