does the by( ) function make growing list

问题

Does the by function make a list that grows one element at a time?

I need to process a data frame with about 4M observations grouped by a factor column. The situation is similar to the example below:

> # Make 4M rows of data
> x = data.frame(col1=1:4000000, col2=10000001:14000000)
> # Make a factor
> x[,"f"] = x[,"col1"] - x[,"col1"] %% 5
>   
> head(x)
  col1     col2 f
1    1 10000001 0
2    2 10000002 0
3    3 10000003 0
4    4 10000004 0
5    5 10000005 5
6    6 10000006 5

Now, a tapply on one of the columns takes a reasonable amount of time:

> t1 = Sys.time()
> z = tapply(x[, 1], x[, "f"], mean)
> Sys.time() - t1
Time difference of 22.14491 secs

But if I do this:

z = by(x[, 1], x[, "f"], mean)

That doesn't finish anywhere near the same time (I gave up after a minute).

Of course, in the above example, tapply could be used, but I actually need to process multiple columns together. What is the better way to do this?

回答1:

by is slower than tapply because it is wrapping by. Let's take a look at some benchmarks: tapply in this situation is more than 3x faster than using by

UPDATED to include @Roland's great recomendation:

library(rbenchmark)
library(data.table)
dt <- data.table(x,key="f")

using.tapply <- quote(tapply(x[, 1], x[, "f"], mean))
using.by <- quote(by(x[, 1], x[, "f"], mean))
using.dtable <- quote(dt[,mean(col1),by=key(dt)])

times <- benchmark(using.tapply, using.dtable, using.by, replications=10, order="relative")
times[,c("test", "elapsed", "relative")] 

#------------------------#
#         RESULTS        # 
#------------------------#

#       COMPARING tapply VS by     #
#-----------------------------------
#              test elapsed relative
#   1  using.tapply   2.453    1.000
#   2      using.by   8.889    3.624

#   COMPARING data.table VS tapply VS by   #
#------------------------------------------#
#             test elapsed relative
#   2  using.dtable   0.168    1.000
#   1  using.tapply   2.396   14.262
#   3      using.by   8.566   50.988

If x$f is a factor, the loss in efficiency between tapply and by is even greater!

Although, notice that they both improve relative to non-factor inputs, while data.table remains approx the same or worse

x[, "f"] <- as.factor(x[, "f"])
dt <- data.table(x,key="f")
times <- benchmark(using.tapply, using.dtable, using.by, replications=10, order="relative")
times[,c("test", "elapsed", "relative")] 

#               test elapsed relative
#   2   using.dtable   0.175    1.000
#   1   using.tapply   1.803   10.303
#   3       using.by   7.854   44.880

As for the why, the short answer is in the documentation itself.

?by :

Description

Function by is an object-oriented wrapper for tapply applied to data frames.

let's take a look at the source for by (or more specificaly, by.data.frame):

by.data.frame
function (data, INDICES, FUN, ..., simplify = TRUE) 
{
    if (!is.list(INDICES)) {
        IND <- vector("list", 1L)
        IND[[1L]] <- INDICES
        names(IND) <- deparse(substitute(INDICES))[1L]
    }
    else IND <- INDICES
    FUNx <- function(x) FUN(data[x, , drop = FALSE], ...)
    nd <- nrow(data)
    ans <- eval(substitute(tapply(seq_len(nd), IND, FUNx, simplify = simplify)), 
        data)
    attr(ans, "call") <- match.call()
    class(ans) <- "by"
    ans
}

We see immediately that there is still a call to tapply plus a lot of extras (including calls to deparse(substitute(.)) and an eval(substitute(.)) both of which are relatively slow). Therefore it makes sense that your tapply will be relatively faster than a similar call to by.

回答2:

Regarding a better way to do this: With 4M rows you should use data.table.

library(data.table)
dt <- data.table(x,key="f")
dt[,mean(col1),by=key(dt)]

dt[,list(mean1=mean(col1),mean2=mean(col2)),by=key(dt)]
dt[,lapply(.SD,mean),by=key(dt)]

来源：https://stackoverflow.com/questions/13705533/does-the-by-function-make-growing-list

标签

benchmarking

tapply