[R] Why a list of NULL's are reduced to NULL?

[William Dunlap]

> -----Original Message-----
> From: r-help-bounces at r-project.org 
> [mailto:r-help-bounces at r-project.org] On Behalf Of Peng Yu
> Sent: Friday, December 11, 2009 9:18 AM
> To: r-help at stat.math.ethz.ch
> Subject: Re: [R] Why a list of NULL's are reduced to NULL?
> 
> On Fri, Dec 11, 2009 at 11:01 AM, William Dunlap 
> <wdunlap at tibco.com> wrote:
> >> -----Original Message-----
> >> From: r-help-bounces at r-project.org
> >> [mailto:r-help-bounces at r-project.org] On Behalf Of Peng Yu
> >> Sent: Friday, December 11, 2009 8:44 AM
> >> To: r-help at stat.math.ethz.ch
> >> Subject: [R] Why a list of NULL's are reduced to NULL?
> >>
> >> The following examples are confusing to me. It is OK, to 
> assigned NULL
> >> to one element in a list. The result is still a list. 
> However, a list
> >> of NULL's are reduced to NULL. I don't understand how this 
> conversion
> >> occurs. Could somebody let me know what is going on?
> >
> > The "simplification" algorithm for reformatting
> > the output of apply and sapply is handy in the
> > common case when you know that FUN will return
> > the same sort of thing each time it it called.
> > The algorithm is not very useful when FUN may return
> > objects of various classes or lengths.  sapply has
> > a simplify=FALSE argument to avoid the simplification
> > (so it acts like lapply) but apply doesn't.
> >
> > I suggest you either change your function to always
> > return one class and length of object or use lapply()
> > or sapply(simplify=FALSE,...) when you must use a function
> > with variable output type.  E.g., instead of
> >   apply(X, 1, function(row){f(row)})
> > use
> >   lapply(seq_len(nrow(X)), function(rowIndex){f(X[rowIndex,])})
> > or
> >   lapply(split(X, row(X)), function(row){f(row)})
> 
> Change my function to always returning one class may not always be
> possible as I may call a third party R package that is not made by me
> and does this kind of wired things of trying to 'simplify'. And I may
> not know all the cases where the third party R package 'simplify' the
> results, which does not always return the same type. In this case, I
> can not be sure the return type is always the same. How do you deal
> with this problem?

I don't understant your constraints.  You say you cannot control
what FUN returns and you cannot control whether apply or lapply
is called.  A reproducible set of examples would help.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

> 
> >> > X=matrix(1:8, nr=4)
> >> > apply(X,1, function(x) {if(x[[1]]==3){NULL}else{x[[1]]}})
> >> [[1]]
> >> [1] 1
> >>
> >> [[2]]
> >> [1] 2
> >>
> >> [[3]]
> >> NULL
> >>
> >> [[4]]
> >> [1] 4
> >>
> >> > apply(X,1, function(x) {NULL})
> >> NULL
> >>
> >> ______________________________________________
> >> 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.
> >>
> >
> 
> ______________________________________________
> 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.
>