Speedy/elegant way to unite many pairs of columns

问题:

Is there an elegant/fastR way to combine all pairs of columns in a data.frame?

For example, using mapply() and paste() we can turn this data.frame:

mydf <- data.frame(a.1 = letters, a.2 = 26:1, b.1 = letters, b.2 = 1:26)
head(mydf)
  a.1 a.2 b.1 b.2
1   a  26   a   1
2   b  25   b   2
3   c  24   c   3
4   d  23   d   4
5   e  22   e   5
6   f  21   f   6

into this data.frame:

mydf2 <- mapply(function(x, y) {
     paste(x, y, sep = ".")},
     mydf[ ,seq(1, ncol(mydf), by = 2)],
     mydf[ ,seq(2, ncol(mydf), by = 2)])
head(mydf2)
     a.1    b.1  
[1,] "a.26" "a.1"
[2,] "b.25" "b.2"
[3,] "c.24" "c.3"
[4,] "d.23" "d.4"
[5,] "e.22" "e.5"
[6,] "f.21" "f.6"

However, this feels clumsy and is a bit slow when applied to big datasets. Any suggestions, perhaps using a Hadley package?

EDIT: The ideal solution would easily scale to large numbers of columns, such that the names of the columns would not need to be included in the function call. Thanks!

回答1:

It's amusing to note that the OP's solution appears to be the fastest one:

f1 <- function(mydf) {
    mapply(function(x, y) {
        paste(x, y, sep = ".")},
        mydf[ ,seq(1, ncol(mydf), by = 2)],
        mydf[ ,seq(2, ncol(mydf), by = 2)])
}

f.thelatemail <- function(mydf) {
    mapply(paste,mydf[c(TRUE,FALSE)],mydf[c(FALSE,TRUE)],sep=".")
}

require(dplyr)

f.on_the_shores_of_linux_sea <- function(mydf) {
    transmute(mydf,x1=paste0( a.1,'.', a.2),x2=paste0( b.1,'.', b.2)) 
}

f.jazurro <- function(mydf) {
    odd <- seq(1, ncol(mydf), 2);
    lapply(odd, function(x) paste(mydf[,x], mydf[,x+1], sep = ".")) %>% 
        do.call(cbind,.)
}

library(data.table) 
f.akrun <- function(mydf) {
    res <- as.data.table(matrix(, ncol=ncol(mydf)/2, nrow=nrow(mydf)))
    indx <- seq(1, ncol(mydf), 2)
    setDT(mydf)
    for(j in seq_along(indx)){
        set(res, i=NULL, j=j, value= paste(mydf[[indx[j]]], 
                                           mydf[[indx[j]+1]], sep='.'))
    }
    res
}

mydf <- data.frame(a.1 = letters, a.2 = 26:1, b.1 = letters, b.2 = 1:26)
mydf <- mydf[rep(1:nrow(mydf),5000),]


library(rbenchmark)
benchmark(f1(mydf),f.thelatemail(mydf),f.on_the_shores_of_linux_sea(mydf),f.jazurro(mydf),f.akrun(mydf))

Results:

#                                 test replications elapsed relative user.self sys.self user.child sys.child
# 5                      f.akrun(mydf)          100  14.000   75.269    13.673    0.296          0         0
# 4                    f.jazurro(mydf)          100   0.388    2.086     0.314    0.071          0         0
# 3 f.on_the_shores_of_linux_sea(mydf)          100  15.585   83.790    15.293    0.280          0         0
# 2                f.thelatemail(mydf)          100  26.416  142.022    25.736    0.639          0         0
# 1                           f1(mydf)          100   0.186    1.000     0.169    0.017          0         0

[Updated Benchmark]

I've added one solution from @thelatemail, which I missed in the original answer, and one solution from @akrun:

f.thelatemail2 <- function(mydf) {
    data.frame(Map(paste,mydf[c(TRUE,FALSE)],mydf[c(FALSE,TRUE)],sep="."))
}

f.akrun2 <- function(mydf) {    
    setDT(mydf)
    indx <- as.integer(seq(1, ncol(mydf), 2))
    mydf2 <- copy(mydf)
    for(j in indx){
        set(mydf2, i=NULL, j=j, value= paste(mydf2[[j]],
                                             mydf2[[j+1]], sep="."))
    }
    mydf2[,indx, with=FALSE]
}

Benchmark:

library(rbenchmark)

benchmark(f1(mydf),f.thelatemail(mydf), f.thelatemail2(mydf), f.on_the_shores_of_linux_sea(mydf),f.jazurro(mydf),f.akrun(mydf),f.akrun2(mydf))
#                                 test replications elapsed relative user.self sys.self user.child sys.child
# 6                      f.akrun(mydf)          100  13.247   69.356    12.897    0.340          0         0
# 7                     f.akrun2(mydf)          100  12.746   66.733    12.405    0.339          0         0
# 5                    f.jazurro(mydf)          100   0.327    1.712     0.254    0.073          0         0
# 4 f.on_the_shores_of_linux_sea(mydf)          100  16.347   85.586    15.838    0.445          0         0
# 2                f.thelatemail(mydf)          100  26.307  137.733    25.536    0.708          0         0
# 3               f.thelatemail2(mydf)          100  15.938   83.445    15.136    0.750          0         0
# 1                           f1(mydf)          100   0.191    1.000     0.156    0.036          0         0

回答2:

I'm not sure this is the best approach. See if the below code gives any speed improvement

require(dplyr)
transmute(mydf,x1=paste0( a.1,'.', a.2),x2=paste0( b.1,'.', b.2)) 

Answer updated based on comment :-)

回答3:

An option using set from data.table. It should be fast for large datasets as it modifies by reference and the overhead of [.data.table is avoided. Assuming that the columns are ordered for each pair of columns.

library(data.table)
res <- as.data.table(matrix(, ncol=ncol(mydf)/2, nrow=nrow(mydf)))
indx <- seq(1, ncol(mydf), 2)
setDT(mydf)
for(j in seq_along(indx)){
   set(res, i=NULL, j=j, value= paste(mydf[[indx[j]]], 
                           mydf[[indx[j]+1]], sep='.'))
 }
head(res)
#    V1  V2
#1: a.26 a.1
#2: b.25 b.2
#3: c.24 c.3
#4: d.23 d.4
#5: e.22 e.5
#6: f.21 f.6

Instead of creating a new result dataset, we can also update the same or a copy of the original dataset. There will be some warnings about type conversion, but I guess this would be a bit faster (not benchmarked)

setDT(mydf)
mydf2 <- copy(mydf)
for(j in indx){
  set(mydf2, i=NULL, j=j, value= paste(mydf2[[j]],
   mydf2[[j+1]], sep="."))
 }
 mydf2[,indx, with=FALSE]

Benchmarks

I tried the benchmarks on a slightly bigger data with many columns.

data

set.seed(24)
d1 <- as.data.frame(matrix(sample(letters,500*10000, replace=TRUE), 
    ncol=500), stringsAsFactors=FALSE)
set.seed(4242)
d2 <- as.data.frame(matrix(sample(1:200,500*10000,
            replace=TRUE), ncol=500))
d3 <- cbind(d1,d2)
mydf <- d3[,order(c(1:ncol(d1), 1:ncol(d2)))]
mydf1 <- copy(mydf) 

Compared f1, f.jazurro (fastest) (from @Marat Talipov's post) with f.akrun2

   microbenchmark(f1(mydf), f.jazurro(mydf), f.akrun2(mydf1),
         unit='relative', times=20L)
   #Unit: relative
   #        expr      min        lq     mean   median       uq      max neval
   #      f1(mydf) 3.420448 2.3217708 2.714495 2.653178 2.819952 2.736376    20
   #f.jazurro(mydf) 1.000000 1.0000000 1.000000 1.000000 1.000000 1.000000    20
   #f.akrun2(mydf1) 1.204488 0.8015648 1.031248 1.042262 1.097136 1.066671    20
   #cld
   #b
   #a 
   #a 

In this, f.jazurro is slighly better than f.akrun2. I think if I increase the group size, nrows etc, it would be an interesting comparison