[R] passing (or obtaining) index or element name of list to FUN in lapply()

[Stephen Tucker]

Hi Professor Ripley,

Thanks for the response. I apologize, my examples were not too real (though
your solutions are indeed clever)... I was trying to ask more generally
whether the element name or index of 'listObj' could be obtained by the
user-function 'myfunction' when used in lapply(X=listObj,FUN=myfunction);
below I illustrate two cases in which I have come across this desire:
(1) In 'Example 1' I essentially take the list element and do some
transformations (optionally some number-crunching), and then plot it with the
element name of the list for the title.
(2) In 'Example 2' I want to read in data from the list element and write the
contents to a file; writing a header line only when operating on the first
element of the list.

## data specification
data1 <- "var1 var2
-0.44 0.17
1.03 0.93
0.85 0.39"
data2 <- "var1 var2
-0.16 0.97
0.93 0.23
0.80 0.42"
L <- list(data1=data1,data2=data2)

##=== Example 1 (want element name) ===
## function definition
plottingfunc <- function(i,x) {
  plot(read.table(textConnection(x[[i]]),header=TRUE),main=names(x)[i])
}
## function application
par(mfrow=c(2,1))
lapply(seq(along=L),plottingfunc,x=L)

##=== Example 2 (want element index) ===
## function definition
readwritefunc <- function(i,x,fout) {
  data <- read.table(textConnection(x[[i]]),header=TRUE)
  if(i==1) cat(paste(colnames(data),collapse=","),"\n",file=fout)
  write.table(data,file=fout,sep=",",col=FALSE,
              row=FALSE,quote=FALSE,append=TRUE)
}
## function application
fout <- file("out.dat",open="w")
lapply(seq(along=L),readwritefunc,x=L,fout=fout)
close(fout)

Since the above code works, I suppose this is more of a question of
aesthetics since I thought the spirit of lapply() was to operate on the
elements of a list and not its indices - I thought perhaps there is a way to
get the index number and element name from within the user-function.

Also, I recall a lesson on 'loop avoidance' from an earlier version of MASS;
this was in the days of S-PLUS dominance and perhaps less applicable now to R
as you mentioned... But old habits die hard; my amygdala still invokes a fear
response at the thought of a loop... (and as of recently, I have been
infatuated with the notion of adhering, albeit loosely, to the 'functional
programming' paradigm which makes me doubly fearful of loops)

Thanks and best regards,

Stephen

--- Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:

> On Tue, 12 Jun 2007, Stephen Tucker wrote:
> 
> > Hello everyone,
> >
> > I wonder if there is a way to pass the index or name of a list to a
> > user-specified function in lapply(). For instance, my desired effect is
> > something like the output of
> >
> >> L <- list(jack=4098,sape=4139)
> >> lapply(seq(along=L),function(i,x) if(i==1) "jack" else "sape",x=L)
> > [[1]]
> > [1] "jack"
> >
> > [[2]]
> > [1] "sape"
> 
> as.list(names(L))
> 
> >> lapply(seq(along=L),function(i,x) if(names(x)[i]=="jack") 1 else 2,x=L)
> > [[1]]
> > [1] 1
> >
> > [[2]]
> > [1] 2
> 
> as.list(seq_along(L))
> 
> lapply() can be faster than a for-loop, but usually not by much: its main 
> advantage is clarity of code.
> 
> I think we need a real-life example to see what you are trying to do.
> 
> > But by passing L as the first argument of lapply(). I thought there was a
> > tangentially-related post on this mailing list in the past but I don't
> recall
> > that it was ever addressed directly (and I can't seem to find it now).
> The
> > examples above are perfectly good alternatives especially if I wrap each
> of
> > the lines in "names<-"() to return lists with appropriate names assigned,
> but
> 
> Try something like
> 
> L[] <- lapply(seq_along(L),function(i,x) if(i==1) "jack" else "sape",x=L)
> 
> > it feels like I am essentially writing a FOR-LOOP - though I was
> surprised to
> > find that speed-wise, it doesn't seem to make much of a difference
> (unless I
> > have not selected a rigorous test):
> >
> >> N <- 10000
> >> y <- runif(N)
> > ## looping through elements of y
> >> system.time(lapply(y,
> > +                    function(x) {
> > +                      set.seed(222)
> > +                      mean(rnorm(1e4,x,1))
> > +                    }))
> > [1] 21.00  0.17 21.29    NA    NA
> > ## looping through indices
> >> system.time(lapply(1:N,
> > +                    function(x,y) {
> > +                      set.seed(222)
> > +                      mean(rnorm(1e4,y[x],1))
> > +                      },y=y))
> > [1] 21.09  0.14 21.26    NA    NA
> >
> > In Python, there are methods for Lists and Dictionaries called
> enumerate(),
> > and iteritems(), respectively. Example applications:
> >
> > ## a list
> > L = ['a','b','c']
> > [x for x in enumerate(L)]
> > ## returns index of list along with the list element
> > [(0, 'a'), (1, 'b'), (2, 'c')]
> >
> > ## a dictionary
> > D = {'jack': 4098, 'sape': 4139}
> > [x for x in D.iteritems()]
> > ## returns element key (name) along with element contents
> > [('sape', 4139), ('jack', 4098)]
> >
> > And this is something of the effect I was looking for...
> >
> > Thanks to all,
> >
> > Stephen
> >
> > ______________________________________________
> > R-help at stat.math.ethz.ch 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.
> >
> 
> -- 
> Brian D. Ripley,                  ripley at stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>