Apply MapThread to all but one variable












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.






list-manipulation







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.






    list-manipulation







    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.






      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 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 41 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

                              How to label and detect the document text images

                              Image
                              1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C
                              Read more

                              Vallis Paradisi

                              Image
                              Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para
                              Read more

                              Tabula Rosettana

                              Image
                              Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a
                              Read more