Split data.frame based on levels of a factor into new data.frames

Since dplyr 0.8.0 , we can also use group_split which has similar behavior as base::split

library(dplyr)
df %>% group_split(g)

#[[1]]
# A tibble: 5 x 3
#       x      y g    
#   <dbl>  <dbl> <fct>
#1 -1.21  -1.45  A    
#2  0.506  1.10  A    
#3 -0.477 -1.17  A    
#4 -0.110  1.45  A    
#5  0.134 -0.969 A    

#[[2]]
# A tibble: 5 x 3
#       x      y g    
#   <dbl>  <dbl> <fct>
#1  0.277  0.575 B    
#2 -0.575 -0.476 B    
#3 -0.998 -2.18  B    
#4 -0.511 -1.07  B    
#5 -0.491 -1.11  B  
#....

It also comes with argument .keep (which is TRUE by default) to specify whether or not the grouped column should be kept.

df %>% group_split(g, .keep = FALSE)

#[[1]]
# A tibble: 5 x 2
#       x      y
#   <dbl>  <dbl>
#1 -1.21  -1.45 
#2  0.506  1.10 
#3 -0.477 -1.17 
#4 -0.110  1.45 
#5  0.134 -0.969

#[[2]]
# A tibble: 5 x 2
#       x      y
#   <dbl>  <dbl>
#1  0.277  0.575
#2 -0.575 -0.476
#3 -0.998 -2.18 
#4 -0.511 -1.07 
#5 -0.491 -1.11 
#....

The difference between base::split and dplyr::group_split is that group_split does not name the elements of the list based on grouping. So

df1 <- df %>% group_split(g)
names(df1) #gives 
NULL

whereas

df2 <- split(df, df$g)
names(df2) #gives
#[1] "A" "B" "C" "D" "E"

data

set.seed(1234)
df <- data.frame(
      x=rnorm(25),
      y=rnorm(25),
      g=rep(factor(LETTERS[1:5]), 5)
)

I think that split does exactly what you want.

Notice that X is a list of data frames, as seen by str:

X <- split(df, df$g)
str(X)

If you want individual object with the group g names you could assign the elements of X from split to objects of those names, though this seems like extra work when you can just index the data frames from the list split creates.

#I used lapply just to drop the third column g which is no longer needed.
Y <- lapply(seq_along(X), function(x) as.data.frame(X[[x]])[, 1:2]) 

#Assign the dataframes in the list Y to individual objects
A <- Y[[1]]
B <- Y[[2]]
C <- Y[[3]]
D <- Y[[4]]
E <- Y[[5]]

#Or use lapply with assign to assign each piece to an object all at once
lapply(seq_along(Y), function(x) {
    assign(c("A", "B", "C", "D", "E")[x], Y[[x]], envir=.GlobalEnv)
    }
)

Edit Or even better than using lapply to assign to the global environment use list2env:

names(Y) <- c("A", "B", "C", "D", "E")
list2env(Y, envir = .GlobalEnv)
A