using DeleteCases to delete vectors with particular index as 0












1














Assume I have a set of vectors 



  set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.


I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.



Can I use DeleteCases in a better way than following? 



 Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]], 

 {j, 1, Length[set]}];



 set =  DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];





matrix vector column deletecases







share|improve this question



























    1














    Assume I have a set of vectors 



      set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.


    I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.



    Can I use DeleteCases in a better way than following? 



     Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]], 
    
 {j, 1, Length[set]}];
    

    
 set =  DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];





    matrix vector column deletecases







    share|improve this question

























      1












      1








      1







      Assume I have a set of vectors 



        set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.


      I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.



      Can I use DeleteCases in a better way than following? 



       Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]], 
      
 {j, 1, Length[set]}];
      

      
 set =  DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];





      matrix vector column deletecases







      share|improve this question













      Assume I have a set of vectors 



        set = {{0,1,1,0,1}, {1,0,1,1,1}, {1,1,1,1,0},{1,0,0,1,0}}.


      I want to delete the vectors which have the last entry as 0. In the above example, it is the 3rd and 4th vectors.



      Can I use DeleteCases in a better way than following? 



       Do[If[set[[j]][[5]] == 0, set[[j]] = ConstantArray[0, Length[set[[1]]]]], 
      
 {j, 1, Length[set]}];
      

      
 set =  DeleteCases[set, ConstantArray[0, Length[set[[1]]]]];





      matrix vector column deletecases




      matrix vector column deletecases






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 6 hours ago









      cleanplaycleanplay

      30918




      30918






















          2 Answers
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          active

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          4














          DeleteCases[set, {___, 0}]
          




          {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}



          In this case Pick also works:



          Pick[set, Last /@ set, 1]
          




          {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}




          You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.



          You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.



          Picking according to the second column:



          Pick[set, set[[All, 2]], 1]
          




          {{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}



          Suppose however that your keep elements are not all the same, like 1 in this example above:



          set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
          

          
Pick[set2, Unitize @ set2[[All, 3]], 1]
          




          {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}



          Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:



          (* Not recommended where performance matters *)
          

          
Pick[set2, Positive @ set2[[All, 3]]]
          

          
Pick[set2, set2[[All, 3]], _?Positive]





          share|improve this answer























          • Thanks. Can you modify this to any index instead of just the last one?
            – cleanplay
            5 hours ago






          • 1




            @cleanplay Please see the addendum to my answer.
            – Mr.Wizard♦
            5 hours ago



















          4














          Or...



          Select[set, Last[#] != 0 &]


          or...



          Select[set, #[[-1]] != 0 &]


          or...



          DeleteCases[set, x_ /; Last[x] == 0]





          share|improve this answer























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            2 Answers
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            active

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            2 Answers
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            active

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            4














            DeleteCases[set, {___, 0}]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}



            In this case Pick also works:



            Pick[set, Last /@ set, 1]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}




            You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.



            You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.



            Picking according to the second column:



            Pick[set, set[[All, 2]], 1]
            




            {{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}



            Suppose however that your keep elements are not all the same, like 1 in this example above:



            set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
            

            
Pick[set2, Unitize @ set2[[All, 3]], 1]
            




            {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}



            Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:



            (* Not recommended where performance matters *)
            

            
Pick[set2, Positive @ set2[[All, 3]]]
            

            
Pick[set2, set2[[All, 3]], _?Positive]





            share|improve this answer























            • Thanks. Can you modify this to any index instead of just the last one?
              – cleanplay
              5 hours ago






            • 1




              @cleanplay Please see the addendum to my answer.
              – Mr.Wizard♦
              5 hours ago
















            4














            DeleteCases[set, {___, 0}]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}



            In this case Pick also works:



            Pick[set, Last /@ set, 1]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}




            You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.



            You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.



            Picking according to the second column:



            Pick[set, set[[All, 2]], 1]
            




            {{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}



            Suppose however that your keep elements are not all the same, like 1 in this example above:



            set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
            

            
Pick[set2, Unitize @ set2[[All, 3]], 1]
            




            {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}



            Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:



            (* Not recommended where performance matters *)
            

            
Pick[set2, Positive @ set2[[All, 3]]]
            

            
Pick[set2, set2[[All, 3]], _?Positive]





            share|improve this answer























            • Thanks. Can you modify this to any index instead of just the last one?
              – cleanplay
              5 hours ago






            • 1




              @cleanplay Please see the addendum to my answer.
              – Mr.Wizard♦
              5 hours ago














            4












            4








            4






            DeleteCases[set, {___, 0}]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}



            In this case Pick also works:



            Pick[set, Last /@ set, 1]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}




            You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.



            You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.



            Picking according to the second column:



            Pick[set, set[[All, 2]], 1]
            




            {{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}



            Suppose however that your keep elements are not all the same, like 1 in this example above:



            set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
            

            
Pick[set2, Unitize @ set2[[All, 3]], 1]
            




            {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}



            Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:



            (* Not recommended where performance matters *)
            

            
Pick[set2, Positive @ set2[[All, 3]]]
            

            
Pick[set2, set2[[All, 3]], _?Positive]





            share|improve this answer














            DeleteCases[set, {___, 0}]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}



            In this case Pick also works:



            Pick[set, Last /@ set, 1]
            




            {{0, 1, 1, 0, 1}, {1, 0, 1, 1, 1}}




            You asked about positions other than last. It is possible to generate a working pattern using Blank, e.g. {_, _, 0, _, _}, and this process can be automated, but that is not usually the first approach I would recommend.



            You can use individual part extraction in either Select or Cases/DeleteCases, or you can create a "mask" for use with Pick as I showed above. Notable differences are that Select only operates at a single level, while Cases and Pick generalize to deeper structures. Pick is often somewhat faster than other methods, especially if one uses fast numeric functions where possible.



            Picking according to the second column:



            Pick[set, set[[All, 2]], 1]
            




            {{0, 1, 1, 0, 1}, {1, 1, 1, 1, 0}}



            Suppose however that your keep elements are not all the same, like 1 in this example above:



            set2 = {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}, {2, 3, 0, 0, 1}, {3, 1, 0, 4, 4}};
            

            
Pick[set2, Unitize @ set2[[All, 3]], 1]
            




            {{3, 1, 2, 1, 4}, {4, 1, 2, 2, 4}}



            Note here that Unitize is needed to make all keep elements into 1. While it is possible to use e.g. Positive instead this is not as efficient because in Mathematica Booleans True and False cannot be packed, while machine-size integers can. It is for this reason that I do not recommend these apparent equivalents:



            (* Not recommended where performance matters *)
            

            
Pick[set2, Positive @ set2[[All, 3]]]
            

            
Pick[set2, set2[[All, 3]], _?Positive]






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 5 hours ago

























            answered 6 hours ago









            Mr.Wizard♦Mr.Wizard

            230k294741038




            230k294741038












            • Thanks. Can you modify this to any index instead of just the last one?
              – cleanplay
              5 hours ago






            • 1




              @cleanplay Please see the addendum to my answer.
              – Mr.Wizard♦
              5 hours ago


















            • Thanks. Can you modify this to any index instead of just the last one?
              – cleanplay
              5 hours ago






            • 1




              @cleanplay Please see the addendum to my answer.
              – Mr.Wizard♦
              5 hours ago
















            Thanks. Can you modify this to any index instead of just the last one?
            – cleanplay
            5 hours ago




            Thanks. Can you modify this to any index instead of just the last one?
            – cleanplay
            5 hours ago




            1




            1




            @cleanplay Please see the addendum to my answer.
            – Mr.Wizard♦
            5 hours ago




            @cleanplay Please see the addendum to my answer.
            – Mr.Wizard♦
            5 hours ago











            4














            Or...



            Select[set, Last[#] != 0 &]


            or...



            Select[set, #[[-1]] != 0 &]


            or...



            DeleteCases[set, x_ /; Last[x] == 0]





            share|improve this answer




























              4














              Or...



              Select[set, Last[#] != 0 &]


              or...



              Select[set, #[[-1]] != 0 &]


              or...



              DeleteCases[set, x_ /; Last[x] == 0]





              share|improve this answer


























                4












                4








                4






                Or...



                Select[set, Last[#] != 0 &]


                or...



                Select[set, #[[-1]] != 0 &]


                or...



                DeleteCases[set, x_ /; Last[x] == 0]





                share|improve this answer














                Or...



                Select[set, Last[#] != 0 &]


                or...



                Select[set, #[[-1]] != 0 &]


                or...



                DeleteCases[set, x_ /; Last[x] == 0]






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 5 hours ago

























                answered 6 hours ago









                David G. StorkDavid G. Stork

                23.5k22051




                23.5k22051






























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