[R] Passing arguments between apply and l(s)apply functions vs. nested for loop

[Charles Berry]

Mark Orr <mo2259 <at> columbia.edu> writes:

> 
> Hi R community, 
> I have a question concerning passing arguments between apply and lapply?
[snip]
> 
> #START CODE SNIPPET
> 
> #LIST AND VECTOR
> rm(list=ls())
> l <- list(1:3,2:3,4:10,7:9)
> v <- 1:3
> 
> #USED IN j loop to catch values
> catch.mat <- matrix(NA,nrow=length(v),ncol=length(l))
> 
> #LOOPS
> for (i in 1:length(v)){
>   for (j in 1:length(l)) {
>     catch.mat[i,j] <- sum(l[[j]]==i)
>   }
> }
> 
> #SIMPLY APPLY OVER catch.mat
> catch.all <- apply(catch.mat,1,sum) 
> 
> catch.mat
> catch.all
> 

Given the constraint you state about uniqueness,

> table(factor(unlist(l),v))

1 2 3 
1 2 2 

gives the same answer as catch.all --- up to names which you can 
remove with unname(table(factor(unlist(l),v)))


[deleted]
> 
> #THEN I ADD A NEW FUNCTION
> f.2 <- function(x,l=l){
>   i <- x
>   rm(x)
>   return(sum(lapply(l,function(x) sum(x==i))))
> }
> 
> mat.1 <- matrix(1:2318,nrow=1,ncol=2318)
> apply(mat.1,1,f.2)
> 
> #BUT GET ERROR
> #> apply(mat.1,1,f.2)
> #Error in lapply(l, function(x) sum(x == i)) : 
> # promise already under evaluation: recursive default argument reference or
earlier problems?


Always helps to ponder the error message --- 'recursive default argument'.

Here is an example:

foo <- function(x=x) x

> foo() ## here it comes again!
Error in foo() : 
  promise already under evaluation: recursive default argument reference 
  or earlier problems?
> foo(1) # this works
[1] 1
> 

The default is what you use when the argument is not given in the call.

And x=x confuses the evaluator. 

You can avoid confusion by choosing a different name like this:

> foo <- function(xx=x) xx
> x <- 3
> foo()
[1] 3
> 


[rest deleted]

HTH,