From huber at ebi.ac.uk  Fri Mar  2 21:01:17 2007
From: huber at ebi.ac.uk (Wolfgang Huber)
Date: Fri, 02 Mar 2007 20:01:17 +0000
Subject: [Rd] extracting rows from a data frame by looping over the row
 names: performance issues
In-Reply-To: <45E87E4B.6070803@gmail.com>
References: <45E86F69.80507@fhcrc.org> <45E87E4B.6070803@gmail.com>
Message-ID: <45E8828D.1060200@ebi.ac.uk>


Hi Herv?

depending on your problem, using "mapply" might help, as in the code 
example below:

a = data.frame(matrix(1:3e4, ncol=3))

print(system.time({
r1 = numeric(nrow(a))
for(i in seq_len(nrow(a))) {
   g = a[i,]
   r1[i] = mean(c(g$X1, g$X2, g$X3))
}}))

print(system.time({
f = function(X1,X2,X3) mean(c(X1, X2, X3))
r2 = do.call("mapply", args=append(f, a))
}))

print(identical(r1, r2))

#   user  system elapsed
   6.049   0.200   6.987
    user  system elapsed
   0.508   0.000   0.509
[1] TRUE

  Best wishes
   Wolfgang

Roger D. Peng wrote:
> Extracting rows from data frames is tricky, since each of the columns could be 
> of a different class.  For your toy example, it seems a matrix would be a more 
> reasonable option.
> 
> R-devel has some improvements to row extraction, if I remember correctly.  You 
> might want to try your example there.
> 
> -roger
> 
> Herve Pages wrote:
>> Hi,
>>
>>
>> I have a big data frame:
>>
>>   > mat <- matrix(rep(paste(letters, collapse=""), 5*300000), ncol=5)
>>   > dat <- as.data.frame(mat)
>>
>> and I need to do some computation on each row. Currently I'm doing this:
>>
>>   > for (key in row.names(dat)) { row <- dat[key, ]; ... do some computation on row... }
>>
>> which could probably considered a very natural (and R'ish) way of doing it
>> (but maybe I'm wrong and the real idiom for doing this is something different).
>>
>> The problem with this "idiomatic form" is that it is _very_ slow. The loop
>> itself + the simple extraction of the rows (no computation on the rows) takes
>> 10 hours on a powerful server (quad core Linux with 8G of RAM)!
>>
>> Looping over the first 100 rows takes 12 seconds:
>>
>>   > system.time(for (key in row.names(dat)[1:100]) { row <- dat[key, ] })
>>      user  system elapsed
>>    12.637   0.120  12.756
>>
>> But if, instead of the above, I do this:
>>
>>   > for (i in nrow(dat)) { row <- sapply(dat, function(col) col[i]) }
>>
>> then it's 20 times faster!!
>>
>>   > system.time(for (i in 1:100) { row <- sapply(dat, function(col) col[i]) })
>>      user  system elapsed
>>     0.576   0.096   0.673
>>
>> I hope you will agree that this second form is much less natural.
>>
>> So I was wondering why the "idiomatic form" is so slow? Shouldn't the idiomatic
>> form be, not only elegant and easy to read, but also efficient?
>>
>>
>> Thanks,
>> H.
>>
>>
>>> sessionInfo()
>> R version 2.5.0 Under development (unstable) (2007-01-05 r40386)
>> x86_64-unknown-linux-gnu
>>
>> locale:
>> LC_CTYPE=en_US;LC_NUMERIC=C;LC_TIME=en_US;LC_COLLATE=en_US;LC_MONETARY=en_US;LC_MESSAGES=en_US;LC_PAPER=en_US;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US;LC_IDENTIFICATION=C
>>
>> attached base packages:
>> [1] "stats"     "graphics"  "grDevices" "utils"     "datasets"  "methods"
>> [7] "base"
>>
>> ______________________________________________
>> R-devel at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
> 


-- 

Best wishes
   Wolfgang

------------------------------------------------------------------
Wolfgang Huber  EBI/EMBL  Cambridge UK  http://www.ebi.ac.uk/huber



[Rd] extracting rows from a data frame by looping over the row names: performance issues
Wolfgang Huber