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 minute. Or, maybe best a bit
before. Hee is the unformatted result whicvh is a tad hard to read but will
be made readable soon:
x <- list(`1` = c(7, 13, 1, 4, 10),
`2` = c(2, 5, 14, 8, 11),
`3` = c(6, 9, 15, 12, 3))
as.integer(unlist(strsplit(as.vector(paste(paste(x$`1`, x$`2`, x$`3`,
sep=","), collapse=",")), split=",")))
The result is: 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
After reading what others wrote, the following is more general one where any
number of vectors in a list can be handled:
as.integer(unlist(strsplit(as.vector(paste(do.call(paste, c(x, sep=",")),
collapse=",")), split=",")))
Perhaps a tad more readable is a version using the new pipe but for obvious
reasons, the dplyr/magrittr pipe works better for me than having to create
silly anonymous functions instead of using a period. You now have a
pipeline:
library(dplyr)
x %>%
c(sep=",") %>%
do.call(paste, .) %>%
paste(collapse=",") %>%
as.vector() %>%
strsplit(split=",") %>%
unlist() %>%
as.integer()
And it returns the right answer!
- You start with x and pipe it as
- the first argument to c() and the second argument already in place is an
option to later use comma as a separator
- that is piped to a do.call() which takes that c() tuple and replaces the
second argument of period with it. You now have taken the original data and
made three text strings like so:
"7,2,6" "13,5,9" "1,14,15" "4,8,12" "10,11,3"
- But you want all those strings collapsed into a single long string with
commas between the parts. Do another paste this time putting the substrings
together and collapsing with a comma. The results is:
"7,2,6,13,5,9,1,14,15,4,8,12,10,11,3"
- But that is not a vector and don't ask why!
- Now split that string at commas:
"7" "2" "6" "13" "5" "9" "1" "14" "15" "4" "8" "12" "10" "11" "3"
- and undo the odd list format it returns to flatten it back into a
character vector:
"7" "2" "6" "13" "5" "9" "1" "14" "15" "4" "8" "12" "10" "11" "3"
- Yep it looks the same but is subtly different. Time to make it into
integers or whatever:
7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
Looked at after the fact, it seems so bloody obvious! And the chance of
someone else trying this approach, justifiably, is low, LOL!
One nice feature of the do.call is this can be extended like so:
x <- list(`1` = c(7, 13, 1, 4, 10),
`2` = c(2, 5, 14, 8, 11),
`3` = c(6, 9, 15, 12, 3),
`4` = c( 101, 102, 103, 104, 105),
`5` = c(-105, -104, -103, -102, -101))
Works fine and does this for the now five columns:
[1] 7 2 6 101 -105 13 5 9 102 -104 1 14 15 103
-103 4 8 12 104 -102
[21] 10 11 3 105 -101
My apologies to all who expected a more serious post. I have been focusing
on Python lately and over there, some things are done differently albeit I
probably would be using the numpy and pandas packages to do this or even a
simple list comprehension using zip:
# Python, not R.
[ (first, second, third) for first, second, third in zip(*x)]
[(7, 2, 6), (13, 5, 9), (1, 14, 15), (4, 8, 12), (10, 11, 3)]
And, of course, that needs to be made into a list of individual 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 still called RSTUDIO and if there
are times one allows a better or at least easier for you, way to do a
transformation, you can often mix and match.
-----Original Message-----
From: R-help <r-help-boun...@r-project.org> 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),
`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
--
Honorary Research Fellow
Department of Statistics
University of Auckland
Stats. Dep't. (secretaries) phone:
+64-9-373-7599 ext. 89622
Home phone: +64-9-480-4619
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