Oh glorious! Thanks Duncan. Fortune cookie nomination!
On 27/09/2024 11:13, Duncan Murdoch wrote:
On 2024-09-26 11:55 p.m., Rolf Turner 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.Don't you find a for loop's naked display of intention to be sexy? Duncan Murdoch
-- Chris Evans (he/him)Visiting Professor, UDLA, Quito, Ecuador & Honorary Professor, University of Roehampton, London, UK.
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