问题
The title says it.
my vector
TF <- c(F,T,T,T,F,F,T,T,F,T,F,T,T,T,T,T,T,T,F)
my desired output
[1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
回答1:
#with(rle(TF), sequence(lengths) * rep(values, lengths))
with(rle(TF), sequence(lengths) * TF) #Like Rich suggested in comments
# [1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
回答2:
You could use rle() along with sequence().
replace(TF, TF, sequence(with(rle(TF), lengths[values])))
# [1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
replace() does the coercion for us.
回答3:
We can use rleid
library(data.table)
ave(TF, rleid(TF), FUN = seq_along)*TF
#[1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
回答4:
By using base R
ave(cumsum(TF==FALSE), cumsum(TF==FALSE), FUN=seq_along)-1
[1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
回答5:
You can also use inverse.seqle from cgwtools:
library(cgwtools)
SEQ = inverse.seqle(rle(TF))
SEQ[!TF] = 0
Or similar to @d.b's answer:
inverse.seqle(rle(TF))*TF
# [1] 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6 7 0
来源：https://stackoverflow.com/questions/46477148/sequentially-index-between-a-boolean-vector-in-r