Repeating a list k times

Question

I have a list, say $\{1,2,3,4\}$ and I would like to repeat this list k times to obtain a list of $4k$ elements that looks like $\{1,2,3,4,1,2,..,1,2,3,4\}$.

I was hoping something like li * k would do this but unfortunately that's pointwise multiplication. What's the right operator that would do something like this for me?

Answer 1

list = {1, 2, 3, 4};

a = Flatten @ Table[list, 4]

{1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4}

b = Round @ Flatten @ ReplicateLayer[4, 1] @ list

{1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4}

c = Round @ Flatten @ ReplicateLayer[4, 2] @ list

{1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4}

ReplicateLayer packs, Table doesn't:

Developer`PackedArrayQ /@ {a, b, c}

{False, True, True}

what-is-a-mathematica-packed-array

One can easily chain layers:

rep = Round @* ReplicateLayer[2,1] @* Flatten @* ReplicateLayer[2, 1];
rep @ list // MatrixForm

Answer 2

"I was hoping something like li * k would do this ..."

ConstantArray + Splice

ClearAll[Star]
Star[l_, k_] := ConstantArray[Splice @ l, k]
{1, 2, 3, 4}⋆3

 {1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4}

Use Table instead of ConstantArray to get the same result.

PadRight

ClearAll[Diamond]
Diamond[l_List, k_Integer] := PadRight[l, k Length@l, l]
{1, 2, 3, 4}⋄3

 {1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4}

Note: You can use any of the Operators without Built-in Meanings instead of \[Star] and \[Diamond].

Note: Although still undocumented, PadRight[l, k Length@l, "Periodic"] also works.

Answer 3

Maybe we can get 10 ways to do this. Just to the add to the list of fun ways to do this using fun function names:

[l0_, k_] := Nest[Join[#, l0] &, {}, k]
[Range@4, 3]
({1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4})

Answer 4

SubstitutionSystem[MapApply[Rule]@Partition[Range[#1],
  2,1,1],1,(#1 #2)-1]&[4,3]//Flatten

(* {1,2,3,4,1,2,3,4,1,2,3,4} *)


SubstitutionSystem[MapApply[Rule]@Partition[Range[#1],
  2,1,1],1,(#1 #2)-1]&[10,3]//Flatten

(* {1,2,3,4,5,6,7,8,9,10,
    1,2,3,4,5,6,7,8,9,10,
    1,2,3,4,5,6,7,8,9,10} *)

Answer 5

Using ConstantArray and Cases:

f[lst_, k_] := Cases[ConstantArray[lst, k], x_ :> Sequence @@ x]
f[list, 4]
({1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4})

Answer 6

list = {1, 2, 3, 4};
k = 4;

ArrayReshape[list, {Length[list] k}, list]

 (* {1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4} *)

Flatten@Array[list &, k]

 (* {1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4} *)
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