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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