grouped operations that result in length not equal to 1 or length of group in dplyr

16
votes

I'm not sure which function to use to do the following:

library(data.table)
dt = data.table(a = 1:4, b = 1:2)

dt[, rep(a[1], 3), by = b]
#   b V1
#1: 1  1
#2: 1  1
#3: 1  1
#4: 2  2
#5: 2  2
#6: 2  2

Both summarise and mutate are unhappy with this length:

library(dplyr)
df = data.frame(a = 1:4, b = 1:2)

df %.% group_by(b) %.% summarise(rep(a[1], 3))
#Error: expecting a single value

df %.% group_by(b) %.% mutate(rep(a[1], 3))
#Error: incompatible size (3), expecting 2 (the group size) or 1
2
rdata.tabledplyr
Don't know if it helps but using your dplyr code with a data.table works and with plyr you can do that too with a data.frame. - dickoa
@dickoa thanks, that's interesting (fwiw this is mostly just an exercise for me to understand how to use dplyr - I don't really see the point of using it with a data.table); sounds like a bug in summarise then - eddi
See github.com/hadley/dplyr/issues/154 - hadley
+1 This is an interesting difference; hopefully the final solution allows arbitrary return lengths for any groups. - BrodieG
In this case df %>% group_by(b) %>% slice(rep(1, 3)) works fine. For rowwise operations, where each row returns an arbitrary number of values, you can use the df %>% mutate(new = map(old, f)) %>% unnest() idiom. - Axeman

2 Answers

13
votes

In dplyr version 0.2 you could do this using the do operator:

> df %>% group_by(b) %>% do(data.frame(a = rep(.$a[1], 3)))
#Source: local data frame [6 x 2]
#Groups: b
#
#  b a
#1 1 1
#2 1 1
#3 1 1
#4 2 2
#5 2 2
#6 2 2
7
votes

While @beginneR's answer does work, it doesn't seem to be a real substitute to the data.table behavior. Consider:

df <- data.frame(a = 1, b = rep(1:1e4, 2))
dt <- data.table(df)
microbenchmark(times=5,
  dt[, rep(a[1], 3), by = b],
  df %>% group_by(b) %>% do(data.frame(a = rep(.$a[1], 3)))
)

has the dplyr implementation >200x slower.

Unit: milliseconds
                                                      expr        min         lq     median         uq
                                dt[, rep(a[1], 3), by = b]   13.14318   13.70248   14.60524   15.26676
 df %>% group_by(b) %>% do(data.frame(a = rep(.$a[1], 3))) 3269.40731 3359.11614 3583.19430 3736.67162

Maybe there is a better way to do this with do that doesn't require calling data.frame each do? Also, the syntax is a bit involved for what is something very simple in data.table.

Otherwise, as per Hadley's issue link, it seems this is expected to be implemented in dplyr in 3.1, which looks to be the next release.