R help - Jul 2012 - Assigning a vector to every element of a list.

I have a vector d of unknown length, and a list b of unknown length. I
would like to replace every element of b with d. Simply writing b<-d does
not work as R tries to fit every element of d to a different element of d,
and b<-rep(d,length(b)) does not work either as it makes a list of
length length(d)*length(b) not a list of length(b). I know how to do this
with a for loop, but I feel that there has to be a more efficient way. Any
suggestions?

	[[alternative HTML version deleted]]

On 2012-07-02 15:16, Spencer Maynes wrote:
> I have a vector d of unknown length, and a list b of unknown length. I
> would like to replace every element of b with d. Simply writing b<-d
does
> not work as R tries to fit every element of d to a different element of d,
> and b<-rep(d,length(b)) does not work either as it makes a list of
> length length(d)*length(b) not a list of length(b). I know how to do this
> with a for loop, but I feel that there has to be a more efficient way. Any
> suggestions?
lapply( b, function(x) x[] <- d )

Peter Ehlers

On Mon, Jul 2, 2012 at 6:16 PM, Spencer Maynes <smaynes89 at gmail.com>
wrote:
> I have a vector d of unknown length, and a list b of unknown length. I
> would like to replace every element of b with d. Simply writing b<-d
does
> not work as R tries to fit every element of d to a different element of d,
> and b<-rep(d,length(b)) does not work either as it makes a list of
> length length(d)*length(b) not a list of length(b). I know how to do this
> with a for loop, but I feel that there has to be a more efficient way. Any
> suggestions?
>
Try this where the first line creates a list, L, whose elements we
want to replace and the second line replaces every element with the
indicated vector:
> L <- list(1, 1:2, "abc")
> L[] <- list(1:4)
> L
[[1]]
[1] 1 2 3 4

[[2]]
[1] 1 2 3 4

[[3]]
[1] 1 2 3 4

-- 
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Hi,
If you want to assign a vector to every element of a list,
vec1<-11:20

list1<-split(LETTERS[1:10],1:length(LETTERS[1:10]))
list2<-lapply(1:10,function(x) vec1)
or,
list3<-lapply(list1,function(x) list1=vec1)
or
list4<-list()
vec2<-1:5
list4[1:length(list1)]<-list(vec2)

# if you want to assign each element of vector to each element of list (assuming
both have the same lengths),

vec1<-11:20
list1<-split(LETTERS[1:10],1:length(LETTERS[1:10]))
newlist<-split(vec1,1:length(vec1))


A.K.


----- Original Message -----
From: Spencer Maynes <smaynes89 at gmail.com>
To: r-help at r-project.org
Cc: 
Sent: Monday, July 2, 2012 6:16 PM
Subject: [R] Assigning a vector to every element of a list.

I have a vector d of unknown length, and a list b of unknown length. I
would like to replace every element of b with d. Simply writing b<-d does
not work as R tries to fit every element of d to a different element of d,
and b<-rep(d,length(b)) does not work either as it makes a list of
length length(d)*length(b) not a list of length(b). I know how to do this
with a for loop, but I feel that there has to be a more efficient way. Any
suggestions?

??? [[alternative HTML version deleted]]

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b <- rep(list(d), length(b))

On 02/07/2012 23:16, Spencer Maynes wrote:
> I have a vector d of unknown length, and a list b of unknown length. I
> would like to replace every element of b with d. Simply writing b<-d
does
> not work as R tries to fit every element of d to a different element of d,
> and b<-rep(d,length(b)) does not work either as it makes a list of
> length length(d)*length(b) not a list of length(b). I know how to do this
> with a for loop, but I feel that there has to be a more efficient way. Any
> suggestions?
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
-- 
Patrick Burns
pburns at pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
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