[R] Is there a sexy way ...?

[Bert Gunter]

How about:
as.vector(do.call(rbind,x))

Cheers,
Bert




However, I much prefer a 2 line version:

On Thu, Sep 26, 2024 at 8:56 PM Rolf Turner <rolfturner using posteo.net> wrote:

>
> 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
>
> --
> Honorary Research Fellow
> Department of Statistics
> University of Auckland
> Stats. Dep't. (secretaries) phone:
>          +64-9-373-7599 ext. 89622
> Home phone: +64-9-480-4619
>
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