Question

Using R, what is the best way I can aggregate rows on a condition that spans multiple rows.
For example to aggregate any rows where z = 0 for n or more times.

What this would look like run on the following sample table with n = 3.

Sample Table x:

x   y   z
0   0   6
5   5   0
40  2   0
4   0   0
10  0   1
0   0   2
11  7   0
0   4   0
0   0   0
0   0   0
0   0   2
18  0   4

Results Table:

x   y   z
0   0   6
49  7   0 <- Above two rows got aggregated
10  0   1
0   0   2
11  11  0 <- Above three rows got aggregated
0   0   2
18  0   4

Answer 1

This is the code I used to produce your result. If you have any questions, fire away.

mmf <- read.table(textConnection("x   y   z # read in your example data
0   0   6
5   5   0
40  2   0
4   0   0
10  0   1
0   0   2
11  7   0
0   4   0
0   0   0
0   0   0
0   0   2
18  0   4"), header = TRUE)

# see where there are zeros in the y column
mmf.rle <- rle(mmf$z) 
mmf.rle <- data.frame(lengths = mmf.rle$lengths, values = mmf.rle$values)

merge.rows <- 3
# select rows that have more or equal to three zeros
mmf.zero <- which(mmf.rle$values == 0 & mmf.rle$lengths >= merge.rows)

for (i in mmf.zero) {
# find which positions are zero, calculate sums and insert the result into a data.frame where the rows in question were turned to NA
    m.mmf <- mmf.rle$lengths[1:i] # select elements from 1 to where the zero appears
    select.rows <- (sum(m.mmf[1:length(m.mmf) - 1])+1):sum(m.mmf) # magic
    mmf.sum <- colSums(mmf[select.rows, ]) # sum values column-wise for rows that have at least three zeros in z
    mmf[select.rows,] <- NA # now that we have a sum by columns, we turn those numbers into NAs...
    mmf[select.rows[1], ] <- mmf.sum # ... and insert summed result into the first NA row       
}

# remove any left over NA rows
mmf <- mmf[complete.cases(mmf),]