I have tested participants at three points in time. I have the dates at which they were tested. I want to make a column which levels are first, second, and third. Each participant has three dates, so they are all different per participant. The data looks like this:
structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L), time_tested = c("2022-02-05", "2022-02-05", "2022-02-05",
"2022-02-08", "2022-02-08", "2022-02-08", "2022-02-11", "2022-02-11",
"2022-02-11", "2022-02-08", "2022-02-08", "2022-02-08", "2022-02-10",
"2022-02-10", "2022-02-10", "2022-02-13", "2022-02-13", "2022-02-13",
"2022-02-05", "2022-02-05", "2022-02-05", "2022-02-08", "2022-02-08",
"2022-02-08", "2022-02-11", "2022-02-11", "2022-02-11")), class = "data.frame", row.names = c(NA,
-27L))
and this is the result I want:
structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L), time_tested = c("2022-02-05", "2022-02-05", "2022-02-05",
"2022-02-08", "2022-02-08", "2022-02-08", "2022-02-11", "2022-02-11",
"2022-02-11", "2022-02-08", "2022-02-08", "2022-02-08", "2022-02-10",
"2022-02-10", "2022-02-10", "2022-02-13", "2022-02-13", "2022-02-13",
"2022-02-05", "2022-02-05", "2022-02-05", "2022-02-08", "2022-02-08",
"2022-02-08", "2022-02-11", "2022-02-11", "2022-02-11"), period = c("first",
"first", "first", "second", "second", "second", "third", "third",
"third", "first", "first", "first", "second", "second", "second",
"third", "third", "third", "first", "first", "first", "second",
"second", "second", "third", "third", "third")), class = "data.frame", row.names = c(NA,
-27L))
Thank you!
CodePudding user response:
Using data.table::rleid
to get the group ids, and ordinal
function from package english to convert it to ordinal.
Base R
df$period <- as.numeric(ave(df$time_tested, df$id, FUN = data.table::rleid))
df$english <- english::ordinal(df$period)
tidyverse
df %>%
group_by(id) %>%
mutate(period = data.table::rleid(time_tested),
english = english::ordinal(period))
output
id time_tested period english
1 1 2022-02-05 1 first
2 1 2022-02-05 1 first
3 1 2022-02-05 1 first
4 1 2022-02-08 2 second
5 1 2022-02-08 2 second
6 1 2022-02-08 2 second
7 1 2022-02-11 3 third
8 1 2022-02-11 3 third
9 1 2022-02-11 3 third
10 2 2022-02-08 1 first
11 2 2022-02-08 1 first
12 2 2022-02-08 1 first
13 2 2022-02-10 2 second
14 2 2022-02-10 2 second
15 2 2022-02-10 2 second
16 2 2022-02-13 3 third
17 2 2022-02-13 3 third
18 2 2022-02-13 3 third
19 3 2022-02-05 1 first
20 3 2022-02-05 1 first
21 3 2022-02-05 1 first
22 3 2022-02-08 2 second
23 3 2022-02-08 2 second
24 3 2022-02-08 2 second
25 3 2022-02-11 3 third
26 3 2022-02-11 3 third
27 3 2022-02-11 3 third