Speedy/elegant way to unite many pairs of columns

Question

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.fr

Answer 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