Cumulative sum until maximum reached, then repeat from zero in the next row

Question

I feel like this is a fairly easy question, but for the life of me I can\'t seem to find the answer. I have a fairly standard dataframe, and what I am trying to do is sum th

Answer 1

Assuming your data.frame is df:

df$difference_sum <- c(0, head(cumsum(df$difference), -1))
# get length of 0's (first keep value gives the actual length)
len <- sum(df$difference_sum %/% 1470 == 0)
df$keep <- (seq_len(nrow(df))-1) %/% len
df <- transform(df, difference_sum = ave(difference, keep, 
          FUN=function(x) c(0, head(cumsum(x), -1))))

#       minutes difference keep difference_sum
# 1  1052991158        180    0              0
# 2  1052991338        180    0            180
# 3  1052991518        180    0            360
# 4  1052991698        180    0            540
# 5  1052991878        180    0            720
# 6  1052992058        180    0            900
# 7  1052992238        180    0           1080
# 8  1052992418        180    0           1260
# 9  1052992598        180    0           1440
# 10 1052992778        180    1              0
# 11 1052992958        180    1            180

Answer 2

I think this is best done with a for loop, can't think of a function that could do so out of the box. The following should do what you want (if I understand you correctly).

current.sum <- 0
for (c in 1:nrow(caribou.sub)) {
    current.sum <- current.sum + caribou.sub[c, "difference"]
    carribou.sub[c, "difference_sum"] <- current.sum
    if (current.sum >= 1470) {
        caribou.sub[c, "keep"] <- 1
        current.sum <- 0
    }
}

Feel free to comment if it does not exactly what you want. But as pointed out by alexwhan, your description is not completely clear.

Answer 3

I still don't understand about when the sum should restart and if it should be zero then. A desired result would help greatly.

Nonetheless, I can't help but think that simply indexing and subtraction would be a straightforward way of doing this. The below code gives the same result as @Henrik's solution.

df$difference_sum <- cumsum(df$difference)
step <- (df$difference_sum %/% 1470) + 1
k <- which(diff(step) > 0) + 1
df$keep <- 0
df$keep[k] <- 1
step[k] <- step[k] - 1
df$difference_sum <- df$difference_sum - c(0, df$difference_sum[k])[step]