[R] Accessing list names in lapply
Bjarke Christensen
Bjarke.Christensen at sydbank.dk
Thu Nov 19 15:43:57 CET 2009
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.Christensen at sydbank.dk>
cc
r-help at 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
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Romain Francois
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