How to use purrr to replace a modifying for loop in R?

Question

I have a dataframe like this:

df <- data.frame(
  Group = c('A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C'),
  Value = c(12, 88, 54, 76, 23, 44, 60, 52, 18)
)

I want to scale each group to a median of 100 and replace the Value column with the new value so the dataframe looks like this:

df_desired <- data.frame(
  Group = c('A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C'),
  Value = c(20, 169.23, 122.73, 126.67, 44.23, 100, 100, 100, 40.91)
)

Using a scale_helper like this:

scale_helper <- function(x, value) x * value / median(x)

I could do this with a for loop, but I want to use purrr instead, if possible. Is there a straightforward way to do it using purrr, or is a for loop the better way to go here?

Do I need to explain that? I’d like an alternative to a for loop. — Natasha R.
– Natasha R., Commented Nov 13, 2021 at 17:11
Are you looking for df %>% group_by(Group) %>% mutate(new_val = scale_helper(Value, 100))? I don't see any neccessarity to use purrr here... — Martin Gal
– Martin Gal, Commented Nov 13, 2021 at 17:14
I'd choose purrr, if you are handling multiple data.frames stored in a list. For handling a single data.frame, dplyr should be sufficient. — Martin Gal
– Martin Gal, Commented Nov 13, 2021 at 17:15
@NatashaR, you are asking whether a for loop is better, but it's not clear to me how to measure that without knowing why you don't want to use one. — user2554330
– user2554330, Commented Nov 13, 2021 at 17:19

Егор Шишунов · Accepted Answer · 2021-11-13 17:15:08Z

2

Loop for is not good way, but I don't understand why you want to use purr. I think, this is good version:

df %>% group_by(Group) %>% mutate(Value = scale_helper(Value, 100)) %>% as.data.frame()

Or you can use data.table. Something like this:

as.data.table(df)[, lapply(.SD, scale_helper, 100), keyby = Group]

answered Nov 13, 2021 at 17:15

Егор Шишунов

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