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

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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
            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





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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
            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


















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