Why is split inefficient on large data frames with many groups?
More an explanation than an answer. Sub-setting a large data.frame is more costly than sub-setting a small data frame
> df100 = df[1:100,]
> idx = c(1, 10, 20)
> microbenchmark(df[idx,], df100[idx,], times=10)
Unit: microseconds
expr min lq mean median uq max neval
df[idx, ] 428.921 441.217 445.3281 442.893 448.022 475.364 10
df100[idx, ] 32.082 32.307 35.2815 34.935 37.107 42.199 10
split()
pays this cost for each group.
The reason can be seen by running Rprof()
> Rprof(); for (i in 1:1000) df[idx,]; Rprof(NULL); summaryRprof()
$by.self
self.time self.pct total.time total.pct
"attr" 1.26 100 1.26 100
$by.total
total.time total.pct self.time self.pct
"attr" 1.26 100 1.26 100
"[.data.frame" 1.26 100 0.00 0
"[" 1.26 100 0.00 0
$sample.interval
[1] 0.02
$sampling.time
[1] 1.26
All of the time is being spent in a call to attr()
. Stepping through the code using debug("[.data.frame")
shows that the pain involves a call like
attr(df, "row.names")
This small example shows a trick that R uses to avoid representing row names that are not present: use c(NA, -5L)
, rather than 1:5
.
> dput(data.frame(x=1:5))
structure(list(x = 1:5), .Names = "x", row.names = c(NA, -5L), class = "data.frame")
Note that attr()
returns a vector -- the row.names are created on the fly, and for a large data.frame a large number of row.names are created
> attr(data.frame(x=1:5), "row.names")
[1] 1 2 3 4 5
So one might expect that even nonsensical row.names would speed the calculation
> dfns = df; rownames(dfns) = rev(seq_len(nrow(dfns)))
> system.time(split(dfns, dfns$x))
user system elapsed
4.048 0.000 4.048
> system.time(split(df, df$x))
user system elapsed
87.772 16.312 104.100
Splitting a vector or matrix would also be fast.
This isn't strictly split.data.frame
issue, there is a more general problem on scalability of data.frame for many groups.
You can get pretty nice speed up if you use split.data.table
. I developed this method on top of regular data.table methods and it seems to scale pretty well here.
system.time(
l1 <- df %>% split(.$x)
)
# user system elapsed
#200.936 0.000 217.496
library(data.table)
dt = as.data.table(df)
system.time(
l2 <- split(dt, by="x")
)
# user system elapsed
# 7.372 0.000 6.875
system.time(
l3 <- split(dt, by="x", sorted=TRUE)
)
# user system elapsed
# 9.068 0.000 8.200
sorted=TRUE
will return the list of the same order as data.frame method, by default data.table method will preserve order present in input data. If you want to stick to data.frame you can at the end use lapply(l2, setDF)
.
PS. split.data.table
was added in 1.9.7, installation of devel version is pretty simple
install.packages("data.table", type="source", repos="http://Rdatatable.github.io/data.table")
More about that in Installation wiki.