Cumulative sum for positive numbers only

Question

I have this vector :

x = c(1,1,1,1,1,0,1,0,0,0,1,1)

And I want to do a cumulative sum for the positive numbers only. I should have the fol

Answer 1

split and lapply version:

x <- c(1,1,1,1,1,0,1,0,0,0,1,1)
unlist(lapply(split(x, cumsum(x==0)), cumsum))

step by step:

a <- split(x, cumsum(x==0)) # divides x into pieces where each 0 starts a new piece
b <- lapply(a, cumsum)  # calculates cumsum in each piece
unlist(b)  # rejoins the pieces 

Result has useless names but is otherwise what you wanted:

# 01 02 03 04 05 11 12  2  3 41 42 43 
#  1  2  3  4  5  0  1  0  0  0  1  2 

Answer 2

Here is another base R solution using aggregate. The idea is to make a data frame with x and a new column named x.1 by which we can apply aggregate functions (cumsum in this case):

x <- c(1,1,1,1,1,0,1,0,0,0,1,1)
r <- rle(x)
df <- data.frame(x, 
x.1=unlist(sapply(1:length(r$lengths), function(i) rep(i, r$lengths[i]))))

# df

   # x x.1
# 1  1   1
# 2  1   1
# 3  1   1
# 4  1   1
# 5  1   1
# 6  0   2
# 7  1   3
# 8  0   4
# 9  0   4
# 10 0   4
# 11 1   5
# 12 1   5

agg <- aggregate(df$x~df$x.1, df, cumsum)
as.vector(unlist(agg$`df$x`))

# [1] 1 2 3 4 5 0 1 0 0 0 1 2

Answer 3

One option is

x1 <- inverse.rle(within.list(rle(x), values[!!values] <- 
                  (cumsum(values))[!!values]))
x[x1!=0] <- ave(x[x1!=0], x1[x1!=0], FUN=seq_along)
x
#[1] 1 2 3 4 5 0 1 0 0 0 1 2

Or a one-line code would be

 x[x>0] <-  with(rle(x), sequence(lengths[!!values]))
 x
 #[1] 1 2 3 4 5 0 1 0 0 0 1 2

Answer 4

Base R, one line solution with Map Reduce :

> Reduce('c', Map(function(u,v) if(v==0) rep(0,u) else 1:u, rle(x)$lengths, rle(x)$values))
 [1] 1 2 3 4 5 0 1 0 0 0 1 2

Or:

unlist(Map(function(u,v) if(v==0) rep(0,u) else 1:u, rle(x)$lengths, rle(x)$values))

Answer 5

Try this one-liner...

Reduce(function(x,y) (x+y)*(y!=0), x, accumulate=T)

Answer 6

Here's a possible solution using data.table v >= 1.9.5 and its new rleid funciton

library(data.table)
as.data.table(x)[, cumsum(x), rleid(x)]$V1
## [1] 1 2 3 4 5 0 1 0 0 0 1 2