Specifying multiple column names inside mutate - r

Question

How do I specify column names inside mutate when multiple columns are generated?
In this example:
set.seed(5)
data.frame(x2 = sample(1:10, 10),
x3 = sample(1:10, 10),
x1 = sample(1:10, 10),
y3 = sample(1:10, 10),
y2 = sample(1:10, 10),
y1 = sample(1:10, 10)) |>
mutate(z1 = x1 - y1,
z2 = x2 - y2,
z3 = x3 - y3) |>
mutate(zz = across(num_range(prefix = 'x',
range = 1:3)) - across(num_range(prefix = 'y',
range = 1:3)))
Resulting in:
x2 x3 x1 y3 y2 y1 z1 z2 z3 zz.x1 zz.x2 zz.x3
1 2 3 9 10 9 6 3 -7 -7 3 -7 -7
2 9 10 6 6 4 5 1 5 4 1 5 4
3 7 6 4 8 8 3 1 -1 -2 1 -1 -2
4 3 2 3 4 10 8 -5 -7 -2 -5 -7 -2
5 1 5 2 5 7 7 -5 -6 0 -5 -6 0
6 6 4 5 3 6 2 3 0 1 3 0 1
7 5 8 10 2 1 4 6 4 6 6 4 6
8 10 7 8 7 3 1 7 7 0 7 7 0
9 4 1 1 9 2 9 -8 2 -8 -8 2 -8
10 8 9 7 1 5 10 -3 3 8 -3 3 8
I want zz.x1 be named zz1, ...

Answer 1

Here's a dplyr-way:
mutate takes its name(s) from the first across and we can change this using the .names-argument. It accepts glue-style input that we can adapt to your needs using str_replace().
library(dplyr)
library(stringr)
df |>
mutate(across(num_range(prefix = 'x', range = 1:3),
.names = "{str_replace(col, 'x', 'z')}")
- across(num_range(prefix = 'y',range = 1:3)))
Output:
x2 x3 x1 y3 y2 y1 z1 z2 z3
1 2 3 9 10 9 6 3 -7 -7
2 9 10 6 6 4 5 1 5 4
3 7 6 4 8 8 3 1 -1 -2
4 3 2 3 4 10 8 -5 -7 -2
5 1 5 2 5 7 7 -5 -6 0
6 6 4 5 3 6 2 3 0 1
7 5 8 10 2 1 4 6 4 6
8 10 7 8 7 3 1 7 7 0
9 4 1 1 9 2 9 -8 2 -8
10 8 9 7 1 5 10 -3 3 8
Data:
set.seed(5)
df <- data.frame(x2 = sample(1:10, 10),
x3 = sample(1:10, 10),
x1 = sample(1:10, 10),
y3 = sample(1:10, 10),
y2 = sample(1:10, 10),
y1 = sample(1:10, 10))
Update: Or similar to OP's desired output
df2 |>
mutate(across(num_range(prefix = 'x', range = 1:3),
.names = "{str_replace(col, 'x', 'zz')}")
- across(num_range(prefix = 'y',range = 1:3)))
Output:
x2 x3 x1 y3 y2 y1 z1 z2 z3 zz1 zz2 zz3
1 2 3 9 10 9 6 3 -7 -7 3 -7 -7
2 9 10 6 6 4 5 1 5 4 1 5 4
3 7 6 4 8 8 3 1 -1 -2 1 -1 -2
4 3 2 3 4 10 8 -5 -7 -2 -5 -7 -2
5 1 5 2 5 7 7 -5 -6 0 -5 -6 0
6 6 4 5 3 6 2 3 0 1 3 0 1
7 5 8 10 2 1 4 6 4 6 6 4 6
8 10 7 8 7 3 1 7 7 0 7 7 0
9 4 1 1 9 2 9 -8 2 -8 -8 2 -8
10 8 9 7 1 5 10 -3 3 8 -3 3 8
Data
set.seed(5)
df2 <- data.frame(x2 = sample(1:10, 10),
x3 = sample(1:10, 10),
x1 = sample(1:10, 10),
y3 = sample(1:10, 10),
y2 = sample(1:10, 10),
y1 = sample(1:10, 10)) |>
mutate(z1 = x1 - y1,
z2 = x2 - y2,
z3 = x3 - y3)

Answer 2

I don't know how to do this with dplyr but in base R it is pretty straightforward. This might partly answer also your previous question.
# hard-coded variable suffixes
suff <- 1:3
# OR suffixes extracted from data
suff <- sort(unique(sub('[a-z]*', '', names(df))))
for (i in suff) {
df[[paste0('zz', i)]] <- df[[paste0('x', i)]] - df[[paste0('y', i)]]
}
df
# x2 x3 x1 y3 y2 y1 zz1 zz2 zz3
# 1 2 3 9 10 9 6 3 -7 -7
# 2 9 10 6 6 4 5 1 5 4
# 3 7 6 4 8 8 3 1 -1 -2
# 4 3 2 3 4 10 8 -5 -7 -2
# 5 1 5 2 5 7 7 -5 -6 0
# 6 6 4 5 3 6 2 3 0 1
# 7 5 8 10 2 1 4 6 4 6
# 8 10 7 8 7 3 1 7 7 0
# 9 4 1 1 9 2 9 -8 2 -8
# 10 8 9 7 1 5 10 -3 3 8
A more efficient way which avoids the loop over suffixes would be like this:
zz <- df[paste0('x', suff)] - df[paste0('y', suff)]
names(zz) <- paste0('zz', suff)
df <- cbind(df, zz)
Data:
set.seed(5)
df <- data.frame(x2 = sample(1:10, 10),
x3 = sample(1:10, 10),
x1 = sample(1:10, 10),
y3 = sample(1:10, 10),
y2 = sample(1:10, 10),
y1 = sample(1:10, 10))