> -----Original Message-----
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Wincent
> Sent: Thursday, April 29, 2010 8:40 PM
> To: r help
> Subject: [R] replace elements in a list
>
> Dear all, I have a list like this: l <-
> list(list(a=1,b=NULL), list(a=2,b=2))
> I want to find out the elements with value of NULL and
> replace them with NA.
I think that rapply(how="replace") ought to be able to do this
but it doesn't do the right thing when the list has a NULL in
it. E.g., see how the following returns a list the shape of
the input list, changing all the non-NULL entries as requested,
but the NULL entries are replaced by empty lists instead of 0's:
> rapply(l, function(x)
+ if(is.null(x)) 0 else paste("*", deparse(x), "*", collapse="/",
sep=""), how="replace")
[[1]]
[[1]]$a
[1] "*1*"
[[1]]$b
list()
[[2]]
[[2]]$a
[1] "*2*"
[[2]]$b
[1] "*2*"
The following function may do what you want:
replaceInList <- function (x, FUN, ...)
{
if (is.list(x)) {
for (i in seq_along(x)) {
x[i] <- list(replaceInList(x[[i]], FUN, ...))
}
x
}
else FUN(x, ...)
}
E.g.,
> replaceInList(l, function(x)if(is.null(x))0 else x)
[[1]]
[[1]]$a
[1] 1
[[1]]$b
[1] 0
[[2]]
[[2]]$a
[1] 2
[[2]]$b
[1] 2
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> The actual case has a very long list, so manually find out and replace
> them is not an option.
> I can use for loop to do this, but I want to know if there is
> vectorized way (or other ways) to do it?
>
> Thanks
> --
> Wincent Rong-gui HUANG
> Doctoral Candidate
> Dept of Public and Social Administration
> City University of Hong Kong
> http://asrr.r-forge.r-project.org/rghuang.html
>
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>
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