Split vector such that means of two groups are nearly equal - r

Question

I have a data frame df with 2 variables A and B. I would like to split A in groups 1 and 2 so that mean(df$B[df$group==1]) as close as possible to mean(df$B[df$group==2])
Or just to express it otherwise, what I would like is to find a cut point (cutp) in df$A that would minimize the abs(mean(df$B[df$A<cutp])-mean(df$B[df$A>=cutp]))
Any ideas?

Answer 1

If you want to find a threshold on variable A, to split the data into two groups, so that the means of B in those two groups be similar, you can compute these means for all possible cut-points, and check when the distance between those means is minimal.
# Sample data
n <- 10
d <- data.frame(
A = rnorm(n),
B = rnorm(n)
)
# The quantity to minimize
# (You can use a loop instead of apply.)
d$differences <- apply(
d, 1,
# Compute the difference of the means for each value of A
function (u) {
i <- d$A <= u[1];
abs( mean( d$B[which(i)]) - mean(d$B[which(!i)] ) )
}
)
# The mean of an empty vector is NaN: discard those values
d$differences[ ! is.finite( d$differences ) ] <- Inf
# Take the minimum
threshold <- d$A[ which.min( d$differences ) ]
# Build the groups
d$group <- ifelse( d$A <= threshold, "group 1", "group 2" )

Answer 2

I'm still not sure how column A factors into it. It seems you want to create a new column that has two levels which create ~= mean values for column B. Column A is obviously associated with the new column created, but does not directly factor into the calculation needed. Am I missing something?
Regardless, here's a start (note this can be made much more robust, but proof of concept should work). Define a tolerance that you find acceptable and then set up a while loop to create new groups until the condition is met, i.e.
FUN <- function(tol){
df$groups <- sample(1:2, nrow(df), TRUE)
while(abs(mean(df$B[df$groups == 1]) - mean(df$B[df$groups == 2])) > tol) {
df$groups <- sample(1:2, nrow(df), TRUE)
}
return(df)
}
set.seed(101)
df <- data.frame(A=runif(20),B=runif(20))
#Test it. Means should be less than .02 different and have roughly equivalent sample sizes.
set.seed(101)
out <- FUN(.02)
library(plyr)
> ddply(out, "groups", summarize, n = length(B), mean = mean(B))
groups n mean
1 1 11 0.5229024
2 2 9 0.5037279
I should note that you could create a runaway function if you set tol super low so don't blame me if your computer crashes.