R subsetting a data frame into multiple data frames based on multiple column values

Question

I am trying to subset a data frame, where I get multiple data frames based on multiple column values. Here is my example

>df
  v1   v2   v3   v4   v5
   A         


        

Answer 1

There's now also nest() from tidyr which is rather nice.

library(tidyr)
nestdf <- df %>% nest(v3:v5)
nestdf$data

> nestdf$data
[[1]]
# A tibble: 2 × 3
     v3    v4    v5
  <int> <int> <int>
1     1    10    12
2     1    10    12

[[2]]
# A tibble: 1 × 3
     v3    v4    v5
  <int> <int> <int>
1    10    12     8

[[3]]
# A tibble: 2 × 3
     v3    v4    v5
  <int> <int> <int>
1     2    12    15
2     2    14    16

Access individual tibbles with nestdf$data[1] and so on.

Answer 2

You are looking for split

split(df, with(df, interaction(v1,v2)), drop = TRUE)
$E.X
  v1 v2 v3 v4 v5
3  E  X  2 12 15
5  E  X  2 14 16

$D.Y
  v1 v2 v3 v4 v5
2  D  Y 10 12  8

$A.Z
  v1 v2 v3 v4 v5
1  A  Z  1 10 12

As noted in the comments

any of the following would work

library(microbenchmark)
microbenchmark(
                split(df, list(df$v1,df$v2), drop = TRUE), 
               split(df, interaction(df$v1,df$v2), drop = TRUE),
               split(df, with(df, interaction(v1,v2)), drop = TRUE))


Unit: microseconds
                                                  expr      min        lq    median       uq      max neval
            split(df, list(df$v1, df$v2), drop = TRUE) 1119.845 1129.3750 1145.8815 1182.119 3910.249   100
     split(df, interaction(df$v1, df$v2), drop = TRUE)  893.749  900.5720  909.8035  936.414 3617.038   100
 split(df, with(df, interaction(v1, v2)), drop = TRUE)  895.150  902.5705  909.8505  927.128 1399.284   100

It appears interaction is slightly faster (probably due the fact that the f = list(...) are just converted to an interaction within the function)

---

Edit

If you just want use the subset data.frames then I would suggest using data.table for ease of coding

library(data.table)

dt <- data.table(df)
dt[, plot(v4, v5), by = list(v1, v2)]