Try this:
g - list(a=1:3, b=4:6, c=7:9)
with(stack(g), split(stack(g), ind))
On 21/02/2008, Lauri Nikkinen [EMAIL PROTECTED] wrote:
R users,
I have a simple lapply question.
g - list(a=1:3, b=4:6, c=7:9)
g - lapply(g, function(x) as.data.frame(x))
lapply(g, function(x) cbind(x, var1 =
R users,
I have a simple lapply question.
g - list(a=1:3, b=4:6, c=7:9)
g - lapply(g, function(x) as.data.frame(x))
lapply(g, function(x) cbind(x, var1 = rep(names(g), each=nrow(x))[1:nrow(x)]))
I get
$a
x var1
1 1a
2 2a
3 3a
$b
x var1
1 4a
2 5a
3 6a
$c
x var1
1
On Thursday 21 February 2008 (19:22:40), Lauri Nikkinen wrote:
R users,
I have a simple lapply question.
g - list(a=1:3, b=4:6, c=7:9)
g - lapply(g, function(x) as.data.frame(x))
And I would like to have
$a
x var1
1 1 a
2 2 a
3 3 a
$b
x var1
1 4 b
2 5 b
3 6
I don't see any difference
--- Lauri Nikkinen [EMAIL PROTECTED] wrote:
R users,
I have a simple lapply question.
g - list(a=1:3, b=4:6, c=7:9)
g - lapply(g, function(x) as.data.frame(x))
lapply(g, function(x) cbind(x, var1 = rep(names(g),
each=nrow(x))[1:nrow(x)]))
I get
$a
x
mapply(data.frame,value=g,nm=lapply(names(g),rep,length(g)),SIMPLIFY=FALSE)
Cheers,
Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of John Kane
Sent: Thursday, February 21, 2008 2:05 PM
To: Lauri Nikkinen; [EMAIL PROTECTED]
Subject:
Jim's solution showed me that mine should be simplified to:
mapply(data.frame,value=g,nm=names(g),SIMPLIFY=FALSE)
This has the slight advantage of automatically naming the list.
Cheers,
Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of jim
Here is one way of doing it:
lapply(names(g), function(z)cbind(x=g[[z]], var1=z))
[[1]]
x var1
1 1a
2 2a
3 3a
[[2]]
x var1
1 4b
2 5b
3 6b
[[3]]
x var1
1 7c
2 8c
3 9c
On Thu, Feb 21, 2008 at 1:22 PM, Lauri Nikkinen [EMAIL PROTECTED] wrote:
R users,
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