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

[@vi@e@gross m@iii@g oii gm@ii@com]

Rolf,

This works, albeit you may not be thrilled:

x <- list(`1` = c(7, 13, 1, 4, 10),
          `2` = c(2, 5,  14, 8, 11),
          `3` = c(6, 9, 15, 12, 3)) 

as.vector(rbind(x[[1]], x[[2]], x[[3]]))

-- output:

> as.vector(rbind(x[[1]], x[[2]], x[[3]]))
 [1]  7  2  6 13  5  9  1 14 15  4  8 12 10 11  3

-----Original Message-----
From: R-help <r-help-bounces using r-project.org> On Behalf Of Rolf Turner
Sent: Thursday, September 26, 2024 11:56 PM
To: r-help using r-project.org
Subject: [R] Is there a sexy way ...?


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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