Hello,
Is this what you had in mind? (I just haven't unpacked the list object yet but you can easily save each seperately with the right name)
library(tidyverse)
df <- data.frame(
stringsAsFactors = FALSE,
ID = c(82950L,32045L,84817L,2857L,
55587L,73103L,5837L,7140L,70450L,34207L,20095L,
30200L,47211L,8932L,82663L),
Department = c("Training","Training","Sales",
"Logistics","Training","Sales","Sales","Sales",
"Logistics","Manufacturing","Manufacturing","Sales",
"Manufacturing","Manufacturing","Manufacturing"),
Strength = c(96L,86L,77L,88L,46L,51L,
62L,81L,56L,85L,58L,91L,36L,75L,39L),
DoD = c(953L,5277L,1633L,5629L,
6710L,1892L,2394L,2854L,5449L,1332L,4589L,1094L,5862L,
4847L,5798L),
Group_ID = c("A","A","C","Z","L","M",
"A","C","A","S","X","W","C","A","A")
)
filter_values <- df %>% select(Department) %>% distinct() %>% unlist()
df_list <- list()
for (i in 1:length(filter_values)) {
df_list[[i]] <- df %>% filter(Department == filter_values[i])
}
df_list[[1]]
#> ID Department Strength DoD Group_ID
#> 1 82950 Training 96 953 A
#> 2 32045 Training 86 5277 A
#> 3 55587 Training 46 6710 L
df_list[[2]]
#> ID Department Strength DoD Group_ID
#> 1 84817 Sales 77 1633 C
#> 2 73103 Sales 51 1892 M
#> 3 5837 Sales 62 2394 A
#> 4 7140 Sales 81 2854 C
#> 5 30200 Sales 91 1094 W
df_list[[3]]
#> ID Department Strength DoD Group_ID
#> 1 2857 Logistics 88 5629 Z
#> 2 70450 Logistics 56 5449 A
df_list[[4]]
#> ID Department Strength DoD Group_ID
#> 1 34207 Manufacturing 85 1332 S
#> 2 20095 Manufacturing 58 4589 X
#> 3 47211 Manufacturing 36 5862 C
#> 4 8932 Manufacturing 75 4847 A
#> 5 82663 Manufacturing 39 5798 A
Created on 2020-10-11 by the reprex package (v0.3.0)