Dear Rolf, dear all,
this was an inspiring challenge :-) This seems to do the task...
--- snip ---
x <- list(`1` = c(7, 13, 1, 4, 10),
`2` = c(2, 5, 14, 8, 11),
`3` = c(6, 9, 15, 12, 3))
f <- factor(rep(1:3,5))
v <- as.vector(unlist(x)[ paste(rep(levels(f), length(x[[1]])),
rep(1:length(x[[1]]), each=length(levels(f))), sep="") ])
v
--- snip ---
I leave it to you, whether this is an elegant solution or not ;-)
Cheers,
Kimmo
Rolf Turner kirjoitti 27.9.2024 klo 6.55:
>
> I have (toy example):
>
> x <- list(`1` = c(7, 13, 1, 4, 10),
> `2` = c(2, 5, 14, 8, 11),
> `3` = c(6, 9, 15, 12, 3))
> and
>
> f <- factor(rep(1:3,5))
>
> I want to create a vector v of length 15 such that the entries of v,
> corresponding to level l of f are the entries of x[[l]]. I.e. I want
> v to equal
>
> c(7, 2, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3)
>
> I can create v "easily enough", using say, a for-loop. It seems to me,
> though, that there should be sexier (single command) way of achieving
> the desired result. However I cannot devise one.
>
> Can anyone point me in the right direction? Thanks.
>
> cheers,
>
> Rolf Turner
>
--
Kimmo Elo
Senior Lecturer | Adjunct professor, Dr.
========================================================
University of Eastern Finland
Department of Geographical and Historical Studies
P.O. Box 111
FIN-80101 Joensuu
Finland
E-mail: kimmo....@uef.fi
ResearchGate: www.researchgate.net/profile/Kimmo_Elo
LAWPOL Consortium (PI): https://lawpol.fi/en
========================================================
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