Rolling queue size - r

Question

I want to calculate number of items waiting or queued over. Let's say, I have fixed capacity of 102 item/hour and different incoming items for 9 hours.
as data table:
dt<-data.table(hour = c(1,2,3,4,5,6,7,8,9),
incoming = c(78,102,115,117,105,99,91,80,71),
capacity = rep(102,9))
I want to calculate queued items in each period.
In 1 and 2 capacity is enough and queue is 0.
In 3, 13 items are queued
In 4, 15+13 backlogged items are queued.
In 6, there were 31 backlogged items and 3 items are deducted so 28 were queued.
I have tried several options but could not figure out how to calculate.
Result should be:

Answer 1

Explicit looping in R won't get you far, and I don't see a vectorized solution for this, but this is trivial to solve using Rcpp:
library(Rcpp)
cppFunction("NumericVector queue(NumericVector x) {
NumericVector res(x.size());
res[0] = std::max<double>(0, x[0]);
for (int i = 1, size = x.size(); i < size; ++i) {
res[i] = std::max<double>(0, res[i-1] + x[i]);
}
return res;
}")
dt[, queued := queue(incoming - capacity)][]
# hour incoming capacity queued
#1: 1 78 102 0
#2: 2 102 102 0
#3: 3 115 102 13
#4: 4 117 102 28
#5: 5 105 102 31
#6: 6 99 102 28
#7: 7 91 102 17
#8: 8 80 102 0
#9: 9 71 102 0

Answer 2

I'd create a separate function to get queued number like #sebastian-c did, but with #R.S. 's logic. Like this
get_queue <- function(x){
n <- length(x)
y <- c(max(0, x[[1]]), rep(0, n - 1))
for(i in 2:n){
y[i] <- max(0, y[i - 1] + x[i])
}
y
}
And then
dt[,incoming_capacity := incoming - capacity]
dt[,queued := get_queue(incoming_capacity)]

Answer 3

Another alternative:
require(data.table)
dt<-data.table(hour = c(1,2,3,4,5,6,7,8,9),
incoming = c(78,102,115,117,105,99,91,80,71),
capacity = rep(102,9))
dt$incoming_capactity<- dt$incoming-dt$capacity
dt$carriedover<- 0
dt$carriedover[1]<- max(0,dt$incoming_capactity[1]) #added
for( i in 2:length(dt$carriedover)) {
dt$carriedover[i]<- max(0,dt$incoming_capactity[i] + dt$carriedover[i-1])
}
dt