R: Quickest way to create dataframe with an alternative to IFELSE

Question

I have a similar question this the one on this thread: Using R, replace all values in a matrix <0.1 with 0?

But in my case I have hypothetically larger dataset and va

Answer 1

One option is data.table

library(data.table)
nm1 <- paste0("IND", rep(letters[1:2], length(seqIndividuals)), 
                    rep(seqIndividuals, each = 2))
setDT(alFrequ)
for(j in seq_along(nm1)) {
      alFrequ[, nm1[j] := A2
             ][runif(.N, 0, 1) < MAF , nm1[j] := A1][]
}

Benchmarks

set.seed(24)
alFrequ <- data.frame(SNP= paste0('rs', sample(600000, 340000, replace=FALSE)),
                   A1 = sample(c("G", "T", "A", "C"), 340000, replace=TRUE),
                   A2 = sample(c("G", "T", "A", "C"), 340000, replace=TRUE),
                   MAF = runif(340000, 0, 1), stringsAsFactors=FALSE)
nm1 <- paste0("IND", rep(letters[1:2], length(seqIndividuals)), 
                          rep(seqIndividuals, each = 2))

system.time({
    setDT(alFrequ)
     for(j in seq_along(nm1)){
     alFrequ[, nm1[j] := A2][runif(.N, 0, 1) < MAF , nm1[j] := A1][]
   }
})
#   user  system elapsed 
#  10.72    1.05   11.76 

and using the OP's code on the original dataset

system.time({
 for(i in seqIndividuals) {
   alFrequ[paste("IND",i,"a",sep="")] = ifelse(runif(length(alFrequ$SNP),0.00,1.00) < 
          alFrequ$MAF, alFrequ$A1, alFrequ$A2)
   alFrequ[paste("IND",i,"b",sep="")] = ifelse(runif(length(alFrequ$SNP),0.00,1.00) < 
             alFrequ$MAF, alFrequ$A1, alFrequ$A2)
 }
})
#    user  system elapsed 
#   72.16    6.82   79.33