R help - Sep 2024 - Is there a sexy way ...?

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
>

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
>>