Re: [R] Accessing list names in lapply
Yet another possibility is to iterate on both values and names simultaneously using mapply(): df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) mapply(function(x, y) plot(x, ylab=y), df1, names(df1)) Enrique -Original Message- Date: Thu, 19 Nov 2009 13:27:08 +0100 From: Bjarke Christensen bjarke.christen...@sydbank.dk Subject: [R] Accessing list names in lapply To: r-help@r-project.org Message-ID: of5533f0b9.99f78ab0-onc1257673.004273ae-c1257673.00446...@bdpnet.dk Content-Type: text/plain; charset=US-ASCII Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ R-help@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.
Re: [R] Accessing list names in lapply
Yes, that is it. Or even simpler: mapply(plot, x=df1, ylab=names(df1), col=1:10, main=paste(Plot, names(df1))) Thank you! Bjarke Christensen Bengoechea Bartolomé Enrique (SIES 73) Til enrique.bengoech r-help@r-project.org e...@credit-suisse. cc com bjarke.christen...@sydbank.dk Emne 20-11-2009 13:19 RE: [R] Accessing list names in lapply Yet another possibility is to iterate on both values and names simultaneously using mapply(): df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) mapply(function(x, y) plot(x, ylab=y), df1, names(df1)) Enrique -Original Message- Date: Thu, 19 Nov 2009 13:27:08 +0100 From: Bjarke Christensen bjarke.christen...@sydbank.dk Subject: [R] Accessing list names in lapply To: r-help@r-project.org Message-ID: of5533f0b9.99f78ab0-onc1257673.004273ae-c1257673.00446...@bdpnet.dk Content-Type: text/plain; charset=US-ASCII Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen -- __ R-help@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.
[R] Accessing list names in lapply
Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ R-help@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.
Re: [R] Accessing list names in lapply
lapply over the list names rather than the list itself: junk - lapply(names(df1), function(nm) plot(df1[[nm]], ylab = nm)) On Thu, Nov 19, 2009 at 7:27 AM, Bjarke Christensen bjarke.christen...@sydbank.dk wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ R-help@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. __ R-help@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.
Re: [R] Accessing list names in lapply
On 19/11/2009 7:27 AM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? No, but you can loop over the indices in lapply, not just in a for loop. For example, lapply(names(df1), function(x) plot(df1[[x]], ylab=x)) Duncan Murdoch A trivial example: df1 - split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen __ R-help@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. __ R-help@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.
Re: [R] Accessing list names in lapply
Maybe this : http://tolstoy.newcastle.edu.au/R/e4/help/08/04/8720.html Romain On 11/19/2009 01:27 PM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/EAD5 : LondonR slides |- http://tr.im/BcPw : celebrating R commit #5 `- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc __ R-help@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.
Re: [R] Accessing list names in lapply
You can try this:
par(mfcol=c(5,2))
lapply(df1, function(x){
nm -
names(eval(as.list(sys.call(-1))[[2]]))[as.numeric(gsub([^0-9],
, deparse(substitute(x]
plot(x, main = nm)
})
On Thu, Nov 19, 2009 at 10:27 AM, Bjarke Christensen
bjarke.christen...@sydbank.dk wrote:
Hi,
When using lapply (or sapply) to loop over a list, can I somehow access the
index of the list from inside the function?
A trivial example:
df1 - split(
x=rnorm(n=100, sd=seq(from=1, to=10, each=10)),
f=letters[seq(from=1, to=10, each=10)]
)
str(df1)
#List of 10
# $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ...
# $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ...
# $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ...
# $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ...
#...
par(mfcol=c(5,2))
lapply(df1, plot)
This plots each element of the list, but the label on the vertical axis is
X[[0L]] (as expected from the documentation in ?lapply). I'd like the
heading for each plot to be the name of that item in the list. This can be
achieved by using a for-loop:
for (i in names(df1)) plot(df1[[i]], ylab=i)
but can it somehow be achieved bu using lapply? I would be hoping for
something like
lapply(df1, function(x) plot(x, ylab=parent.index()))
or some way to parse the index number out of the call, using match.call()
or something like that.
Thanks in advance for any comments,
Bjarke Christensen
__
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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.
--
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Re: [R] Accessing list names in lapply
Thanks to everybody who replied - I got three distinct, very useful suggestions. Bjarke Christensen Romain Francois romain.francois@ dbmail.com Til Bjarke Christensen 19-11-2009 14:33 bjarke.christen...@sydbank.dk cc r-help@r-project.org Emne Re: [R] Accessing list names in lapply Maybe this : http://tolstoy.newcastle.edu.au/R/e4/help/08/04/8720.html Romain On 11/19/2009 01:27 PM, Bjarke Christensen wrote: Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num [1:10] -1.14 -1.773 2.512 -2.072 -0.904 ... # $ d: num [1:10] 2.109 1.243 0.627 -2.343 -6.071 ... #... par(mfcol=c(5,2)) lapply(df1, plot) This plots each element of the list, but the label on the vertical axis is X[[0L]] (as expected from the documentation in ?lapply). I'd like the heading for each plot to be the name of that item in the list. This can be achieved by using a for-loop: for (i in names(df1)) plot(df1[[i]], ylab=i) but can it somehow be achieved bu using lapply? I would be hoping for something like lapply(df1, function(x) plot(x, ylab=parent.index())) or some way to parse the index number out of the call, using match.call() or something like that. Thanks in advance for any comments, Bjarke Christensen -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/EAD5 : LondonR slides |- http://tr.im/BcPw : celebrating R commit #5 `- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc __ R-help@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.
