Avoid using a loop to get sum of rows in R, where I want to start and stop the sum on different columns for each row - r

Question

I am relatively new to R from Stata. I have a data frame that has 100+ columns and thousands of rows. Each row has a start value, stop value, and 100+ columns of numerical values. The goal is to get the sum of each row from the column that corresponds to the start value to the column that corresponds to the stop value. This is direct enough to do in a loop, that looks like this (data.frame is df, start is the start column, stop is the stop column):
for(i in 1:nrow(df)) {
df$out[i] <- rowSums(df[i,df$start[i]:df$stop[i]])
}
This works great, but it is taking 15 minutes or so. Does anyone have any suggestions on a faster way to do this?

Answer 1

You can do this using some algebra (if you have a sufficient amount of memory):
DF <- data.frame(start=3:7, end=4:8)
DF <- cbind(DF, matrix(1:50, nrow=5, ncol=10))
# start end 1 2 3 4 5 6 7 8 9 10
#1 3 4 1 6 11 16 21 26 31 36 41 46
#2 4 5 2 7 12 17 22 27 32 37 42 47
#3 5 6 3 8 13 18 23 28 33 38 43 48
#4 6 7 4 9 14 19 24 29 34 39 44 49
#5 7 8 5 10 15 20 25 30 35 40 45 50
take <- outer(seq_len(ncol(DF)-2)+2, DF$start-1, ">") &
outer(seq_len(ncol(DF)-2)+2, DF$end+1, "<")
diag(as.matrix(DF[,-(1:2)]) %*% take)
#[1] 7 19 31 43 55

Answer 2

If you are dealing with values of all the same types, you typically want to do things in matrices. Here is a solution in matrix form:
rows <- 10^3
cols <- 10^2
start <- sample(1:cols, rows, replace=T)
end <- pmin(cols, start + sample(1:(cols/2), rows, replace=T))
# first 2 cols of matrix are start and end, the rest are
# random data
mx <- matrix(c(start, end, runif(rows * cols)), nrow=rows)
# use `apply` to apply a function to each row, here the
# function sums each row excluding the first two values
# from the value in the start column to the value in the
# end column
apply(mx, 1, function(x) sum(x[-(1:2)][x[[1]]:x[[2]]]))
# df version
df <- as.data.frame(mx)
df$out <- apply(df, 1, function(x) sum(x[-(1:2)][x[[1]]:x[[2]]]))
You can convert your data.frame to a matrix with as.matrix. You can also run the apply directly on your data.frame as shown, which should still be reasonably fast. The real problem with your code is that your are modifying a data frame nrow times, and modifying data frames is very slow. By using apply you get around that by generating your answer (the $out column), which you can then cbind back to your data frame (and that means you modify your data frame just once).