Create a list for each pattern of grep

General
1

Hellow i have the following df and pattern

df <- data.frame(class=c(5,8,10,15),student=c("US_NORTH","US_SOUTH","UK_EAST","BRAZIL"))
pattern <- c("US","UK")

I would identify US and UK students without using a loop.
The US student should be together in a list and the UK students in sublist as shown below:

target <- list(US=c(5,8),UK=10)

I know a loop could be used but i would like to have a vector solution.
Thanks

2

Here is one solution.

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(purrr)
df <- data.frame(class=c(5,8,10,15),student=c("US_NORTH","US_SOUTH","UK_EAST","BRAZIL"))
pattern <- c("US","UK")
MyFunc <- function(KEY){
  tmp <- filter(df,grepl(KEY,x = student))
  tmp$class
}
target <- map(pattern,MyFunc)
names(target) <- pattern
target
#> $US
#> [1] 5 8
#> 
#> $UK
#> [1] 10

Created on 2021-11-08 by the reprex package (v2.0.1)

3

Thank a lot.
It works.
But do you have a native R solution?
My program should work on any laptop, even for those who have problems with package installation.

4

You can rewrite my solution by replacing the filter function with subsetting of df using square brackets, e.g. df[grepl(),]. You can replace the map function with lapply.

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