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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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                              Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
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