Cumulative sum that resets when 0 is encountered
Another late idea:
ff = function(x)
{
cs = cumsum(x)
cs - cummax((x == 0) * cs)
}
ff(c(0, 1, 3, 0, 0, 5, 2))
#[1] 0 1 4 0 0 5 7
And to compare:
library(data.table)
ffdt = function(x)
data.table(x)[, whatiwant := cumsum(x), by = rleid(x == 0L)]$whatiwant
x = as.numeric(x) ##because 'cumsum' causes integer overflow
identical(ff(x), ffdt(x))
#[1] TRUE
microbenchmark::microbenchmark(ff(x), ffdt(x), times = 25)
#Unit: milliseconds
# expr min lq median uq max neval
# ff(x) 315.8010 362.1089 372.1273 386.3892 405.5218 25
# ffdt(x) 374.6315 407.2754 417.6675 447.8305 534.8153 25
You could use the Reduce
function with a custom function that returns 0 when the new value encountered is 0 and otherwise adds the new value to the accumulated value:
Reduce(function(x, y) if (y == 0) 0 else x+y, c(1, 0, 1, 1), accumulate=TRUE)
# [1] 1 0 1 2
Another base would be just
with(df, ave(b, cumsum(b == 0), FUN = cumsum))
## [1] 1 0 1 2
This will just divide column b
to groups according to 0
appearances and compute the cumulative sum of b
per these groups
Another solution using the latest data.table
version (v 1.9.6+)
library(data.table) ## v 1.9.6+
setDT(df)[, whatiwant := cumsum(b), by = rleid(b == 0L)]
# campaign date b whatiwant
# 1: a jan 1 1
# 2: b feb 0 0
# 3: c march 1 1
# 4: d april 1 2
Some benchmarks per comments
set.seed(123)
x <- sample(0:1e3, 1e7, replace = TRUE)
system.time(res1 <- ave(x, cumsum(x == 0), FUN = cumsum))
# user system elapsed
# 1.54 0.24 1.81
system.time(res2 <- Reduce(function(x, y) if (y == 0) 0 else x+y, x, accumulate=TRUE))
# user system elapsed
# 33.94 0.39 34.85
library(data.table)
system.time(res3 <- data.table(x)[, whatiwant := cumsum(x), by = rleid(x == 0L)])
# user system elapsed
# 0.20 0.00 0.21
identical(res1, as.integer(res2))
## [1] TRUE
identical(res1, res3$whatiwant)
## [1] TRUE