Why doesn't my function save my assignment? - r

Question

f1 <- function(x) {
zx1 <- sample(1:nrow(zone4[[x]]), nrow(zone4[[x]]), replace=F)
zone4[[x]]$randnums <- zx1
}
f1(1)
## DOESN'T UPDATE zone4[[1]]
zx2 <- sample(1:nrow(zone4[[1]]), nrow(zone4[[1]]), replace=F)
zone4[[1]]$randnums <- zx2
## DOES UPDATE zone[[1]]
If I make a function f1() like shown above, the object 'zone4[[x]]' is not updated. However, if I run the same command as above but explicitly state 'x', as shown below, then the object 'zone4[[x]]' is updated. Why could this be? I want to know because I want to run iterations of the code. If within the definition of the function f1() above I write "names(zone4[[x]])", then the output I get tells me that the function did what it was supposed to, but when queried again, zone[[x]] appears to be unchanged. Thank you for your help. The idea is to make random numbers for each subset of a data set for a given year and another variable, zone. The data set was originally a single data frame, but I used the split() function to separate the data according to year and zone, of which there are 4. Maybe there is a better way to assign random numbers to specific subsets of data without using the split() function?

Answer 1

R functions don't usually have side effects (ie. changing things in global objects)
This is a good thing (most of the time as we don't want unintended consequences)
The idiomatic approach is to assign the results to a new object (it can be the same name to overwrite the original)
f1 <- function(x) {
zx1 <- sample(1:nrow(zone4[[x]]), nrow(zone4[[x]]), replace=F)
zone4[[x]]$randnums <- zx1
# usually a good idea to return the complete object
# especially when a replacement function (in your case `[[<-`)
# is the last one called
return(zone4)
}
zone4 <- f1(1)
An alternative would be to use data.table
library(data.table)
zone4 <- lapply(zone4, as.data.table)
f1 <- function(x) {
zone4[[x]][,randnums := sample(.N)]
invisible(NULL)
}