Can I aggregate with parameters taken from data frame? - r

Question

I'd like to perform different aggregations in a loop to be applied to different row subsets of my data, but it seems tricky to achieve (if possible at all):
t <- data.frame(agg=c(list("field1"=field1, "field2"=field2), ...),
fun=c(mean, ...))
f <- function(x) {
for (i in 1:nrow(t) {
y <- aggregate(x, by=t$agg[i], FUN=t$fun[i])
# do something with y
}
}
One problem is that the field list agg triggers an error when trying to build the data frame ("object 'field1' not found"), and the other problem is that R does not like to assign a function value to fun ("cannot coerce class ""function"" to a data.frame").
Appendix:
A concrete example for my data (just to match the definitions above) could be:
> d <- data.frame(field1=round(rnorm(5, 10, 1)),field2=letters[round(rnorm(5, 10, 1))], field3=1:5)
> d
field1 field2 field3
1 11 j 1
2 11 i 2
3 10 j 3
4 12 i 4
5 11 j 5
> with(d, aggregate(d$field3,by=list(field1, field2),FUN=mean))
Group.1 Group.2 x
1 11 i 2
2 12 i 4
3 10 j 3
4 11 j 3
Playing tricks with the variable names in the data frame, I still get this:
> with(d,t <- data.frame(agg=c(list("field1"=field1, "field2"=field2)),fun=c(mean)))
Error in as.data.frame.default(x[[i]], optional = TRUE) :
cannot coerce class ""function"" to a data.frame

Answer 1

The problems were several, mostly caused by R making exceptions to general processing:
First a vector cannot be nested, but only lists can. Still all the elements are required to have the same type.
Second, data.frame does some magic treatment when constructing the variables (causing the inability to assign closures), so it cannot be used.
Finally I had to refer to variables to aggregate by name
So the definition looks like this (where , ... means "add more similar items"):
t <- list(agg=list(c("field1", "field2"), ...),
fun=list(mean, ...))
f <- function(x) {
for (i in 1:length(t$agg)) {
agg <- t$agg[[i]]
aggList <- lapply(agg, FUN=function(e) x[[e]])
names(aggList) <- agg
y <- aggregate(x, by=aggList, FUN=t$fun[[i]])
# do something with y
}
}
Note: In the actual solution I added another list holding the names of the columns to select for the aggregated data frame to avoid warnings about mean returning NA.