Basis enumeration of large lists

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I want to combine two list in the following way without using Table or any loop structure.



genList[n_] := Table[RandomInteger[1, 10, 4], n] (* list generating function*)


The two lists that needs to be combined are:-



list1 = genList[3]
list2 = genList[4]


What I want to achieve is as follows:-



Partition[
Flatten[
Table[list1[[i]], list2[[j]], i, 1
Length[list1],j,1 Length[list2]]
]
,8]


So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.



What would be an efficient way of doing this with large lists?



Let's say with Length[list1] = 100 and Length[list2] = 200.



Also Length[list2] > Length[list1]










share|improve this question



























    up vote
    3
    down vote

    favorite
    1












    I want to combine two list in the following way without using Table or any loop structure.



    genList[n_] := Table[RandomInteger[1, 10, 4], n] (* list generating function*)


    The two lists that needs to be combined are:-



    list1 = genList[3]
    list2 = genList[4]


    What I want to achieve is as follows:-



    Partition[
    Flatten[
    Table[list1[[i]], list2[[j]], i, 1
    Length[list1],j,1 Length[list2]]
    ]
    ,8]


    So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.



    What would be an efficient way of doing this with large lists?



    Let's say with Length[list1] = 100 and Length[list2] = 200.



    Also Length[list2] > Length[list1]










    share|improve this question

























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      I want to combine two list in the following way without using Table or any loop structure.



      genList[n_] := Table[RandomInteger[1, 10, 4], n] (* list generating function*)


      The two lists that needs to be combined are:-



      list1 = genList[3]
      list2 = genList[4]


      What I want to achieve is as follows:-



      Partition[
      Flatten[
      Table[list1[[i]], list2[[j]], i, 1
      Length[list1],j,1 Length[list2]]
      ]
      ,8]


      So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.



      What would be an efficient way of doing this with large lists?



      Let's say with Length[list1] = 100 and Length[list2] = 200.



      Also Length[list2] > Length[list1]










      share|improve this question















      I want to combine two list in the following way without using Table or any loop structure.



      genList[n_] := Table[RandomInteger[1, 10, 4], n] (* list generating function*)


      The two lists that needs to be combined are:-



      list1 = genList[3]
      list2 = genList[4]


      What I want to achieve is as follows:-



      Partition[
      Flatten[
      Table[list1[[i]], list2[[j]], i, 1
      Length[list1],j,1 Length[list2]]
      ]
      ,8]


      So I simply need to enumerate each of the elements of list 1 combined with all of the elements of list 2.



      What would be an efficient way of doing this with large lists?



      Let's say with Length[list1] = 100 and Length[list2] = 200.



      Also Length[list2] > Length[list1]







      list-manipulation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 3 at 8:01









      Henrik Schumacher

      47k466134




      47k466134










      asked Dec 3 at 7:20









      Hubble07

      2,915719




      2,915719




















          2 Answers
          2






          active

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          up vote
          3
          down vote



          accepted










          Join @@@ Tuples[list1, list2] (* or *)
          Flatten /@ Tuples[list1, list2] (* or *)
          Join @@ Outer[Join, list1, list2, 1]



          5, 3, 5, 4, 9, 2, 10, 6, 5, 3, 5, 4, 10, 6, 4, 7,

          5, 3, 5, 4, 2,
          1, 7, 1, 5, 3, 5, 4, 6, 6, 1, 9,

          2, 6, 9, 5, 9, 2, 10, 6, 2,
          6, 9, 5, 10, 6, 4, 7,

          2, 6, 9, 5, 2, 1, 7, 1, 2, 6, 9, 5, 6, 6,
          1, 9, 10, 8, 9, 6, 9, 2, 10, 6,

          10, 8, 9, 6, 10, 6, 4, 7, 10,
          8, 9, 6, 2, 1, 7, 1, 10, 8, 9, 6, 6, 6, 1, 9







          share|improve this answer





























            up vote
            3
            down vote













            This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.



            genList[n_] := RandomInteger[1, 10, n, 4];
            m = 100;
            n = 100;
            list1 = genList[m];
            list2 = genList[n];

            a = Partition[Flatten[Table[list1[[i]], list2[[j]], i, 1 Length[list1], j, 1 Length[list2]]], 8]; // RepeatedTiming // First
            b = Flatten /@ Tuples[list1, list2]; // RepeatedTiming // First
            c = ArrayReshape[
            Tuples[list1, list2],
            Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]
            ]; // RepeatedTiming // First
            a == b == c



            0.026



            0.0012



            0.000095



            True







            share|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote



              accepted










              Join @@@ Tuples[list1, list2] (* or *)
              Flatten /@ Tuples[list1, list2] (* or *)
              Join @@ Outer[Join, list1, list2, 1]



              5, 3, 5, 4, 9, 2, 10, 6, 5, 3, 5, 4, 10, 6, 4, 7,

              5, 3, 5, 4, 2,
              1, 7, 1, 5, 3, 5, 4, 6, 6, 1, 9,

              2, 6, 9, 5, 9, 2, 10, 6, 2,
              6, 9, 5, 10, 6, 4, 7,

              2, 6, 9, 5, 2, 1, 7, 1, 2, 6, 9, 5, 6, 6,
              1, 9, 10, 8, 9, 6, 9, 2, 10, 6,

              10, 8, 9, 6, 10, 6, 4, 7, 10,
              8, 9, 6, 2, 1, 7, 1, 10, 8, 9, 6, 6, 6, 1, 9







              share|improve this answer


























                up vote
                3
                down vote



                accepted










                Join @@@ Tuples[list1, list2] (* or *)
                Flatten /@ Tuples[list1, list2] (* or *)
                Join @@ Outer[Join, list1, list2, 1]



                5, 3, 5, 4, 9, 2, 10, 6, 5, 3, 5, 4, 10, 6, 4, 7,

                5, 3, 5, 4, 2,
                1, 7, 1, 5, 3, 5, 4, 6, 6, 1, 9,

                2, 6, 9, 5, 9, 2, 10, 6, 2,
                6, 9, 5, 10, 6, 4, 7,

                2, 6, 9, 5, 2, 1, 7, 1, 2, 6, 9, 5, 6, 6,
                1, 9, 10, 8, 9, 6, 9, 2, 10, 6,

                10, 8, 9, 6, 10, 6, 4, 7, 10,
                8, 9, 6, 2, 1, 7, 1, 10, 8, 9, 6, 6, 6, 1, 9







                share|improve this answer
























                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  Join @@@ Tuples[list1, list2] (* or *)
                  Flatten /@ Tuples[list1, list2] (* or *)
                  Join @@ Outer[Join, list1, list2, 1]



                  5, 3, 5, 4, 9, 2, 10, 6, 5, 3, 5, 4, 10, 6, 4, 7,

                  5, 3, 5, 4, 2,
                  1, 7, 1, 5, 3, 5, 4, 6, 6, 1, 9,

                  2, 6, 9, 5, 9, 2, 10, 6, 2,
                  6, 9, 5, 10, 6, 4, 7,

                  2, 6, 9, 5, 2, 1, 7, 1, 2, 6, 9, 5, 6, 6,
                  1, 9, 10, 8, 9, 6, 9, 2, 10, 6,

                  10, 8, 9, 6, 10, 6, 4, 7, 10,
                  8, 9, 6, 2, 1, 7, 1, 10, 8, 9, 6, 6, 6, 1, 9







                  share|improve this answer














                  Join @@@ Tuples[list1, list2] (* or *)
                  Flatten /@ Tuples[list1, list2] (* or *)
                  Join @@ Outer[Join, list1, list2, 1]



                  5, 3, 5, 4, 9, 2, 10, 6, 5, 3, 5, 4, 10, 6, 4, 7,

                  5, 3, 5, 4, 2,
                  1, 7, 1, 5, 3, 5, 4, 6, 6, 1, 9,

                  2, 6, 9, 5, 9, 2, 10, 6, 2,
                  6, 9, 5, 10, 6, 4, 7,

                  2, 6, 9, 5, 2, 1, 7, 1, 2, 6, 9, 5, 6, 6,
                  1, 9, 10, 8, 9, 6, 9, 2, 10, 6,

                  10, 8, 9, 6, 10, 6, 4, 7, 10,
                  8, 9, 6, 2, 1, 7, 1, 10, 8, 9, 6, 6, 6, 1, 9








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 3 at 7:54

























                  answered Dec 3 at 7:45









                  kglr

                  175k9197402




                  175k9197402




















                      up vote
                      3
                      down vote













                      This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.



                      genList[n_] := RandomInteger[1, 10, n, 4];
                      m = 100;
                      n = 100;
                      list1 = genList[m];
                      list2 = genList[n];

                      a = Partition[Flatten[Table[list1[[i]], list2[[j]], i, 1 Length[list1], j, 1 Length[list2]]], 8]; // RepeatedTiming // First
                      b = Flatten /@ Tuples[list1, list2]; // RepeatedTiming // First
                      c = ArrayReshape[
                      Tuples[list1, list2],
                      Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]
                      ]; // RepeatedTiming // First
                      a == b == c



                      0.026



                      0.0012



                      0.000095



                      True







                      share|improve this answer


























                        up vote
                        3
                        down vote













                        This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.



                        genList[n_] := RandomInteger[1, 10, n, 4];
                        m = 100;
                        n = 100;
                        list1 = genList[m];
                        list2 = genList[n];

                        a = Partition[Flatten[Table[list1[[i]], list2[[j]], i, 1 Length[list1], j, 1 Length[list2]]], 8]; // RepeatedTiming // First
                        b = Flatten /@ Tuples[list1, list2]; // RepeatedTiming // First
                        c = ArrayReshape[
                        Tuples[list1, list2],
                        Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]
                        ]; // RepeatedTiming // First
                        a == b == c



                        0.026



                        0.0012



                        0.000095



                        True







                        share|improve this answer
























                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.



                          genList[n_] := RandomInteger[1, 10, n, 4];
                          m = 100;
                          n = 100;
                          list1 = genList[m];
                          list2 = genList[n];

                          a = Partition[Flatten[Table[list1[[i]], list2[[j]], i, 1 Length[list1], j, 1 Length[list2]]], 8]; // RepeatedTiming // First
                          b = Flatten /@ Tuples[list1, list2]; // RepeatedTiming // First
                          c = ArrayReshape[
                          Tuples[list1, list2],
                          Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]
                          ]; // RepeatedTiming // First
                          a == b == c



                          0.026



                          0.0012



                          0.000095



                          True







                          share|improve this answer














                          This is a variant of kglr's approach that employs ArrayFlatten instead of mapping Flatten. Notice also that I changed genList so that it produces packed arrays; this is crucial for performance.



                          genList[n_] := RandomInteger[1, 10, n, 4];
                          m = 100;
                          n = 100;
                          list1 = genList[m];
                          list2 = genList[n];

                          a = Partition[Flatten[Table[list1[[i]], list2[[j]], i, 1 Length[list1], j, 1 Length[list2]]], 8]; // RepeatedTiming // First
                          b = Flatten /@ Tuples[list1, list2]; // RepeatedTiming // First
                          c = ArrayReshape[
                          Tuples[list1, list2],
                          Length[list1] Length[list2], Dimensions[list1][[2]] + Dimensions[list2][[2]]
                          ]; // RepeatedTiming // First
                          a == b == c



                          0.026



                          0.0012



                          0.000095



                          True








                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Dec 3 at 9:00

























                          answered Dec 3 at 7:59









                          Henrik Schumacher

                          47k466134




                          47k466134



























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