[R] substitute "x" for "pattern" in a list, while preservign list "structure". lapply, gsub, list...?

[Marc Schwartz]

On Wed, 2007-05-16 at 13:18 -0400, Gabor Grothendieck wrote:
> On 5/16/07, Marc Schwartz <marc_schwartz at comcast.net> wrote:
> > On Wed, 2007-05-16 at 09:25 -0700, new ruser wrote:
> > > I am experimenting with some of the common r functions.
> > > I had a question re:using "gsub" (or some similar functions) on the
> > > contents of a list.
> > >
> > > I want to design a function that looks at "everything" contained din a
> > > list, and anytime it finds the text string "pattern" replace it with
> > > "x".  I also wish to preserve the "structure" of the original list.
> > > What is a good way to accomplish this?
> > >
> > > I tried :
> > >
> > > a = matrix(data=c(23,45,'red',78),nrow=2)
> > > b = c('red','green',1,2,3)
> > > d = data.frame( test1=c(223,445,'red',78,56) , test2=
> > > c('red',NA,NA,NA,NA) )
> > > e= list(a,b,d)
> > > list1 = list(a,b,d,e)
> > >
> > > list2 = lapply(list1,function(list)(gsub("red","green",list)))
> > >
> > > str(list1)
> > > str(list2)
> > >
> > > but the structue fo the list changed.
> >
> > I suspect that you will need to use rapply(), which is a recursive
> > version of lapply().
> >
> > For example:
> >
> > > str(list1)
> > List of 4
> >  $ : chr [1:2, 1:2] "23" "45" "red" "78"
> >  $ : chr [1:5] "red" "green" "1" "2" ...
> >  $ :'data.frame':       5 obs. of  2 variables:
> >  ..$ test1: Factor w/ 5 levels "223","445","56",..: 1 2 5 4 3
> >  ..$ test2: Factor w/ 1 level "red": 1 NA NA NA NA
> >  $ :List of 3
> >  ..$ : chr [1:2, 1:2] "23" "45" "red" "78"
> >  ..$ : chr [1:5] "red" "green" "1" "2" ...
> >  ..$ :'data.frame':    5 obs. of  2 variables:
> >  .. ..$ test1: Factor w/ 5 levels "223","445","56",..: 1 2 5 4 3
> >  .. ..$ test2: Factor w/ 1 level "red": 1 NA NA NA NA
> >
> >
> >
> > list3 <- rapply(list1, function(x) gsub("red", "green", x),
> >                how = "replace")
> >
> > > str(list3)
> > List of 4
> >  $ : chr [1:2, 1:2] "23" "45" "green" "78"
> >  $ : chr [1:5] "green" "green" "1" "2" ...
> >  $ :List of 2
> >  ..$ test1: chr [1:5] "223" "445" "green" "78" ...
> >  ..$ test2: chr [1:5] "green" NA NA NA ...
> >  $ :List of 3
> >  ..$ : chr [1:2, 1:2] "23" "45" "green" "78"
> >  ..$ : chr [1:5] "green" "green" "1" "2" ...
> >  ..$ :List of 2
> >  .. ..$ test1: chr [1:5] "223" "445" "green" "78" ...
> >  .. ..$ test2: chr [1:5] "green" NA NA NA ...
> >
> >
> > Note however, the impact of using gsub(), which is that factors are
> > coerced to characters. So consider what you want the end game to be.
> >
> > See ?rapply for more information.
> >
> 
> Also note that from your str's above it appears that data.frames do not
> survive rapply but rather are changed to plain lists.  I don't know whether
> that is intentional on the part of the developers or not.

Right, which I noted and would be consistent with the behavior of
lapply() on data frames, which also simplifies the result to a normal
list.

Since rapply() is documented to be recursive version of lapply(), the
result above would be expected, at least implicitly.

Thanks for raising the point Gabor.

Regards,

Marc