grouping a character vector into new groups using dplyr - r

Question

I have a data frame that looks like this:
# A tibble: 5 x 5
# Groups: Trial [1]
GID Trial pop `1A-1145442` `1A-1158042`
<chr> <chr> <chr> <int> <int>
GID421213 ES1 ES1-5 12 11
GID419903 ES1 ES1-5 22 12
GID3881 ES1 ES1-5 22 22
GID13646 ES1 ES1-5 12 12
GID418846 ES1 ES1-5 22 11
Here is a dput of it :
structure(list(GID = c("GID421213", "GID419903", "GID3881", "GID13646",
"GID418846"), Trial = c("ES1", "ES1", "ES1", "ES1", "ES1"), pop = c("ES1-5",
"ES1-5", "ES1-5", "ES1-5", "ES1-5"), `1A-1145442` = c(12L, 22L,
22L, 12L, 22L), `1A-1158042` = c(11L, 12L, 22L, 12L, 11L)), row.names =
c(NA, -5L), class = c("grouped_df", "tbl_df", "tbl", "data.frame"), vars =
"Trial", drop = TRUE, indices = list(0:4), group_sizes = 5L,
biggest_group_size = 5L, labels = structure(list(Trial = "ES1"), row.names
= c(NA, -1L), class = "data.frame", vars = "Trial", drop = TRUE))
I want to perform a regrouping transformation into a new column from the Trial column just as I did in the past with the pop column using regex operations but now with dplyr. The Trial column consists of ES values from 1 to 38: I would like to group in this fashion ES1-3,ES3-6,ES7-9 and so forth using the dplyr package. I know I could start with df >%> group_by(df,Trial) but from there on I have no idea how I could operate.

Answer 1

library(dplyr)
df %>%
mutate(pop2 = case_when(
Trial == "ES1" | Trial == "ES2" | Trial == "ES3" ~ "ES1-3",
Trial == "ES4" | Trial == "ES5" | Trial == "ES6" ~ "ES4-6"
))
Will return
# A tibble: 5 x 6
# Groups: Trial [1]
GID Trial pop `1A-1145442` `1A-1158042` pop2
<chr> <chr> <chr> <int> <int> <chr>
1 GID421213 ES1 ES1-5 12 11 ES1-3
2 GID419903 ES1 ES1-5 22 12 ES1-3
3 GID3881 ES1 ES1-5 22 22 ES1-3
4 GID13646 ES1 ES1-5 12 12 ES1-3
5 GID418846 ES1 ES1-5 22 11 ES1-3

Answer 2

Given
(df <- data.frame(Trial = paste0("ES", 1:10)))
# Trial
# 1 ES1
# 2 ES2
# 3 ES3
# 4 ES4
# 5 ES5
# 6 ES6
# 7 ES7
# 8 ES8
# 9 ES9
# 10 ES10
We may, using base R, do
size <- 3
groups <- (as.numeric(substring(df$Trial, 3)) - 1) %/% size
(df$newCol <- sprintf("ES%d-%d", 1 + groups * size, size * (1 + groups)))
# [1] "ES1-3" "ES1-3" "ES1-3" "ES4-6" "ES4-6" "ES4-6" "ES7-9" "ES7-9"
# [9] "ES7-9" "ES10-12"
Here as.numeric(substring(df$Trial, 3)) gets the numeric part of df$Trial and converts it to a numeric vector. Subtracting 1 and using %/% then returns the group number for each element of df$Trial, starting from 0. Given a group number, we can easily construct a new column with sprintf.
size is the size of groups. E.g., setting size <- 5 would give values ES1-5, ES6-10, and so on.

Answer 3

Here's a solution that uses parse_number from readr.
df %>%
mutate(grp = cut(parse_number(Trial),
breaks = seq(1, 38, by = 3),
right = FALSE)) %>%
group_by(grp)
This pulls out the number from Trial then cuts to create a grouping variable, which it then groups by. right=FALSE indicates that the interval is closed on the left.
An edit based on a comment below.
df %>%
mutate(grp = cut(parse_number(Trial),
breaks = c(seq(1, 34, by = 3) 38),
right = FALSE),
include.lowest = TRUE) %>%
group_by(grp)