Generating a list with duplicate entries












5












$begingroup$


How can I get a function that gives the following output?



listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX]



{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}







list-manipulation







share|improve this question









New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    Wolfram Challenges.
    $endgroup$
    – J. M. is computer-less♦
    11 hours ago
















5












$begingroup$


How can I get a function that gives the following output?



listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX]



{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}







list-manipulation







share|improve this question









New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    Wolfram Challenges.
    $endgroup$
    – J. M. is computer-less♦
    11 hours ago














5












5








5





$begingroup$


How can I get a function that gives the following output?



listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX]



{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}







list-manipulation







share|improve this question









New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




How can I get a function that gives the following output?



listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX]



{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}







list-manipulation




list-manipulation






share|improve this question









New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 12 hours ago









MarcoB

37.2k556113




37.2k556113






New contributor




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Check out our Code of Conduct.









asked 12 hours ago









lisalisa

261




261




New contributor




lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    $begingroup$
    Wolfram Challenges.
    $endgroup$
    – J. M. is computer-less♦
    11 hours ago














  • 4




    $begingroup$
    Wolfram Challenges.
    $endgroup$
    – J. M. is computer-less♦
    11 hours ago








4




4




$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
11 hours ago




$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
11 hours ago










7 Answers
7






active

oldest

votes


















5












$begingroup$

Alternatively:



listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]



{{a, a, a}, { b, b, b}, {c, c, c}}




Or:



listX = {a, b, c, d};

numberX = 3;



Transpose@Table[i, {numberX}, {i, listX}]





share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Or ConstantArray[#, numberX] & /@ listX
    $endgroup$
    – Bob Hanlon
    9 hours ago



















4












$begingroup$

I like using KroneckerProduct for problems like this:



KroneckerProduct[listX, ConstantArray[1, numberX]]



{{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




The KroneckerProduct approach should be much faster than the others for large vectors.






share|improve this answer









$endgroup$





















    3












    $begingroup$

    one way might be



    listX    = {a, b, c, d}
    
numberX  = 3
    
Transpose[{listX}].{Table[1, {numberX}]}


    Mathematica graphics






    share|improve this answer









    $endgroup$





















      3












      $begingroup$

      Transpose@ConstantArray[listX, numberX]



      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







      share|improve this answer









      $endgroup$





















        2












        $begingroup$

        Array:



        Array[listX &, numberX, 1, Transpose[{##}] &]



        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




        ArrayPad:



        ArrayPad[List /@ listX, {0, {0, numberX - 1}}, "Fixed"]



        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




        PadRight:



        PadRight[{#}, numberX, "Fixed"] & /@ listX  )* or *)
        
PadRight[List /@ listX, {Automatic, numberX}, "Fixed"]



        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




        TensorProduct:



        TensorProduct[listX, Array[1 &, numberX]]



        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




        ArrayResample:



        ArrayResample[listX, Scaled @ numberX , "Bin", Resampling -> "NearestLeft"]



        {a, a, a, b, b, b, c, c, c, d, d, d}




        Partition[%, numberX]



        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







        share|improve this answer











        $endgroup$





















          1












          $begingroup$

          Yet Another Way:



          Flatten[ConstantArray[{listX}, numberX], {2, 3}]


          And another (the last argument 1 is necessary only when listX is not a flat list):



          Outer[Times, listX, ConstantArray[1, numberX], 1]





          share|improve this answer









          $endgroup$





















            0












            $begingroup$

            Transpose[{listX}[[ConstantArray[1, numberX]]]]





            share|improve this answer









            $endgroup$













              Your Answer





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              7 Answers
              7






              active

              oldest

              votes








              7 Answers
              7






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              Alternatively:



              listX = {a, b, c};
              
numberX = 3;
              
Table[ConstantArray[i, numberX], {i, listX}]



              {{a, a, a}, { b, b, b}, {c, c, c}}




              Or:



              listX = {a, b, c, d};
              
numberX = 3;
              

              
Transpose@Table[i, {numberX}, {i, listX}]





              share|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Or ConstantArray[#, numberX] & /@ listX
                $endgroup$
                – Bob Hanlon
                9 hours ago
















              5












              $begingroup$

              Alternatively:



              listX = {a, b, c};
              
numberX = 3;
              
Table[ConstantArray[i, numberX], {i, listX}]



              {{a, a, a}, { b, b, b}, {c, c, c}}




              Or:



              listX = {a, b, c, d};
              
numberX = 3;
              

              
Transpose@Table[i, {numberX}, {i, listX}]





              share|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Or ConstantArray[#, numberX] & /@ listX
                $endgroup$
                – Bob Hanlon
                9 hours ago














              5












              5








              5





              $begingroup$

              Alternatively:



              listX = {a, b, c};
              
numberX = 3;
              
Table[ConstantArray[i, numberX], {i, listX}]



              {{a, a, a}, { b, b, b}, {c, c, c}}




              Or:



              listX = {a, b, c, d};
              
numberX = 3;
              

              
Transpose@Table[i, {numberX}, {i, listX}]





              share|improve this answer











              $endgroup$



              Alternatively:



              listX = {a, b, c};
              
numberX = 3;
              
Table[ConstantArray[i, numberX], {i, listX}]



              {{a, a, a}, { b, b, b}, {c, c, c}}




              Or:



              listX = {a, b, c, d};
              
numberX = 3;
              

              
Transpose@Table[i, {numberX}, {i, listX}]






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 10 hours ago

























              answered 11 hours ago









              MarcoBMarcoB

              37.2k556113




              37.2k556113








              • 1




                $begingroup$
                Or ConstantArray[#, numberX] & /@ listX
                $endgroup$
                – Bob Hanlon
                9 hours ago














              • 1




                $begingroup$
                Or ConstantArray[#, numberX] & /@ listX
                $endgroup$
                – Bob Hanlon
                9 hours ago








              1




              1




              $begingroup$
              Or ConstantArray[#, numberX] & /@ listX
              $endgroup$
              – Bob Hanlon
              9 hours ago




              $begingroup$
              Or ConstantArray[#, numberX] & /@ listX
              $endgroup$
              – Bob Hanlon
              9 hours ago











              4












              $begingroup$

              I like using KroneckerProduct for problems like this:



              KroneckerProduct[listX, ConstantArray[1, numberX]]



              {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




              The KroneckerProduct approach should be much faster than the others for large vectors.






              share|improve this answer









              $endgroup$


















                4












                $begingroup$

                I like using KroneckerProduct for problems like this:



                KroneckerProduct[listX, ConstantArray[1, numberX]]



                {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                The KroneckerProduct approach should be much faster than the others for large vectors.






                share|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  I like using KroneckerProduct for problems like this:



                  KroneckerProduct[listX, ConstantArray[1, numberX]]



                  {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                  The KroneckerProduct approach should be much faster than the others for large vectors.






                  share|improve this answer









                  $endgroup$



                  I like using KroneckerProduct for problems like this:



                  KroneckerProduct[listX, ConstantArray[1, numberX]]



                  {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                  The KroneckerProduct approach should be much faster than the others for large vectors.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 10 hours ago









                  Carl WollCarl Woll

                  69.6k394180




                  69.6k394180























                      3












                      $begingroup$

                      one way might be



                      listX    = {a, b, c, d}
                      
numberX  = 3
                      
Transpose[{listX}].{Table[1, {numberX}]}


                      Mathematica graphics






                      share|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        one way might be



                        listX    = {a, b, c, d}
                        
numberX  = 3
                        
Transpose[{listX}].{Table[1, {numberX}]}


                        Mathematica graphics






                        share|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          one way might be



                          listX    = {a, b, c, d}
                          
numberX  = 3
                          
Transpose[{listX}].{Table[1, {numberX}]}


                          Mathematica graphics






                          share|improve this answer









                          $endgroup$



                          one way might be



                          listX    = {a, b, c, d}
                          
numberX  = 3
                          
Transpose[{listX}].{Table[1, {numberX}]}


                          Mathematica graphics







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 12 hours ago









                          NasserNasser

                          58.1k489206




                          58.1k489206























                              3












                              $begingroup$

                              Transpose@ConstantArray[listX, numberX]



                              {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                              share|improve this answer









                              $endgroup$


















                                3












                                $begingroup$

                                Transpose@ConstantArray[listX, numberX]



                                {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                                share|improve this answer









                                $endgroup$
















                                  3












                                  3








                                  3





                                  $begingroup$

                                  Transpose@ConstantArray[listX, numberX]



                                  {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                                  share|improve this answer









                                  $endgroup$



                                  Transpose@ConstantArray[listX, numberX]



                                  {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}








                                  share|improve this answer












                                  share|improve this answer



                                  share|improve this answer










                                  answered 11 hours ago









                                  RomanRoman

                                  2,735717




                                  2,735717























                                      2












                                      $begingroup$

                                      Array:



                                      Array[listX &, numberX, 1, Transpose[{##}] &]



                                      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                      ArrayPad:



                                      ArrayPad[List /@ listX, {0, {0, numberX - 1}}, "Fixed"]



                                      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                      PadRight:



                                      PadRight[{#}, numberX, "Fixed"] & /@ listX  )* or *)
                                      
PadRight[List /@ listX, {Automatic, numberX}, "Fixed"]



                                      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                      TensorProduct:



                                      TensorProduct[listX, Array[1 &, numberX]]



                                      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                      ArrayResample:



                                      ArrayResample[listX, Scaled @ numberX , "Bin", Resampling -> "NearestLeft"]



                                      {a, a, a, b, b, b, c, c, c, d, d, d}




                                      Partition[%, numberX]



                                      {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                                      share|improve this answer











                                      $endgroup$


















                                        2












                                        $begingroup$

                                        Array:



                                        Array[listX &, numberX, 1, Transpose[{##}] &]



                                        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                        ArrayPad:



                                        ArrayPad[List /@ listX, {0, {0, numberX - 1}}, "Fixed"]



                                        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                        PadRight:



                                        PadRight[{#}, numberX, "Fixed"] & /@ listX  )* or *)
                                        
PadRight[List /@ listX, {Automatic, numberX}, "Fixed"]



                                        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                        TensorProduct:



                                        TensorProduct[listX, Array[1 &, numberX]]



                                        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                        ArrayResample:



                                        ArrayResample[listX, Scaled @ numberX , "Bin", Resampling -> "NearestLeft"]



                                        {a, a, a, b, b, b, c, c, c, d, d, d}




                                        Partition[%, numberX]



                                        {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                                        share|improve this answer











                                        $endgroup$
















                                          2












                                          2








                                          2





                                          $begingroup$

                                          Array:



                                          Array[listX &, numberX, 1, Transpose[{##}] &]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          ArrayPad:



                                          ArrayPad[List /@ listX, {0, {0, numberX - 1}}, "Fixed"]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          PadRight:



                                          PadRight[{#}, numberX, "Fixed"] & /@ listX  )* or *)
                                          
PadRight[List /@ listX, {Automatic, numberX}, "Fixed"]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          TensorProduct:



                                          TensorProduct[listX, Array[1 &, numberX]]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          ArrayResample:



                                          ArrayResample[listX, Scaled @ numberX , "Bin", Resampling -> "NearestLeft"]



                                          {a, a, a, b, b, b, c, c, c, d, d, d}




                                          Partition[%, numberX]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}







                                          share|improve this answer











                                          $endgroup$



                                          Array:



                                          Array[listX &, numberX, 1, Transpose[{##}] &]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          ArrayPad:



                                          ArrayPad[List /@ listX, {0, {0, numberX - 1}}, "Fixed"]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          PadRight:



                                          PadRight[{#}, numberX, "Fixed"] & /@ listX  )* or *)
                                          
PadRight[List /@ listX, {Automatic, numberX}, "Fixed"]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          TensorProduct:



                                          TensorProduct[listX, Array[1 &, numberX]]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}




                                          ArrayResample:



                                          ArrayResample[listX, Scaled @ numberX , "Bin", Resampling -> "NearestLeft"]



                                          {a, a, a, b, b, b, c, c, c, d, d, d}




                                          Partition[%, numberX]



                                          {{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}








                                          share|improve this answer














                                          share|improve this answer



                                          share|improve this answer








                                          edited 4 hours ago

























                                          answered 4 hours ago









                                          kglrkglr

                                          187k10203421




                                          187k10203421























                                              1












                                              $begingroup$

                                              Yet Another Way:



                                              Flatten[ConstantArray[{listX}, numberX], {2, 3}]


                                              And another (the last argument 1 is necessary only when listX is not a flat list):



                                              Outer[Times, listX, ConstantArray[1, numberX], 1]





                                              share|improve this answer









                                              $endgroup$


















                                                1












                                                $begingroup$

                                                Yet Another Way:



                                                Flatten[ConstantArray[{listX}, numberX], {2, 3}]


                                                And another (the last argument 1 is necessary only when listX is not a flat list):



                                                Outer[Times, listX, ConstantArray[1, numberX], 1]





                                                share|improve this answer









                                                $endgroup$
















                                                  1












                                                  1








                                                  1





                                                  $begingroup$

                                                  Yet Another Way:



                                                  Flatten[ConstantArray[{listX}, numberX], {2, 3}]


                                                  And another (the last argument 1 is necessary only when listX is not a flat list):



                                                  Outer[Times, listX, ConstantArray[1, numberX], 1]





                                                  share|improve this answer









                                                  $endgroup$



                                                  Yet Another Way:



                                                  Flatten[ConstantArray[{listX}, numberX], {2, 3}]


                                                  And another (the last argument 1 is necessary only when listX is not a flat list):



                                                  Outer[Times, listX, ConstantArray[1, numberX], 1]






                                                  share|improve this answer












                                                  share|improve this answer



                                                  share|improve this answer










                                                  answered 2 hours ago









                                                  Michael E2Michael E2

                                                  148k12198477




                                                  148k12198477























                                                      0












                                                      $begingroup$

                                                      Transpose[{listX}[[ConstantArray[1, numberX]]]]





                                                      share|improve this answer









                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        Transpose[{listX}[[ConstantArray[1, numberX]]]]





                                                        share|improve this answer









                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          Transpose[{listX}[[ConstantArray[1, numberX]]]]





                                                          share|improve this answer









                                                          $endgroup$



                                                          Transpose[{listX}[[ConstantArray[1, numberX]]]]






                                                          share|improve this answer












                                                          share|improve this answer



                                                          share|improve this answer










                                                          answered 11 hours ago









                                                          Henrik SchumacherHenrik Schumacher

                                                          55.7k576154




                                                          55.7k576154






















                                                              lisa is a new contributor. Be nice, and check out our Code of Conduct.










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