Using paste to create logical expression for data frame subset - r

Question

I have two dataframes, remove and dat (the actual dataframe). remove specifies various combinations of the factor variables found in dat, and how many to sample (remove$cases).
Reproducible example:
set.seed(83)
dat <- data.frame(RateeGender=sample(c("Male", "Female"), size = 1500, replace = TRUE),
RateeAgeGroup=sample(c("18-39", "40-49", "50+"), size = 1500, replace = TRUE),
Relationship=sample(c("Direct", "Manager", "Work Peer", "Friend/Family"), size = 1500, replace = TRUE),
X=rnorm(n=1500, mean=0, sd=1),
y=rnorm(n=1500, mean=0, sd=1),
z=rnorm(n=1500, mean=0, sd=1))
What I am trying to accomplish is to read in a row from remove and use it to subset dat. My current approach looks like:
remove <- expand.grid(RateeGender = c("Male", "Female"),
RateeAgeGroup = c("18-39","40-49", "50+"),
Relationship = c("Direct", "Manager", "Work Peer", "Friend/Family"))
remove$cases <- c(36,34,72,58,47,38,18,18,15,22,17,10,24,28,11,27,15,25,72,70,52,43,21,27)
# For each row of remove (combination of factor levels:)
for (i in 1:nrow(remove)) {
selection <- character()
# For each column of remove (particular selection):
for (j in 1:(ncol(remove)-1)){
add <- paste0("dat$", names(remove)[j], ' == "', remove[i,j], '" & ')
selection <- paste0(selection, add)
}
selection <- sub(' & $', '', selection) # Remove trailing ampersand
cat(selection, sep = "\n") # What does selection string look like?
tmp <- sample(dat[selection, ], size = remove$cases[i], replace = TRUE)
}
The output from cat() while the loop runs looks right, for example: dat$RateeGender == "Male" & dat$RateeAgeGroup == "18-39" & dat$Relationship == "Direct" and if I paste that into dat[dat$RateeGender == "Male" & dat$RateeAgeGroup3 == "18-39" & dat$Relationship == "Direct" ,], I get the right subset.
However, if I run the loop as written with dat[selection, ], each subset only returns NAs. I get the same outcome if I use subset(). Note, I have replace = TRUE in the above solely because of the random sampling. In the actual application, there will always be more cases per combination than required.
I know I can dynamically construct formulas for lm() and other functions using paste() in this way, but am obviously missing something in translating this into working with [,].
Any advice would be really appreciated!

Answer 1

You cannot use character expressions as you describe to subset either with [ or subset. If you wanted to do that you would have to construct the entire expression, and then use eval. That said, there is a better solution using merge. For example, let's get all the entries in dat that match the first two rows from remove:
merge(dat, remove[1:2,])
If we want all the rows that don't match those two, then:
subset(merge(dat, remove[1:2,], all.x=TRUE), is.na(cases))
This is assuming you want to join on the columns with the same names across the two tables. If you have a lot of data you should consider using data.table as it is very fast for this type of operation.

Answer 2

I upvoted BrodieG's answer before I realized it doesn't do what you wanted in situations wehre the size of the category is smaller than the number of samples desired. (In fact his method doesn't really do sampling at all, but I think it is is an elegant solution to a different question so I'm not reversing my vote. And you could use a similar split strategy as illustrated below with that data.frame as the input.).
sub <- lapply( split(dat, with(dat, paste(RateeGender, # split vector
RateeAgeGroup,
Relationship, sep="_")) ),
function (d) { n= with(remove, remove[
RateeGender==d$RateeGender[1]&
RateeAgeGroup==d$RateeAgeGroup[1]&
Relationship==d$Relationship[1],
"cases"])
cat(n);
sample(d, n, repl=TRUE) } )