concatenate a list with itself

Question

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.

This makes me think there is a duplicate somewhere ArrayPad[{1, 2}, 2, "Periodic"] — Kuba, Apr 04 '17 at 20:05
Is this close enough for a duplicate? Generate cyclic list from a list — Kuba, Apr 04 '17 at 20:10
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:-)! — Dimitris, Apr 05 '17 at 20:13
Maybe just because you encounter two "non-normal" answers.So you will think this.:-) — yode, Apr 05 '17 at 20:19

score 4 · Accepted Answer · edited Apr 13 '17 at 12:55

4

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.

edited Apr 13 '17 at 12:55

Community

1

answered Apr 04 '17 at 20:13

yode

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Anton Antonov · Answer 2 · 2017-04-05T20:31:12.367

4

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.} *)

edited Apr 05 '17 at 20:31

answered Apr 04 '17 at 20:17

Anton Antonov

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Upvote for trick.. – yode Apr 04 '17 at 20:18
@yode It is related to your answer, using a dedicated function. – Anton Antonov Apr 04 '17 at 20:19

concatenate a list with itself

2 Answers2