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 1 hour ago









      zakkzakk

      458514




      458514






















          2 Answers
          2






          active

          oldest

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          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
            17 mins 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
            17 mins ago











          Your Answer





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          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
            17 mins 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
            17 mins 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 1 hour 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
            17 mins ago


















          • $begingroup$
            Exactly what I was looking for! Thanks!
            $endgroup$
            – zakk
            17 mins ago
















          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          17 mins ago




          $begingroup$
          Exactly what I was looking for! Thanks!
          $endgroup$
          – zakk
          17 mins 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
            17 mins 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
            17 mins 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 1 hour ago









          kglrkglr

          180k9200413




          180k9200413












          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            17 mins ago


















          • $begingroup$
            Thank you very much!
            $endgroup$
            – zakk
            17 mins ago
















          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          17 mins ago




          $begingroup$
          Thank you very much!
          $endgroup$
          – zakk
          17 mins ago


















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