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 <b...@xs4all.nl> wrote:
On 31-12-2014, at 08:40, Karim Mezhoud <kmezh...@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/
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