Apply MapThread to all but one variableHow do you efficiently return all of a List but one element?All values...

I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?

Is there a way to generate a list of distinct numbers such that no two subsets ever have an equal sum?

Critique of timeline aesthetic

What's the polite way to say "I need to urinate"?

What makes accurate emulation of old systems a difficult task?

How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?

Do I have an "anti-research" personality?

Is the claim "Employers won't employ people with no 'social media presence'" realistic?

Checks user level and limit the data before saving it to mongoDB

Don’t seats that recline flat defeat the purpose of having seatbelts?

What is the most expensive material in the world that could be used to create Pun-Pun's lute?

How to pronounce 'c++' in Spanish

How to not starve gigantic beasts

How exactly does Hawking radiation decrease the mass of black holes?

What happens in the secondary winding if there's no spark plug connected?

"You've called the wrong number" or "You called the wrong number"

"Hidden" theta-term in Hamiltonian formulation of Yang-Mills theory

Why did C use the -> operator instead of reusing the . operator?

Why does Mind Blank stop the Feeblemind spell?

Aligning equation numbers vertically

Could the terminal length of components like resistors be reduced?

Was there a shared-world project before "Thieves World"?

Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?

Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?



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













1












$begingroup$


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.










share|improve this question











$endgroup$

















    1












    $begingroup$


    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.










    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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.










      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      corey979

      20.9k64382




      20.9k64382










      asked 2 hours ago









      SmerdjakovSmerdjakov

      1255




      1255






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          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
            1 hour 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









          $endgroup$





















            1












            $begingroup$

            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









            $endgroup$














              Your Answer








              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "387"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f197144%2fapply-mapthread-to-all-but-one-variable%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              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
                1 hour ago
















              3












              $begingroup$

              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
                1 hour ago














              3












              3








              3





              $begingroup$

              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 1 hour ago









              Sjoerd SmitSjoerd Smit

              4,600817




              4,600817








              • 1




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




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









                  corey979corey979

                  20.9k64382




                  20.9k64382























                      1












                      $begingroup$

                      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 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












                          1








                          1





                          $begingroup$

                          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 44 mins ago









                          Carl WollCarl Woll

                          75.9k3100198




                          75.9k3100198






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematica Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f197144%2fapply-mapthread-to-all-but-one-variable%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

                              How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

                              變成蝙蝠會怎樣? 參考資料 外部連結 导航菜单Thomas Nagel, "What is it like to be a...