Yes, unsplit() it is. I was messing around with ave() (which can be hammered
into submission, sort of).
My overlooking unsplit() is somewhat impressive in view of "svn diff -c
18591"
-pd
> On 27 Sep 2024, at 17:08 , Martin Maechler wrote:
>
>> Chris Evans via R-help
>>on Fri
gt; Sent: Saturday, September 28, 2024 10:11 PM
> To: CALUM POLWART
> Cc: avi.e.gr...@gmail.com; Lennart Kasserra ;
> r-help@r-project.org
> Subject: Re: [R] Is there a sexy way ...?
>
>
>
> On Sat, 28 Sep 2024 10:26:31 +0100
> CALUM POLWART wrote:
>
> > Avi
>
ringr
packages to make my silly version ever less silly! LOL!
-Original Message-
From: Rolf Turner
Sent: Saturday, September 28, 2024 10:11 PM
To: CALUM POLWART
Cc: avi.e.gr...@gmail.com; Lennart Kasserra ;
r-help@r-project.org
Subject: Re: [R] Is there a sexy way ...?
On Sat, 28 Sep 2024
On Sat, 28 Sep 2024 10:26:31 +0100
CALUM POLWART wrote:
> Avi
>
> I fear this was all a huge social experiment.
>
> Testing if a post titled "sexy way" would increase engagement...
I conjecture that this conjecture was tongue-in-cheek. Be that as it
were 😊️, let me assure everyone that s
ct sense. Others are happier with a while(True) construct
that makes clear the contents will decide when to break out. To expect
people to agree on what is "sexy" is not a reasonable expectation. LOL!
-Original Message-----
From: Sorkin, John
Sent: Saturday, September 28, 2024 12:01
On Sat, 28 Sep 2024, J C Nash wrote:
On 2024-09-28 13:57, avi.e.gr...@gmail.com wrote:
Python users often ask if a solution is “pythonic”. But I am not aware
of R users having any special name like “R-thritic” and that may be a
good thing.
Nice, added on R-Forge :-)
Achim
__
On 2024-09-28 13:57, avi.e.gr...@gmail.com wrote:
Python users often ask if a solution is “pythonic”. But I am not aware
of R users having any special name like “R-thritic” and that may be a
good thing.
__
R-help@r-project.org mailing list -- To UN
of R users having any special name like “R-thritic” and that
may be a good thing.
From: CALUM POLWART
Sent: Saturday, September 28, 2024 5:27 AM
To: avi.e.gr...@gmail.com
Cc: Lennart Kasserra ; Rolf Turner
; r-help@r-project.org
Subject: Re: [R] Is there a sexy way ...?
Avi
I fear
problem space as quite vast.
>
> -Original Message-
> From: R-help On Behalf Of Lennart Kasserra
> Sent: Saturday, September 28, 2024 1:59 AM
> To: Rolf Turner ; r-help@r-project.org;
> lennart.kasse...@gmail.com
> Subject: Re: [R] Is there a sexy way ...?
>
> Sorry
om
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 o
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 in
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 1
0 11 12
c 13 14 15 16 17 18
d 19 20 21 22 23 24
-Original Message-
From: Sorkin, John
Sent: Saturday, September 28, 2024 12:01 AM
To: avi.e.gr...@gmail.com; 'Rolf Turner' ;
r-help@r-project.org
Subject: Re: [R] Is there a sexy way ...?
"Sexy code&q
: 'Rolf Turner'; r-help@r-project.org
Subject: Re: [R] Is there a sexy way ...?
Rold,
We need to be clear on what makes an answer sexy! LOL!
I decided it was sexy to do it in a way that nobody (normal) would and had
not suggested yet.
Here is an original version I will explain in a
ual items
# Python, not R.
[num
for elem in [(first, second, third) for first, second, third in zip(*x)]
for num in elem]
[7, 2, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3]
For any interested, you can combine python and R in the same program back
and forth on the same data inside what is st
This post obviously beats a fossilized horse, so feel free to ignore.
So here's my problem with my "solution" to Rolf's query as well as
several others: it assumes that one knows the details of how matrices
are stored as vectors with a 'dim' attribute. Although this might be
considered elementary
, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3]
---END OUTPUT ---
I am not saying to use python, just showing other ways people find
comfortable. But choosing the right data structure can make things easy, or
at least easier.
-Original Message-
From: R-help On Behalf Of Rolf Turner
Sent: Th
> Chris Evans via R-help
> on Fri, 27 Sep 2024 12:20:47 +0200 writes:
> Oh glorious! Thanks Duncan.
> Fortune cookie nomination!
I don't disagree with the nomination -- thank you, Duncan!
However, please note that I'm sure Rolf's was challenged /
question was ment to work
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 <- facto
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 l
Rolf can tell us for sure but I thought the goal was to use v ?
Maybe not ? Either way, I think Bert wins for shortest and Kimmo
wins for longest. IMHO, elegance is in the eye of the
beholder.
On Fri, Sep 27, 2024 at 4:35 AM Stephen Berman via R-help <
r-help@r-project.org> wrote:
> Yet ano
Yet another way (not as sexy as Deepayan's):
as.vector(t(sapply(x, c)))
Steve Berman
On Fri, 27 Sep 2024 10:45:06 +0300 Eric Berger wrote:
> v <- as.numeric(matrix(unlist(x),ncol=5,byrow=TRUE))
> v
> [1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
>
> On Fri, Sep 27, 2024 at 8:33 AM Deepayan
Dear Rolf, dear all,
this was an inspiring challenge :-) This seems to do the task...
--- snip ---
x <- list(`1` = c(7, 13, 1, 4, 10),
`2` = c(2, 5, 14, 8, 11),
`3` = c(6, 9, 15, 12, 3))
f <- factor(rep(1:3,5))
v <- as.vector(unlist(x)[ paste(rep(levels(f), length(x[[1]]
v <- as.numeric(matrix(unlist(x),ncol=5,byrow=TRUE))
v
[1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
On Fri, Sep 27, 2024 at 8:33 AM Deepayan Sarkar
wrote:
>
> > unsplit(x, f)
> [1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
>
> Is more general (works if the subgroups are imbalanced), and
> unsplit(x, f)
[1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
Is more general (works if the subgroups are imbalanced), and hopefully more
sexy as well :-)
Best,
-Deepayan
On Fri, 27 Sept 2024 at 10:11, Bert Gunter wrote:
> ... And, in fact, I just realized that
>
> c(do.call(rbind, x))
>
... And, in fact, I just realized that
c(do.call(rbind, x))
is even better.
-- Bert
On Thu, Sep 26, 2024 at 9:26 PM Bert Gunter wrote:
> Sorry, hit send by accident.
> The 2-line version is:
>
> x <- do.call(rbind, x)
> dim(x) <- NULL
>
> Cheers,
> Bert
>
> On Thu, Sep 26, 2024 at 9:23 PM Be
Sorry, hit send by accident.
The 2-line version is:
x <- do.call(rbind, x)
dim(x) <- NULL
Cheers,
Bert
On Thu, Sep 26, 2024 at 9:23 PM Bert Gunter wrote:
> How about:
> as.vector(do.call(rbind,x))
>
> Cheers,
> Bert
>
>
>
>
> However, I much prefer a 2 line version:
>
> On Thu, Sep 26, 2024 at
How about:
as.vector(do.call(rbind,x))
Cheers,
Bert
However, I much prefer a 2 line version:
On Thu, Sep 26, 2024 at 8:56 PM 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))
> an
4 8 12 10 11 3
-Original Message-
From: R-help On Behalf Of Rolf Turner
Sent: Thursday, September 26, 2024 11:56 PM
To: r-help@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),
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
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