I would like to produce a column in a data.frame that counts the consecutive id of the groups (s column in dummy df)
dummy_df = data.frame(s = c("a", "a", "b","b", "b", "c","c", "a", "a", "c", "c","a","a"),
desired_output= c(1,1,1,1,1,1,1,2,2,2,2,3,3))
dummy_df$rleid_output = rleid(dummy_df$s)
dummy_df
s desired_output rleid_output
1 a 1 1
2 a 1 1
3 b 1 2
4 b 1 2
5 b 1 2
6 c 1 3
7 c 1 3
8 a 2 4
9 a 2 4
10 c 2 5
11 c 2 5
12 a 3 6
13 a 3 6
I would say it's similar to what rleid()
does but restarting the counting when a new group is seen. However, I can't find a way to do it in such straight way. Thanks.
You can do:
dummy_df$out <- with(rle(dummy_df$s), rep(ave(lengths, values, FUN = seq_along), lengths))
Result:
s desired_output out
1 a 1 1
2 a 1 1
3 b 1 1
4 b 1 1
5 b 1 1
6 c 1 1
7 c 1 1
8 a 2 2
9 a 2 2
10 c 2 2
11 c 2 2
12 a 3 3
13 a 3 3
If you are willing to use data.table
( rleid
is part of the package), you can do it in two steps as follows:
library(data.table)
dummy_df = data.frame(s = c("a", "a", "b", "b", "b", "c", "c", "a", "a", "c", "c", "a", "a"))
# cast data.frame to data.table
setDT(dummy_df)
# create auxiliary variable
dummy_df[, rleid_output := rleid(s)]
# obtain desired output
dummy_df[, desired_output := rleid(rleid_output), by = "s"]
# end result
dummy_df
#> s rleid_output desired_output
#> 1: a 1 1
#> 2: a 1 1
#> 3: b 2 1
#> 4: b 2 1
#> 5: b 2 1
#> 6: c 3 1
#> 7: c 3 1
#> 8: a 4 2
#> 9: a 4 2
#> 10: c 5 2
#> 11: c 5 2
#> 12: a 6 3
#> 13: a 6 3
Created on 2020-10-16 by the reprex package (v0.3.0)
you can try a tidyverse
in combination with the base R rle
function
library(tidyverse)
rle(dummy_df$s) %>%
with(., data.frame(a=.$length, b=.$value)) %>%
group_by(b) %>%
mutate(n = 1:n()) %>%
with(., rep(n, times=a)) %>%
bind_cols(dummy_df, res=.)
s desired_output res
1 a 1 1
2 a 1 1
3 b 1 1
4 b 1 1
5 b 1 1
6 c 1 1
7 c 1 1
8 a 2 2
9 a 2 2
10 c 2 2
11 c 2 2
12 a 3 3
13 a 3 3
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