Operation acting on arbitrary number of matrices, element-wise












3












$begingroup$


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









$endgroup$

















    3












    $begingroup$


    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









    $endgroup$















      3












      3








      3





      $begingroup$


      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 Jan 22 at 16:14









      zakkzakk

      463514




      463514






















          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









          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            Jan 22 at 17:10



















          4












          $begingroup$

          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
            Jan 22 at 17:11











          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









          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            Jan 22 at 17:10
















          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









          $endgroup$













          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            Jan 22 at 17:10














          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









          $endgroup$



          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












          share|improve this answer



          share|improve this answer










          answered Jan 22 at 16:23









          or1426or1426

          1562




          1562












          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            Jan 22 at 17:10


















          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            Jan 22 at 17:10
















          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          Jan 22 at 17:10




          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          Jan 22 at 17:10











          4












          $begingroup$

          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
            Jan 22 at 17:11
















          4












          $begingroup$

          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
            Jan 22 at 17:11














          4












          4








          4





          $begingroup$

          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 Jan 22 at 16:25









          kglrkglr

          183k10202417




          183k10202417












          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            Jan 22 at 17:11


















          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            Jan 22 at 17:11
















          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          Jan 22 at 17:11




          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          Jan 22 at 17:11


















          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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more