Count of a string of values between values - r

Question

I have a simple dataframe that is a set of ID columns and values of 0 or 1, for an example:
data.frame(replicate(10,sample(0:1,1000,rep=TRUE)))
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1 1 0 1 0 0 1 1 1 0
2 0 0 0 1 0 1 0 0 1 0
3 0 1 1 1 1 0 1 1 1 1
4 0 0 0 1 1 1 1 1 1 0
5 1 0 1 0 1 1 0 1 1 0
6 0 1 1 1 1 1 0 1 1 1
I want to write a code or loop that for every column, counts the number of 0's until encountering another 1, and continues down the column. So ideally the output is a new dataframe with the same ID column head, and a list of counts:
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 3 1 2 1 2 1 1 1 NA 2
2 1 2 1 1 NA 1 2 NA NA 2
I'm not sure how to do this and also the row outcome may be of different lengths. If each column has to create a new dataframe that's fine.

Answer 1

Here's a base R solution. I used a size-10 example instead of a size 1000 example so we can actually see what's going on and make sure it looks right.
set.seed(47)
d = data.frame(replicate(10,sample(0:1,10,rep=TRUE)))
d
# X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
# 1 0 0 0 0 0 0 1 1 0 0
# 2 0 1 0 1 0 0 0 0 0 0
# 3 1 1 1 0 1 0 0 0 1 0
# 4 0 0 0 0 0 1 1 1 1 1
# 5 1 1 0 1 0 0 1 1 1 0
# 6 0 1 1 1 1 1 1 1 0 1
# 7 1 1 0 0 1 0 0 1 1 0
# 8 0 0 1 0 1 0 1 0 0 0
# 9 0 0 0 1 1 1 0 0 1 1
# 10 1 1 1 0 1 0 1 1 0 0
results = lapply(d, function(x) with(rle(x), lengths[values == 0]))
max_length = max(lengths(results))
results = lapply(results, function(x) {length(x) = max_length; x})
results = do.call(cbind, results)
results
# X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
# [1,] 2 1 2 1 2 3 2 2 2 3
# [2,] 1 1 2 2 2 1 1 2 1 1
# [3,] 1 2 1 2 NA 2 1 NA 1 2
# [4,] 2 NA 1 1 NA 1 NA NA 1 1

Answer 2

One dplyr and purrr option could be:
map(.x = names(df),
~ df %>%
mutate(rleid = with(rle(get(.x)), rep(seq_along(lengths), lengths))) %>%
filter(get(.x) == 0) %>%
group_by(rleid = cumsum(!duplicated(rleid))) %>%
summarise(!!.x := n())) %>%
reduce(full_join, by = c("rleid" = "rleid"))
rleid X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
<int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
1 1 1 1 2 2 9 2 3 4 1 1
2 2 1 1 NA 3 NA 1 1 2 1 1
3 3 1 3 NA NA NA 2 1 NA 2 2
4 4 1 NA NA NA NA 1 NA NA 1 2
Sample data:
set.seed(123)
df <- data.frame(replicate(10, sample(0:1, 10, rep = TRUE)))
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 0 1 1 1 0 0 1 1 0 0
2 1 0 1 1 0 0 0 1 1 1
3 0 1 1 1 0 1 0 1 0 0
4 1 1 1 1 0 0 0 0 1 1
5 1 0 1 0 0 1 1 0 0 0
6 0 1 1 0 0 0 0 0 0 0
7 1 0 1 1 0 0 1 0 1 1
8 1 0 1 0 0 1 1 1 1 0
9 1 0 0 0 0 1 1 0 1 0
10 0 1 0 0 1 0 0 0 0 1

Answer 3

Here's an alternate approach that uses the indices of the 1 values to determine the runs of zero (using Gregor's data):
library(purrr)
map(df, ~ {
y <- diff(c(0, which(.x == 1), nrow(df) + 1)) - 1
y[y != 0]
}) %>%
map_df(`length<-`, max(lengths(.)))
# A tibble: 4 x 10
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 2 1 2 1 2 3 2 2 2 3
2 1 1 2 2 2 1 1 2 1 1
3 1 2 1 2 NA 2 1 NA 1 2
4 2 NA 1 1 NA 1 NA NA 1 1
Or same in base R:
res <- lapply(df, function(x) {
y <- diff(c(0, which(x == 1), nrow(df) + 1)) - 1
y[y != 0]})
data.frame(do.call(cbind, lapply(res, `length<-`, max(lengths(res)))))