Create groups to identify repeating numbers-CodePudding

So I have a vector of 1's and 0's and I want to identify different "groups" of repeating 1's.

What I want:

v1 group
1  a
1  a
1  a
0  NA
0  NA
1  b
1  b
0  NA
1  c

What's the best way to do this, preferably in base R?

CodePudding user response：

Using diff and cumsum.

letters[cumsum((c(0, diff(x)) > 0))   1] |>
  replace(x == 0, NA)
# [1] "a" "a" "a" NA  NA  "b" "b" NA  "c"

Data:

    x <- c(1, 1, 1, 0, 0, 1, 1, 0, 1)

CodePudding user response：

There is a convenient function in data.table for this, i.e.

replace(data.table::rleid(df$v1), df$v1 == 0, NA)
#[1]  1  1  1 NA NA  3  3 NA  5

If you want the letters, suggestion from @sindri_baldur works UNTIL THE ALPHABET RUNS OUT, i.e.

letters[data.table::frank(replace(data.table::rleid(df$v1), df$v1 == 0, NA_real_), ties.method = 'dense', na.last = 'keep')]
#[1] "a" "a" "a" NA  NA  "b" "b" NA  "c"

CodePudding user response：

Using rle:

v1 <- c(1,1,1,0,0,1,1,0,1,0,1,1,0,1,1,1,1,1)
d <- data.frame(v1)

d[d$v1 == 1, "group"] <- letters[with(rle(d$v1), rep(cumsum(values[values == 1]), lengths[values == 1]))]

   v1 group
1   1     a
2   1     a
3   1     a
4   0  <NA>
5   0  <NA>
6   1     b
7   1     b
8   0  <NA>
9   1     c
10  0  <NA>
11  1     d
12  1     d
13  0  <NA>
14  1     e
15  1     e
16  1     e
17  1     e
18  1     e

CodePudding user response：

We can try

with(
    rle(v1),
    letters[replace(
        rep(values * cumsum(values), lengths),
        v1 == 0,
        NA
    )]
)

CodePudding user response：

One way might be to use rle.

. <- rle(v1)
i <- .$values == 1
.$values[i] <- letters[factor(.$lengths[i])]
.$values[!i] <- NA
data.frame(v1, group=inverse.rle(.))
#   v1 group
#1   1     c
#2   1     c
#3   1     c
#4   0  <NA>
#5   0  <NA>
#6   1     b
#7   1     b
#8   0  <NA>
#9   1     a
#10  0  <NA>
#11  1     b
#12  1     b
#13  0  <NA>
#14  1     d
#15  1     d
#16  1     d
#17  1     d
#18  1     d

Data:

v1 <- c(1,1,1,0,0,1,1,0,1,0,1,1,0,1,1,1,1,1)