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Apply MapThread to all but one variable


How do you efficiently return all of a List but one element?All values for a function with two arguments without OuterEfficiently finding the maximum value of a column in a matrixnested use of Apply/Map/MapThread in pure functionsMapThread AlternativesFinding neighbors from listMapThread problemapply binary operation to all adjacent pairsFlip sign of one variable in listFind numbers from Mean, Variance and Correlation coefficient













2












$begingroup$


I would like to know what is the most efficient to implement the following computation. Given three lists



    a = {a_1,a_2, a_3, …, a_n}

    b = {b_1,b_2, b_3, …, b_n}

    c = {c_1,c_2, c_3, …, c_n}


and a function $f(x_1,x_2,x_3)$, obtain 



     f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  

     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 

     .....            .....            .....   .....

     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n)


I cannot find a solution not using For.






list-manipulation







share|improve this question











$endgroup$

















    2












    $begingroup$


    I would like to know what is the most efficient to implement the following computation. Given three lists



        a = {a_1,a_2, a_3, …, a_n}
    
    b = {b_1,b_2, b_3, …, b_n}
    
    c = {c_1,c_2, c_3, …, c_n}


    and a function $f(x_1,x_2,x_3)$, obtain 



         f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  
    
     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 
    
     .....            .....            .....   .....
    
     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n)


    I cannot find a solution not using For.






    list-manipulation







    share|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I would like to know what is the most efficient to implement the following computation. Given three lists



          a = {a_1,a_2, a_3, …, a_n}
      
    b = {b_1,b_2, b_3, …, b_n}
      
    c = {c_1,c_2, c_3, …, c_n}


      and a function $f(x_1,x_2,x_3)$, obtain 



           f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  
      
     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 
      
     .....            .....            .....   .....
      
     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n)


      I cannot find a solution not using For.






      list-manipulation







      share|improve this question











      $endgroup$




      I would like to know what is the most efficient to implement the following computation. Given three lists



          a = {a_1,a_2, a_3, …, a_n}
      
    b = {b_1,b_2, b_3, …, b_n}
      
    c = {c_1,c_2, c_3, …, c_n}


      and a function $f(x_1,x_2,x_3)$, obtain 



           f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  
      
     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 
      
     .....            .....            .....   .....
      
     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n)


      I cannot find a solution not using For.






      list-manipulation




      list-manipulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      corey979

      20.9k64382




      20.9k64382










      asked 2 hours ago









      SmerdjakovSmerdjakov

      1305




      1305






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Here's one way to do it with Outer:



          n = 3;
          
l1 = Array[a, n];
          
l2 = Array[b, n];
          
l3 = Array[c, n];
          

          
Outer[
          
  f[#1[[1]], #1[[2]], #2] &,
          
  Transpose @ {l1, l2},
          
  l3,
          
  1
          
]



          Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
          f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
          f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
          f[a[3], b[3], c[3]]}}







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
            $endgroup$
            – Roman
            2 hours ago



















          2












          $begingroup$

          a = {a1, a2, a3, a4, a5};
          
b = {b1, b2, b3, b4, b5};
          
c = {c1, c2, c3, c4, c5};
          

          
Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]


          enter image description here






          share|improve this answer









          $endgroup$





















            2












            $begingroup$

            Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:



            n = 3;
            
l1 = Array[a,n];
            
l2 = Array[b,n];
            
l3 = Array[c,n];


            Using Thread:



            Thread /@ Thread[f[l1, l2, l3], List, 2]



            {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
            f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
            f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
            f[a[3], b[3], c[3]]}}







            share|improve this answer









            $endgroup$














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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              Here's one way to do it with Outer:



              n = 3;
              
l1 = Array[a, n];
              
l2 = Array[b, n];
              
l3 = Array[c, n];
              

              
Outer[
              
  f[#1[[1]], #1[[2]], #2] &,
              
  Transpose @ {l1, l2},
              
  l3,
              
  1
              
]



              Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
              f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
              f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
              f[a[3], b[3], c[3]]}}







              share|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
                $endgroup$
                – Roman
                2 hours ago
















              4












              $begingroup$

              Here's one way to do it with Outer:



              n = 3;
              
l1 = Array[a, n];
              
l2 = Array[b, n];
              
l3 = Array[c, n];
              

              
Outer[
              
  f[#1[[1]], #1[[2]], #2] &,
              
  Transpose @ {l1, l2},
              
  l3,
              
  1
              
]



              Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
              f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
              f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
              f[a[3], b[3], c[3]]}}







              share|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
                $endgroup$
                – Roman
                2 hours ago














              4












              4








              4





              $begingroup$

              Here's one way to do it with Outer:



              n = 3;
              
l1 = Array[a, n];
              
l2 = Array[b, n];
              
l3 = Array[c, n];
              

              
Outer[
              
  f[#1[[1]], #1[[2]], #2] &,
              
  Transpose @ {l1, l2},
              
  l3,
              
  1
              
]



              Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
              f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
              f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
              f[a[3], b[3], c[3]]}}







              share|improve this answer









              $endgroup$



              Here's one way to do it with Outer:



              n = 3;
              
l1 = Array[a, n];
              
l2 = Array[b, n];
              
l3 = Array[c, n];
              

              
Outer[
              
  f[#1[[1]], #1[[2]], #2] &,
              
  Transpose @ {l1, l2},
              
  l3,
              
  1
              
]



              Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
              f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
              f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
              f[a[3], b[3], c[3]]}}








              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 2 hours ago









              Sjoerd SmitSjoerd Smit

              4,610817




              4,610817








              • 1




                $begingroup$
                Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
                $endgroup$
                – Roman
                2 hours ago














              • 1




                $begingroup$
                Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
                $endgroup$
                – Roman
                2 hours ago








              1




              1




              $begingroup$
              Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
              $endgroup$
              – Roman
              2 hours ago




              $begingroup$
              Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
              $endgroup$
              – Roman
              2 hours ago











              2












              $begingroup$

              a = {a1, a2, a3, a4, a5};
              
b = {b1, b2, b3, b4, b5};
              
c = {c1, c2, c3, c4, c5};
              

              
Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]


              enter image description here






              share|improve this answer









              $endgroup$


















                2












                $begingroup$

                a = {a1, a2, a3, a4, a5};
                
b = {b1, b2, b3, b4, b5};
                
c = {c1, c2, c3, c4, c5};
                

                
Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]


                enter image description here






                share|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  a = {a1, a2, a3, a4, a5};
                  
b = {b1, b2, b3, b4, b5};
                  
c = {c1, c2, c3, c4, c5};
                  

                  
Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]


                  enter image description here






                  share|improve this answer









                  $endgroup$



                  a = {a1, a2, a3, a4, a5};
                  
b = {b1, b2, b3, b4, b5};
                  
c = {c1, c2, c3, c4, c5};
                  

                  
Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 2 hours ago









                  corey979corey979

                  20.9k64382




                  20.9k64382























                      2












                      $begingroup$

                      Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:



                      n = 3;
                      
l1 = Array[a,n];
                      
l2 = Array[b,n];
                      
l3 = Array[c,n];


                      Using Thread:



                      Thread /@ Thread[f[l1, l2, l3], List, 2]



                      {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
                      f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
                      f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
                      f[a[3], b[3], c[3]]}}







                      share|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:



                        n = 3;
                        
l1 = Array[a,n];
                        
l2 = Array[b,n];
                        
l3 = Array[c,n];


                        Using Thread:



                        Thread /@ Thread[f[l1, l2, l3], List, 2]



                        {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
                        f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
                        f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
                        f[a[3], b[3], c[3]]}}







                        share|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:



                          n = 3;
                          
l1 = Array[a,n];
                          
l2 = Array[b,n];
                          
l3 = Array[c,n];


                          Using Thread:



                          Thread /@ Thread[f[l1, l2, l3], List, 2]



                          {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
                          f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
                          f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
                          f[a[3], b[3], c[3]]}}







                          share|improve this answer









                          $endgroup$



                          Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:



                          n = 3;
                          
l1 = Array[a,n];
                          
l2 = Array[b,n];
                          
l3 = Array[c,n];


                          Using Thread:



                          Thread /@ Thread[f[l1, l2, l3], List, 2]



                          {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
                          f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
                          f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
                          f[a[3], b[3], c[3]]}}








                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 1 hour ago









                          Carl WollCarl Woll

                          75.9k3100198




                          75.9k3100198






























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