问题
I have a data frame with a binary vector that I want to do a cumulative count of. However I would like to count the 'groups of 1's' rather than each individual 1 and create a new vector of this count while retaining the 0 separating values. i.e.
df1 <- data.frame(c(0,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,1,1,1)
n bin
1 0
2 1
3 1
4 1
5 1
6 0
7 0
8 0
9 1
10 1
11 1
12 1
13 1
14 0
15 0
16 0
17 1
18 1
19 1
becomes
n bin cumul
1 0 0
2 1 1
3 1 1
4 1 1
5 1 1
6 0 0
7 0 0
8 0 0
9 1 2
10 1 2
11 1 2
12 1 2
13 1 2
14 0 0
15 0 0
16 0 0
17 1 3
18 1 3
19 1 3
how do I go about this?
回答1:
You can use the rleid function from package data.table:
df1 <- data.frame(bin = c(0,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,1,1,1))
library(data.table)
setDT(df1)
df1[, cumul := rleid(bin)]
df1[bin == 0, cumul := 0]
df1[bin == 1, cumul := rleid(cumul)]
# bin cumul
# 1: 0 0
# 2: 1 1
# 3: 1 1
# 4: 1 1
# 5: 1 1
# 6: 0 0
# 7: 0 0
# 8: 0 0
# 9: 1 2
#10: 1 2
#11: 1 2
#12: 1 2
#13: 1 2
#14: 0 0
#15: 0 0
#16: 0 0
#17: 1 3
#18: 1 3
#19: 1 3
回答2:
Although somehow manual:
l <- rle(df1$c1)$lengths
v <- rle(df1$c1)$values
v2 <- cumsum(v)
v2[duplicated(v2)] <- 0
df1$cumul <- rep(v2, times = l)
df1
c1 cumul
1 0 0
2 1 1
3 1 1
4 1 1
5 1 1
6 0 0
7 0 0
8 0 0
9 1 2
10 1 2
11 1 2
12 1 2
13 1 2
14 0 0
15 0 0
16 0 0
17 1 3
18 1 3
19 1 3
回答3:
Yet another
x<-c(0,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,1,1,1)
d<-cumsum(diff(c(0,x))>0)
d[x==0]<-0
cbind(x,d)
x d
[1,] 0 0
[2,] 1 1
[3,] 1 1
[4,] 1 1
[5,] 1 1
[6,] 0 0
[7,] 0 0
[8,] 0 0
[9,] 1 2
[10,] 1 2
[11,] 1 2
[12,] 1 2
[13,] 1 2
[14,] 0 0
[15,] 0 0
[16,] 0 0
[17,] 1 3
[18,] 1 3
[19,] 1 3
来源：https://stackoverflow.com/questions/35227312/cumulative-count-of-blocks-of-1-with-0-separators-in-a-binary-vector-in-r