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 5 hours ago









      cleanplaycleanplay

      30918




      30918






















          2 Answers
          2






          active

          oldest

          votes


















          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            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%2f189089%2fusing-deletecases-to-delete-vectors-with-particular-index-as-0%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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 5 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 5 hours ago









                David G. StorkDavid G. Stork

                23.5k22051




                23.5k22051






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f189089%2fusing-deletecases-to-delete-vectors-with-particular-index-as-0%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

                    Fluorita

                    Image
                    Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
                    Read more

                    Hulsita

                    Image
                    Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
                    Read more

                    Península de Txukotka

                    Image
                    Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
                    Read more