Re: [R] How to iteratively extract elements out of a list
Hi
try to do it without loop
lapply(m,function(x) x[x2])
HTH
Petr
On 25 Aug 2006 at 13:52, xpRt.wannabe wrote:
Date sent: Fri, 25 Aug 2006 13:52:51 -0500
From: xpRt.wannabe [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] How to iteratively extract elements out of a list
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as the size of 'n') the values that are greater than 2 in each
component of 'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch 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.
Petr Pikal
[EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
sub.m - lapply(m, function(x)x[x2])
sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
sub4.m - lapply(m, function(x)x[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
xpRt.wannabe wrote:
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch 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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch 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.
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
lapply(m,function(x)x[x2]) [[1]] [1] 3 4 [[2]] [1] 4 5 [[3]] [1] 4 __ R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
Thank you!
On 8/26/06, Patrick Burns [EMAIL PROTECTED] wrote:
sub.m - lapply(m, function(x)x[x2])
sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
sub4.m - lapply(m, function(x)x[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
xpRt.wannabe wrote:
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch 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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch 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.
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
| sub.m - lapply(m, function(x)x[x2])
| sub.m
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| [[3]]
| [1] 4
|
| sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| sub4.m - lapply(m, function(x)x[x4])
| sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
| [[1]]
| [1] 5
Or slightly shorter in this case:
sub.m[sapply(sub.m, function(x) length(x) == 2)]
etc.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
On 8/26/06, Patrick Connolly [EMAIL PROTECTED] wrote: On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote: | sub.m - lapply(m, function(x)x[x2]) | sub.m | [[1]] | [1] 3 4 | | [[2]] | [1] 4 5 | | [[3]] | [1] 4 | | sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))] | [[1]] | [1] 3 4 | | [[2]] | [1] 4 5 | | sub4.m - lapply(m, function(x)x[x4]) | sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))] | [[1]] | [1] 5 Or slightly shorter in this case: sub.m[sapply(sub.m, function(x) length(x) == 2)] or even shorter: sub.m[sapply(sub.m, length) == 2] __ R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch 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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch 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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch 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.
Re: [R] How to iteratively extract elements out of a list
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch 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.
