[R] Fwd: which is faster "for" or "apply"

Karim Mezhoud kmezhoud at gmail.com
Wed Dec 31 16:37:57 CET 2014


Many Many Many thanks!
it is a demonstrative lesson. I need time to  test all examples :)
Thank you for your time and support.
Happy and Healthy New Year

  Ô__
 c/ /'_;~~~~kmezhoud
(*) \(*)   ⴽⴰⵔⵉⵎ  ⵎⴻⵣⵀⵓⴷ
http://bioinformatics.tn/



On Wed, Dec 31, 2014 at 2:38 PM, Martin Morgan <mtmorgan at fredhutch.org>
wrote:

> On 12/31/2014 12:22 AM, Karim Mezhoud wrote:
>
>> Thanks,
>> It seems for loop spends less time ;)
>>
>> with
>> dim(DataFrame)
>> [1] 338  70
>>
>> For loop has
>>     user  system elapsed
>>    0.012   0.000   0.012
>>
>> and apply has
>>    user  system elapsed
>>    0.020   0.000   0.021
>>
>
> The timings are so short that the answer in terms of speed is 'it does not
> matter'.
>
> Here is a selection of approaches
>
> f0 <- function(df) {
>     for (i in seq_along(df))
>         df[,i] <- as.numeric(df[,i])
>     df
> }
>
> f0a <- function(df) {
>     ## data.frame is a list-of-equal-length vectors; access each
>     ## column with "[["
>     for (i in seq_along(df))
>         df[[i]] <- as.numeric(df[[i]])
>     df
> }
>
> f0c <- compiler::cmpfun(f0)  ## loops sometimes benefit from compilation
>
> f1 <- function(df)
>     as.data.frame(apply(df, 2, as.numeric))
>
> f2 <- function(df) {
>     ## replace all columns of df with list-of-vectors
>     df[] <- lapply(df, as.numeric)
>     df
> }
>
> f3 <- function(df) {
>     ## coerce to matrix to avoid the explicit loop, use mode<- to
>     ## change storage of elements
>     m <- as.matrix(df)
>     mode(m) <- "numeric"
>     as.data.frame(m)
> }
>
> f4 <- function(df) {
>     ## if it's a matrix, why are we returning a data.frame?
>     m <- as.matrix(df)
>     mode(m) <- "numeric"
>     m
> }
>
> f4a <- function(df)
>     ## unlist to single vector, coerce, then format as matrix
>     matrix(as.numeric(unlist(df, use.names=FALSE)), nrow(df),
>            dimnames=dimnames(df))
>
> It's important to test that different methods return the same result
> (perhaps allowing for differences in attributes such as row or column
> names). The microbenchmark package repeats timings across multiple trials
> (default 100 times).
>
> library(microbenchmark)
> test <- function(df) {
>     stopifnot(
>         identical(f0(df), f0a(df)),
>         identical(f0(df), f0c(df)),
>         identical(f0(df), f1(df)),
>         identical(f0(df), f2(df)),
>         identical(f0(df), f3(df)),
>         identical(as.matrix(f0(df)), f4(df)),
>         all.equal(f4(df), f4a(df), check.attributes=FALSE))
>     microbenchmark(f0(df), f0a(df), f1(df), f2(df), f3(df), f4(df),
> f4a(df))
> }
>
> Here are some data sets
>
> m <- matrix(rnorm(338 * 70), 338)
> df <- as.data.frame(m)
> dfc <- as.data.frame(lapply(df, as.character), stringsAsFactors=FALSE)
> dff <- as.data.frame(lapply(df, as.character))
>
> and results
>
> > test(df)
> Unit: microseconds
>     expr      min        lq      mean    median        uq      max neval
>   f0(df) 6208.956 6270.5500 6367.4138 6306.7110 6362.2225 7731.281   100
>  f0a(df) 2917.973 2975.2090 3024.8623 3002.3805 3036.5365 3951.618   100
>  f0c(df) 6078.399 6150.1085 6264.0998 6188.3690 6244.5725 7684.116   100
>   f1(df) 2698.074 2743.2905 2821.8453 2769.3655 2805.5345 4033.229   100
>   f2(df) 1989.057 2041.0685 2066.1830 2055.0020 2083.8545 2267.732   100
>   f3(df) 1532.435 1572.9810 1609.7378 1597.6245 1624.2305 2003.584   100
>   f4(df)  808.593  828.5445  852.2626  847.5355  864.6665 1180.977   100
>  f4a(df)  422.657  437.2705  458.9845  455.2470  465.5815  695.443   100
> > test(dfc)
> Unit: milliseconds
>     expr       min        lq      mean    median        uq       max neval
>   f0(df) 11.416532 11.647858 11.915287 11.767647 12.016276 14.239622   100
>  f0a(df)  8.095709  8.211116  8.380638  8.289895  8.454948  9.529026   100
>  f0c(df) 11.339293 11.577811 11.772087 11.702341 11.896729 12.674766   100
>   f1(df)  8.227371  8.277147  8.422412  8.331403  8.490411  9.145499   100
>   f2(df)  6.907888  7.010828  7.162529  7.147198  7.239048  7.763758   100
>   f3(df)  6.608107  6.688232  6.845936  6.792066  6.892635  8.359274   100
>   f4(df)  5.859482  5.939680  6.046976  5.993804  6.105388  6.968601   100
>  f4a(df)  5.372214  5.460987  5.556687  5.521542  5.614482  6.107081   100
> > test(dff)
> Error: identical(f0(df), f1(df)) is not TRUE
>
> Except when dealing with factors, the use of explicit loops is the
> slowest. With factors, matrix-based methods coerce the level labels to
> numeric, whereas vector-based methods coerce the underlying codes (level
> values) of the factor; obviously great care needs to be taken.
>
> > f0(dff)[1:5, 1:5]
>    V1  V2  V3  V4  V5
> 1 150 232 294  88  56
> 2 159   8  89  59  10
> 3 132 171  40 205 119
> 4 214 273  26 262 216
> 5 281  49 255  31 233
> > f1(dff)[1:5, 1:5]
>           V1          V2         V3         V4          V5
> 1 -1.7092463 0.50234009  0.8492982 -0.5636901 -0.38545566
> 2 -2.3020854 -0.05580931 -0.5963673 -0.3671748 -0.09408031
> 3 -1.2915110 -2.46181533 -0.2470108 0.3301129 -1.06810225
> 4  0.3065989 0.89263099 -0.1717432  0.7721411 0.35856334
> 5  0.8795616 -0.43049898  0.4560515 -0.1722099  0.46125149
>
> In terms of 'best practice', I would represent my data in the appropriate
> data structure in the first place (as a matrix of appropriate type, rather
> than data.frame, so the entire coercion is irrelevant). If faced with a
> data.frame with specific columns to coerce I would use the approach
>
>     cidx <- sapply(df, is.character)      # index of columns to coerce
>     df[cidx] <- lapply(df[cidx], as.numeric)
>
> which seems to be reasonably correct, expressive, compact, and speedy.
>
> Martin Morgan
>
>
>
>>    Ô__
>>   c/ /'_;~~~~kmezhoud
>> (*) \(*)   ⴽⴰⵔⵉⵎ  ⵎⴻⵣⵀⵓⴷ
>> http://bioinformatics.tn/
>>
>>
>>
>> On Wed, Dec 31, 2014 at 8:54 AM, Berend Hasselman <bhh at xs4all.nl> wrote:
>>
>>
>>>  On 31-12-2014, at 08:40, Karim Mezhoud <kmezhoud at gmail.com> wrote:
>>>>
>>>> Hi All,
>>>> I would like to choice between these two data frame convert. which is
>>>> faster?
>>>>
>>>>    for(i in 1:ncol(DataFrame)){
>>>>
>>>>                     DataFrame[,i] <- as.numeric(DataFrame[,i])
>>>>                 }
>>>>
>>>>
>>>> OR
>>>>
>>>> DataFrame <- as.data.frame(apply(DataFrame,2 ,function(x)
>>>> as.numeric(x)))
>>>>
>>>>
>>>>
>>> Try it and use system.time.
>>>
>>> Berend
>>>
>>>  Thanks
>>>> Karim
>>>>   Ô__
>>>> c/ /'_;~~~~kmezhoud
>>>> (*) \(*)   ⴽⴰⵔⵉⵎ  ⵎⴻⵣⵀⵓⴷ
>>>> http://bioinformatics.tn/
>>>>
>>>>        [[alternative HTML version deleted]]
>>>>
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>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>>
>>> http://www.R-project.org/posting-guide.html
>>>
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>>>
>>>
>>         [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
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>
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