Hi Rolf,
this topic is probably already saturated, but here is a tidyverse solution:
```
library(purrr)
x <- list(
`1` = c(7, 13, 1, 4, 10),
`2` = c(2, 5, 14, 8, 11),
`3` = c(6, 9, 15, 12, 3)
)
x |>
pmap(~ c(..1, ..2, ..3)) |>
reduce(c)
#> [1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
```
Here, we map over the elements of the list in parallel (hence pmap),
always combining the elements at the current position into a vector,
which will result in a list like this:
```
[[1]]
[1] 7 2 6
[[2]]
[1] 13 5 9
...
```
And then we reduce this resulting list into a vector by successively
combining its elements with `c()`. I think the formula syntax is a bit
idiosyncratic, you could also do this with an anonymous function like
pmap(\(`1`, `2`, `3`) c(`1`, `2`, `3`)), or if the list was unnamed as
pmap(\(x, y, z) c(x, y, z)).
I personally find the tidyverse-esque code to be very explicit &
readable, but given base R can do this very concisely one might argue
that it is superfluous to bring in an extra library for this. I think
Bert's solution (
`c(do.call(rbind, x))`) is great if `f` has no substantive meaning, and
Deepayan's solution (`unsplit(x, f)`) is perfect in case it does - does
not get much sexier than that, I am afraid.
Best,
Lennart
Am 27.09.24 um 05:55 schrieb Rolf Turner:
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
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