R help - Nov 2009 - 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

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.Christensen at 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 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.
>

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 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.
>

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
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`- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc

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.Christensen at 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 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.
>


-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

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.Christensen at sydbank.dk>
Subject: [R] Accessing list names in lapply
To: r-help at r-project.org
Message-ID:
	<OF5533F0B9.99F78AB0-ONC1257673.004273AE-C1257673.004466FF at 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