Create new scope inside for loop in R [duplicate] - r

Question

So consider the following chunk of code which does not work as most people might expect it to
#cartoon example
a <- c(3,7,11)
f <- list()
#manual initialization
f[[1]]<-function(x) a[1]+x
f[[2]]<-function(x) a[2]+x
f[[3]]<-function(x) a[3]+x
#desired result for the rest of the examples
f[[1]](1)
# [1] 4
f[[3]](1)
# [1] 12
#attempted automation
for(i in 1:3) {
f[[i]] <- function(x) a[i]+x
}
f[[1]](1)
# [1] 12
f[[3]](1)
# [1] 12
Note that we get 12 both times after we attempt to "automate". The problem is, of course, that i isn't being enclosed in the function's private environment. All the functions refer to the same i in the global environment (which can only have one value) since a for loop does not seem to create different environment for each iteration.
sapply(f, environment)
# [[1]]
# <environment: R_GlobalEnv>
# [[2]]
# <environment: R_GlobalEnv>
# [[3]]
# <environment: R_GlobalEnv>
So I though I could get around with with the use of local() and force() to capture the i value
for(i in 1:3) {
f[[i]] <- local({force(i); function(x) a[i]+x})
}
f[[1]](1)
# [1] 12
f[[3]](1)
# [1] 12
but this still doesn't work. I can see they all have different environments (via sapply(f, environment)) however they appear to be empty (ls.str(envir=environment(f[[1]]))). Compare this to
for(i in 1:3) {
f[[i]] <- local({ai<-i; function(x) a[ai]+x})
}
f[[1]](1)
# [1] 4
f[[3]](1)
# [1] 12
ls.str(envir=environment(f[[1]]))
# ai : int 1
ls.str(envir=environment(f[[3]]))
# ai : int 3
So clearly the force() isn't working like I was expecting. I was assuming it would capture the current value of i into the current environment. It is useful in cases like
#bad
f <- lapply(1:3, function(i) function(x) a[i]+x)
#good
f <- lapply(1:3, function(i) {force(i); function(x) a[i]+x})
where i is passed as a parameter/promise, but this must not be what's happening in the for-loop.
So my question is: is possible to create this list of functions without local() and variable renaming? Is there a more appropriate function than force() that will capture the value of a variable from a parent frame into the local/current environment?

Answer 1

This isn't a complete answer, partly because I'm not sure exactly what the question is (even though I found it quite interesting!).
Instead, I'll just present two alternative for-loops that do work. They've helped clarify the issues in my mind (in particular by helping me to understand for the first time why force() does anything at all in a call to lapply()). I hope they'll help you as well.
First, here is one that's a much closer equivalent of your properly function lapply() call, and which works for the same reason that it does:
a <- c(3,7,11)
f <- list()
## `for` loop equivalent of:
## f <- lapply(1:3, function(i) {force(i); function(x) a[i]+x})
for(i in 1:3) {
f[[i]] <- {X <- function(i) {force(i); function(x) a[i]+x}; X(i)}
}
f[[1]](1)
# [1] 4
Second, here is one that does use local() but doesn't (strictly- or literally-speaking) rename i. It does, though, "rescope" it, by adding a copy of it to the local environment. In one sense, it's only trivially different from your functioning for-loop, but by focusing attention on i's scope, rather than its name, I think it helps shed light on the real issues underlying your question.
a <- c(3,7,11)
f <- list()
for(i in 1:3) {
f[[i]] <- local({i<-i; function(x) a[i]+x})
}
f[[1]](1)
# [1] 4

Answer 2

Will this approach work for you?
ff<-list()
for(i in 1:3) {
fillit <- parse(text=paste0('a[',i,']+x' ))
ff[[i]] <- function(x) ''
body(ff[[i]])[1]<-fillit
}
It's sort of a lower-level way to construct a function, but it does work "as you want it to."

Answer 3

An alternative for force() that would work in a for-loop local environment is
capture <- function(...) {
vars<-sapply(substitute(...()), deparse);
pf <- parent.frame();
Map(assign, vars, mget(vars, envir=pf, inherits = TRUE), MoreArgs=list(envir=pf))
}
Which can then be used like
for(i in 1:3) {
f[[i]] <- local({capture(i); function(x) a[i]+x})
}
(Taken from the comments in Josh's answer above)