Operation acting on arbitrary number of matrices, element-wise












2












$begingroup$


I have a certain number of $N times M$ matrices:



$$ M_1 = begin{pmatrix} a_{11} & a_{12} & ... & a_{1M} \
a_{21} & & & \
vdots & & ddots & \
a_{N1} & & & a_{NM}
end{pmatrix}
$$



$$ M_2 = begin{pmatrix} b_{11} & b_{12} & ... & b_{1M} \
b_{21} & & & \
vdots & & ddots & \
b_{N1} & & & b_{NM}
end{pmatrix}
$$



and I want to create a new matrix, applying a certain operation $f$ element by element, obtaining something like



$$ M = begin{pmatrix} f(a_{11},b_{11},...) & f(a_{12},b_{12},...) & ... & f(a_{1M},b_{1M},...) \
f(a_{21},b_{21},...) & & & \
vdots & & ddots & \
f(a_{N1},b_{N1},...) & & & f(a_{NM},b_{NM},...)
end{pmatrix}
$$



where the $f$ takes as many arguments as the number of matrices.



As of now I am implementing this using a Table,



With[{dims = Dimensions[dataA]}, Table[f[dataA[[x, y]], dataB[[x, y]], dataC[[x, y]]], {x, 1, dims[[1]]}, {y, 1, dims[[2]]}]]]


I was wondering if there's some more idiomatic Mathematica way of doing this.






matrix







share|improve this question









$endgroup$

















    2












    $begingroup$


    I have a certain number of $N times M$ matrices:



    $$ M_1 = begin{pmatrix} a_{11} & a_{12} & ... & a_{1M} \
    a_{21} & & & \
    vdots & & ddots & \
    a_{N1} & & & a_{NM}
    end{pmatrix}
    $$



    $$ M_2 = begin{pmatrix} b_{11} & b_{12} & ... & b_{1M} \
    b_{21} & & & \
    vdots & & ddots & \
    b_{N1} & & & b_{NM}
    end{pmatrix}
    $$



    and I want to create a new matrix, applying a certain operation $f$ element by element, obtaining something like



    $$ M = begin{pmatrix} f(a_{11},b_{11},...) & f(a_{12},b_{12},...) & ... & f(a_{1M},b_{1M},...) \
    f(a_{21},b_{21},...) & & & \
    vdots & & ddots & \
    f(a_{N1},b_{N1},...) & & & f(a_{NM},b_{NM},...)
    end{pmatrix}
    $$



    where the $f$ takes as many arguments as the number of matrices.



    As of now I am implementing this using a Table,



    With[{dims = Dimensions[dataA]}, Table[f[dataA[[x, y]], dataB[[x, y]], dataC[[x, y]]], {x, 1, dims[[1]]}, {y, 1, dims[[2]]}]]]


    I was wondering if there's some more idiomatic Mathematica way of doing this.






    matrix







    share|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I have a certain number of $N times M$ matrices:



      $$ M_1 = begin{pmatrix} a_{11} & a_{12} & ... & a_{1M} \
      a_{21} & & & \
      vdots & & ddots & \
      a_{N1} & & & a_{NM}
      end{pmatrix}
      $$



      $$ M_2 = begin{pmatrix} b_{11} & b_{12} & ... & b_{1M} \
      b_{21} & & & \
      vdots & & ddots & \
      b_{N1} & & & b_{NM}
      end{pmatrix}
      $$



      and I want to create a new matrix, applying a certain operation $f$ element by element, obtaining something like



      $$ M = begin{pmatrix} f(a_{11},b_{11},...) & f(a_{12},b_{12},...) & ... & f(a_{1M},b_{1M},...) \
      f(a_{21},b_{21},...) & & & \
      vdots & & ddots & \
      f(a_{N1},b_{N1},...) & & & f(a_{NM},b_{NM},...)
      end{pmatrix}
      $$



      where the $f$ takes as many arguments as the number of matrices.



      As of now I am implementing this using a Table,



      With[{dims = Dimensions[dataA]}, Table[f[dataA[[x, y]], dataB[[x, y]], dataC[[x, y]]], {x, 1, dims[[1]]}, {y, 1, dims[[2]]}]]]


      I was wondering if there's some more idiomatic Mathematica way of doing this.






      matrix







      share|improve this question









      $endgroup$




      I have a certain number of $N times M$ matrices:



      $$ M_1 = begin{pmatrix} a_{11} & a_{12} & ... & a_{1M} \
      a_{21} & & & \
      vdots & & ddots & \
      a_{N1} & & & a_{NM}
      end{pmatrix}
      $$



      $$ M_2 = begin{pmatrix} b_{11} & b_{12} & ... & b_{1M} \
      b_{21} & & & \
      vdots & & ddots & \
      b_{N1} & & & b_{NM}
      end{pmatrix}
      $$



      and I want to create a new matrix, applying a certain operation $f$ element by element, obtaining something like



      $$ M = begin{pmatrix} f(a_{11},b_{11},...) & f(a_{12},b_{12},...) & ... & f(a_{1M},b_{1M},...) \
      f(a_{21},b_{21},...) & & & \
      vdots & & ddots & \
      f(a_{N1},b_{N1},...) & & & f(a_{NM},b_{NM},...)
      end{pmatrix}
      $$



      where the $f$ takes as many arguments as the number of matrices.



      As of now I am implementing this using a Table,



      With[{dims = Dimensions[dataA]}, Table[f[dataA[[x, y]], dataB[[x, y]], dataC[[x, y]]], {x, 1, dims[[1]]}, {y, 1, dims[[2]]}]]]


      I was wondering if there's some more idiomatic Mathematica way of doing this.






      matrix




      matrix






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 56 mins ago









      zakkzakk

      458514




      458514






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          MapThread is designed to do exactly this. Example:



          A = {{a11, a12}, {a21, a22}}; 
          
B = {{b11, b12}, {b21, b22}}; 
          
MapThread[f, {A, B}, 2]


          Gives



          {{f[a11, b11], f[a12, b12]}, {f[a21, b21], f[a22, b22]}}


          The 2 is because you want to apply to elements of lists of lists. The arguments to "f" are 2 levels deep.






          share|improve this answer








          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            38 secs ago



















          3












          $begingroup$

          m1 = Array[a, {3, 3}];
          
m2 = Array[b, {3, 3}];
          
SetAttributes[foo, Listable]
          
foo[m1, m2] // MatrixForm // TeXForm



          $left(
          begin{array}{ccc}
          text{foo}(a(1,1),b(1,1)) & text{foo}(a(1,2),b(1,2)) & text{foo}(a(1,3),b(1,3)) \
          text{foo}(a(2,1),b(2,1)) & text{foo}(a(2,2),b(2,2)) & text{foo}(a(2,3),b(2,3)) \
          text{foo}(a(3,1),b(3,1)) & text{foo}(a(3,2),b(3,2)) & text{foo}(a(3,3),b(3,3)) \
          end{array}
          right)$







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            11 secs ago











          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%2f190007%2foperation-acting-on-arbitrary-number-of-matrices-element-wise%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












          $begingroup$

          MapThread is designed to do exactly this. Example:



          A = {{a11, a12}, {a21, a22}}; 
          
B = {{b11, b12}, {b21, b22}}; 
          
MapThread[f, {A, B}, 2]


          Gives



          {{f[a11, b11], f[a12, b12]}, {f[a21, b21], f[a22, b22]}}


          The 2 is because you want to apply to elements of lists of lists. The arguments to "f" are 2 levels deep.






          share|improve this answer








          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            38 secs ago
















          4












          $begingroup$

          MapThread is designed to do exactly this. Example:



          A = {{a11, a12}, {a21, a22}}; 
          
B = {{b11, b12}, {b21, b22}}; 
          
MapThread[f, {A, B}, 2]


          Gives



          {{f[a11, b11], f[a12, b12]}, {f[a21, b21], f[a22, b22]}}


          The 2 is because you want to apply to elements of lists of lists. The arguments to "f" are 2 levels deep.






          share|improve this answer








          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            38 secs ago














          4












          4








          4





          $begingroup$

          MapThread is designed to do exactly this. Example:



          A = {{a11, a12}, {a21, a22}}; 
          
B = {{b11, b12}, {b21, b22}}; 
          
MapThread[f, {A, B}, 2]


          Gives



          {{f[a11, b11], f[a12, b12]}, {f[a21, b21], f[a22, b22]}}


          The 2 is because you want to apply to elements of lists of lists. The arguments to "f" are 2 levels deep.






          share|improve this answer








          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          MapThread is designed to do exactly this. Example:



          A = {{a11, a12}, {a21, a22}}; 
          
B = {{b11, b12}, {b21, b22}}; 
          
MapThread[f, {A, B}, 2]


          Gives



          {{f[a11, b11], f[a12, b12]}, {f[a21, b21], f[a22, b22]}}


          The 2 is because you want to apply to elements of lists of lists. The arguments to "f" are 2 levels deep.







          share|improve this answer








          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer






          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 47 mins ago









          or1426or1426

          1562




          1562




          New contributor




          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          or1426 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.












          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            38 secs ago


















          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            38 secs ago
















          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          38 secs ago




          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          38 secs ago











          3












          $begingroup$

          m1 = Array[a, {3, 3}];
          
m2 = Array[b, {3, 3}];
          
SetAttributes[foo, Listable]
          
foo[m1, m2] // MatrixForm // TeXForm



          $left(
          begin{array}{ccc}
          text{foo}(a(1,1),b(1,1)) & text{foo}(a(1,2),b(1,2)) & text{foo}(a(1,3),b(1,3)) \
          text{foo}(a(2,1),b(2,1)) & text{foo}(a(2,2),b(2,2)) & text{foo}(a(2,3),b(2,3)) \
          text{foo}(a(3,1),b(3,1)) & text{foo}(a(3,2),b(3,2)) & text{foo}(a(3,3),b(3,3)) \
          end{array}
          right)$







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            11 secs ago
















          3












          $begingroup$

          m1 = Array[a, {3, 3}];
          
m2 = Array[b, {3, 3}];
          
SetAttributes[foo, Listable]
          
foo[m1, m2] // MatrixForm // TeXForm



          $left(
          begin{array}{ccc}
          text{foo}(a(1,1),b(1,1)) & text{foo}(a(1,2),b(1,2)) & text{foo}(a(1,3),b(1,3)) \
          text{foo}(a(2,1),b(2,1)) & text{foo}(a(2,2),b(2,2)) & text{foo}(a(2,3),b(2,3)) \
          text{foo}(a(3,1),b(3,1)) & text{foo}(a(3,2),b(3,2)) & text{foo}(a(3,3),b(3,3)) \
          end{array}
          right)$







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            11 secs ago














          3












          3








          3





          $begingroup$

          m1 = Array[a, {3, 3}];
          
m2 = Array[b, {3, 3}];
          
SetAttributes[foo, Listable]
          
foo[m1, m2] // MatrixForm // TeXForm



          $left(
          begin{array}{ccc}
          text{foo}(a(1,1),b(1,1)) & text{foo}(a(1,2),b(1,2)) & text{foo}(a(1,3),b(1,3)) \
          text{foo}(a(2,1),b(2,1)) & text{foo}(a(2,2),b(2,2)) & text{foo}(a(2,3),b(2,3)) \
          text{foo}(a(3,1),b(3,1)) & text{foo}(a(3,2),b(3,2)) & text{foo}(a(3,3),b(3,3)) \
          end{array}
          right)$







          share|improve this answer









          $endgroup$



          m1 = Array[a, {3, 3}];
          
m2 = Array[b, {3, 3}];
          
SetAttributes[foo, Listable]
          
foo[m1, m2] // MatrixForm // TeXForm



          $left(
          begin{array}{ccc}
          text{foo}(a(1,1),b(1,1)) & text{foo}(a(1,2),b(1,2)) & text{foo}(a(1,3),b(1,3)) \
          text{foo}(a(2,1),b(2,1)) & text{foo}(a(2,2),b(2,2)) & text{foo}(a(2,3),b(2,3)) \
          text{foo}(a(3,1),b(3,1)) & text{foo}(a(3,2),b(3,2)) & text{foo}(a(3,3),b(3,3)) \
          end{array}
          right)$








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 45 mins ago









          kglrkglr

          180k9200413




          180k9200413












          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            11 secs ago


















          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            11 secs ago
















          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          11 secs ago




          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          11 secs ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematica Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f190007%2foperation-acting-on-arbitrary-number-of-matrices-element-wise%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

          SQL Server 17 - Attemping to backup to remote NAS but Access is denied

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 0 I was attemping to use Ola Hallengrens scripts to backup to a remote NAS, but it was failing and not giving any direct errors, so I tried doing it manually using Server Management Studio and got the following error: TITLE: Microsoft SQL Server Management Studio Failed to retrieve data for this request. (Microsoft.SqlServer.Management.Sdk.Sfc) For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft%20SQL%20Server&LinkId=20476 ADDITIONAL INFORMATION: An exception occurred while executing a Transact-SQL statement or batch. (Microsoft.SqlServer.ConnectionInfo) The operating system returned the error '0x80070005(Access is denied.)&#
          Read more

          Always On Availability groups resolving state after failover - Remote harden of transaction...

          Image
          0 My explanation will be opinion based as it is based on my findings. If more information is needed, I am always happy to add it. The setup We have 4 SQL Server 2016 Instances, and all 4 underlying servers are in a WSFC (In my eyes a strange setup since Always On AG's are 2 by 2. But that is a different question.) 4 nodes + Fileshare witness Version: 13.0.5026.0 (all nodes) The instances that are having the failover issues are instances 1 & 2. They have 3 AG's between them. AG1 is primary on instance 1 and is having the issue. AG2 and AG3 are primary on instance 2. Instances 3 and 4 have their own separate AG's (2). AG1's settings The week before Last week, after checking for long running queries (none) and the synchronization state (green accross the board) we initiated a
          Read more

          Restoring from pg_dump with foreign key constraints

          Image
          5 2 In restoring a database from a pg_dump , a number of errors are being generated and the whole table is subsequently being ignored. An example: ERROR: insert or update on table "channelproducts" violates foreign key constraint "fk_rails_dfaae373a5" DETAIL: Key (channel_id)=(1) is not present in table "channels". Interestingly enough, I've noted that all these instances are popping up because of the sequence of loading. channels is after channelproducts both alphabetically and in the file and thus I can understand why postgres complains about having to create a child without a parent. caveat: the foreign key is being generated automatically by rails 4.2: I could remove the problem at the source but that still does not really solve the problem... version: PostgreSQL
          Read more