使 for 循环在 R 中运行得更快

Question

I want to create a model where I duplicate a sentence several times, introducing random error each time. The duplicates of the sentence also get duplicated. So, in cycle one, I start with just "example_sentence". In cycle two, I have two copies of that sentence. In cycle three, I have 4 copies of that sentence. I want to do this for 25 cycles with 20k sentences. The code I wrote to do that works way too slowly, and I am wondering if there is a way to make my nested for loops more efficient? Here is the part of the code that is the slowest:

    alphabet <- c("a","b","d","j")
    modr1 <- "sentencetoduplicate"

    errorRate <- c()
    errorRate <- append(errorRate, rep(1,1))
    errorRate <- append(errorRate, rep(0,999))

    duplicate <- c(modr1)
    for (q in 1:25) {
      collect <- c()
      for (z in seq_along(duplicate)) {
        modr1 = duplicate[z]
        compile1 <- c()
        for (k in 1:nchar(modr1)) {
          error <- sample(errorRate, 1, replace = TRUE)
          if (error == 1) {
            compile1 <- append(compile1, sub(substring(modr1,k,k),sample(alphabet,1,replace=TRUE),substring(modr1,k,k)))
          } else {
            compile1 <- append(compile1, substring(modr1,k,k))
          }
        }
        modr1 <- paste(compile1, collapse='')
        collect <- append(collect, modr1)
      }
      duplicate <- append(duplicate, collect)
    }

Answer 1

Here is a faster approach to your loop, but I think the problem of applying this to your problem of 20K sentences remains!

f <- function(let, alphabet = c("a","b","c","d","j"),error_rate=1/1000) {
  lenlet=length(let)
  let = unlist(let)
  k <- rbinom(length(let),1,prob = error_rate)
  let[k==1] <- sample(alphabet,size = sum(k==1), replace=T)
  return(as.list(as.data.frame(matrix(let, ncol=lenlet))))
}

modr1 <- "sentencetoduplicate"
k <- data.table(list(strsplit(modr1,"")[[1]]))

for(q in 1:25) {
    k[, V1:=list(f(V1))]
    k <- k[rep(1:nrow(k),2)]
}

Updated with slightly faster version! (Notice this is no longer by=1:nrow(k) )