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

                    What visual should I use to simply compare current year value vs last year in Power BI desktop

                    Image
                    0 I have column SubToBind that shows current year ratio and column PY SubToBind which shows previous year ratio. How can I simply create a card that would be displaying them both in a nice way? Something like that: I cannot achieve that with regular card, I also tried Card with States by OKViz but also unable to see both values. In KPI visual - I dont like the word "Goal", it will confuse end users. Are there any cards or other visuals that would do the job? Thanks powerbi dax powerbi-embedded share | improve this question edited Jun 11 '18 at 20:11 Oleg ...
                    Read more

                    How to ignore python UserWarning in pytest?

                    Image
                    up vote 1 down vote favorite I am using openpyxl for parsing .xlsm files, and pytest for testing. When I open a file, I get the: OpenPyxl -> UserWarning: Data Validation extension is not supported and will be removed That is not really a problem bcs. program works and I can't change the .xlsm file to fix that. But... When I run pytest with something like: def test_wrong_file_format(): assert check_excel(open_excel('file.xlsm')) == True I will get the warning i mentioned altought check_excel(open_excel('file.xlsm')) returns True and the test should suceed... Is there a nice way to tell the pytest that "It's not a bug it's a feature" and tests should pass even when they get this warning? Is there other way than using something like: with pytest.warns(UserWarning): ...
                    Read more

                    Alexandru Averescu

                    Image
                    Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center ...
                    Read more