Apply MapThread to all but one variable












4












$begingroup$


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











$endgroup$

















    4












    $begingroup$


    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











    $endgroup$















      4












      4








      4





      $begingroup$


      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 7 hours ago









      corey979

      20.9k64382




      20.9k64382










      asked 7 hours ago









      SmerdjakovSmerdjakov

      1405




      1405






















          4 Answers
          4






          active

          oldest

          votes


















          6












          $begingroup$

          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
            7 hours ago



















          5












          $begingroup$

          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









          $endgroup$





















            3












            $begingroup$

            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









            $endgroup$





















              1












              $begingroup$

              Another way with Curry and Through.



              Through /@ Apply[Curry[f, {2, 3, 1}] /@ c] /@ Transpose@{a, b}



              Mathematica graphics




              Hope this helps.






              share|improve this answer









              $endgroup$














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                4 Answers
                4






                active

                oldest

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                4 Answers
                4






                active

                oldest

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                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                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
                  7 hours ago
















                6












                $begingroup$

                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
                  7 hours ago














                6












                6








                6





                $begingroup$

                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 7 hours ago









                Sjoerd SmitSjoerd Smit

                4,630817




                4,630817








                • 1




                  $begingroup$
                  Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
                  $endgroup$
                  – Roman
                  7 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
                  7 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
                7 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
                7 hours ago











                5












                $begingroup$

                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









                $endgroup$


















                  5












                  $begingroup$

                  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









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    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 7 hours ago









                    corey979corey979

                    20.9k64382




                    20.9k64382























                        3












                        $begingroup$

                        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









                        $endgroup$


















                          3












                          $begingroup$

                          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









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            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 6 hours ago









                            Carl WollCarl Woll

                            75.9k3100198




                            75.9k3100198























                                1












                                $begingroup$

                                Another way with Curry and Through.



                                Through /@ Apply[Curry[f, {2, 3, 1}] /@ c] /@ Transpose@{a, b}



                                Mathematica graphics




                                Hope this helps.






                                share|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Another way with Curry and Through.



                                  Through /@ Apply[Curry[f, {2, 3, 1}] /@ c] /@ Transpose@{a, b}



                                  Mathematica graphics




                                  Hope this helps.






                                  share|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Another way with Curry and Through.



                                    Through /@ Apply[Curry[f, {2, 3, 1}] /@ c] /@ Transpose@{a, b}



                                    Mathematica graphics




                                    Hope this helps.






                                    share|improve this answer









                                    $endgroup$



                                    Another way with Curry and Through.



                                    Through /@ Apply[Curry[f, {2, 3, 1}] /@ c] /@ Transpose@{a, b}



                                    Mathematica graphics




                                    Hope this helps.







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 4 hours ago









                                    EdmundEdmund

                                    26.8k330103




                                    26.8k330103






























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                                        Ida-Boy-Ed-Garten

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                                        Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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