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I guess this is rather trivial. Given, e.g., 

L={1,2}

what are some common efficient ways to obtain

L
(*{1,2,1,2,1,2}*)

that is, the list concatenated three times with itself and obtained the new value. Thanks in advance.

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    $\begingroup$ Catenate@ConstantArray[L, 3]? $\endgroup$
    – Szabolcs
    Apr 4, 2017 at 19:59
  • $\begingroup$ closely related: 111863, 113327 related: 123743 $\endgroup$
    – Kuba ♦
    Apr 4, 2017 at 20:01
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    $\begingroup$ Outer[Times, ConstantArray[1, 5], {1, 2}] // Flatten $\endgroup$
    – bill s
    Apr 4, 2017 at 20:05
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    $\begingroup$ Is this close enough for a duplicate? Generate cyclic list from a list $\endgroup$
    – Kuba ♦
    Apr 4, 2017 at 20:10
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    $\begingroup$ Thanks for the comments and suggested workarounds. I accept the duplicate nature of the question but just as a side remark. I wrote: ""I guess this is rather trivial". After having seen the responses I think I should not have written this phrase. I use Mathematica for several years and still I could not find anything to do it on my oown. I use Python for one month and I am able to simply write 3*L to get the desired output:-)! $\endgroup$
    – Dimitris
    Apr 5, 2017 at 20:13

2 Answers 2

Reset to default
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L = {1, 2};
Charting`padList[L, 6]

{1,2,1,2,1,2}

Ps:Function Charting`CommonDump`listPad,Catenate,Flatten,ReplicateLayer,PaddingLayer,Table,ConstantArray,ArrayReshape and Array is relevant.Also ArrayPad from Kuba's comment here.

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There is a dedicated function in R:

Needs["RLink`"]    
InstallR[]    
REvaluate["rep(c(1,2),5)"]

(* {1., 2., 1., 2., 1., 2., 1., 2., 1., 2.} *)

Another possibility is to use length.out:

REvaluate["rep(c(1,2),length.out=5)"]    

(* {1., 2., 1., 2., 1.} *)
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  • $\begingroup$ Upvote for trick.. $\endgroup$
    – yode
    Apr 4, 2017 at 20:18
  • $\begingroup$ @yode It is related to your answer, using a dedicated function. $\endgroup$
    – Anton Antonov
    Apr 4, 2017 at 20:19

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