"I thought: why make this overly complicated,..."
Indeed, though "complicated" is in the eyes of the beholder.
One wonders whether any of this is necessary, though: see ?apply , as in
apply(a, 1, whatever...)
to do things rowwise.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind i
On 2020-04-09 18:00 +, aiguo li wrote:
| That is awesome! Thanks.
Dear AiGuo,
I thought: why make this overly
complicated, when this is also
possible:
a <- matrix(LETTERS[1:16], nrow=4)
X <- split(x=a[,-1], f=a[,1])
lapply(X=X, FUN=as.factor)
Best,
Rasmus
_
On 2020-04-09 18:00 +, aiguo li wrote:
| That is awesome! Thanks.
I'm glad this was helpful for you!
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On 2020-04-09 18:50 +0100, Rui Barradas wrote:
| Hello,
|
| Your post is unreadable, please repost in
| *plain text*, not HTML.
Hi! It was not so bad? I was able to
extract out the core parts at least to
prepare an answer ... maybe a bit hard with
no line breaks, but ...
Best,
Rasmus
___
Hello,
Your post is unreadable, please repost in *plain text*, not HTML.
Rui Barradas
Às 16:00 de 09/04/20, aiguo li via R-help escreveu:
Hello allI need to create a r list with each row as a list object and named with the
element in the first column. Illustrated below:> a<-
as.data.frame(m
On 2020-04-09 15:00 +, aiguo li via R-help wrote:
| Hello allI need to create a r list with
| each row as a list object and named with
| the element in the first column.
Dear aiguo,
Perhaps this fits your bill?
a <- matrix(LETTERS[1:16], nrow = 4)
FUN <- function(x) { as.factor(x[-1]
Hello allI need to create a r list with each row as a list object and named
with the element in the first column. Illustrated below:> a<-
as.data.frame(matrix(LETTERS[1:16],nrow = 4))> a V1 V2 V3 V41 A E I M2 B
F J N3 C G K O4 D H L P
I want the list looks like$A[1] E I MLevels
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