Avi
I fear this was all a huge social experiment.
Testing if a post titled "sexy way" would increase engagement...
On Sat, 28 Sep 2024, 07:21 , <avi.e.gross at gmail.com> wrote:
> I see a book coming:
> "666 ways to do the same thing in R ranked by sexiness."
>
> Kidding aside, if you look under the covers of some of the functions we
> are using, we may find we are taking steps back as some of them use others
> and perhaps more functionality than we need.
>
> But for a new reader , looking at many approaches may open up other ways
> and ideas and see the problem space as quite vast.
>
> -----Original Message-----
> From: R-help <r-help-bounces at r-project.org> On Behalf Of Lennart
Kasserra
> Sent: Saturday, September 28, 2024 1:59 AM
> To: Rolf Turner <rolfturner at posteo.net>; r-help at r-project.org;
> lennart.kasserra at gmail.com
> Subject: Re: [R] Is there a sexy way ...?
>
> Sorry to append, but I just realised that of course
>
> ```
>
> x |>
> pmap(c) |>
> reduce(c) |>
> unname()
>
> ```
>
> also works and is a general solution in case your list has more than
> three elements. Here, we map in parallel over all elements of the list,
> always combining the current set of elements into a vector, and then
> reduce the resulting list into a vector by combining the elements in
> order. This yields a named vector which we can un-name given this was
> not desired.n
>
> All the best,
>
> Lennart
>
> Am 28.09.24 um 07:52 schrieb Lennart Kasserra:
> > 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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>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> https://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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