Replace a set of pattern matches with corresponding replacement strings in R

Question

The str_replace (and preg_replace) function in PHP replaces all occurrences of the search string with the replacement string. What interests me the most here, is that if search and replace args are arrays (in R we call that vectors), then str_replace takes a value from each array (vector) and uses them to search and replace on subject.

In other words, does R (or some R package) have a function to perform the following:

string <- "The quick brown fox jumped over the lazy dog."
patterns     <- c("quick", "brown", "fox")
replacements <- c("slow",  "black", "bear")
xxx_replace_xxx(string, patterns, replacements)          ## ???
## [1] "The slow black bear jumped over the lazy dog."

So I am seeking for something like chartr, but for search patterns and replacement strings of arbitrary number of characters. This cannot be done via one call to gsub() as its replacement argument can be a single string only, see ?gsub. So my current implementation is like:

xxx_replace_xxx <- function(string, patterns, replacements) {
   for (i in seq_along(patterns))
      string <- gsub(patterns[i], replacements[i], string, fixed=TRUE)
   string
}

However, I am looking for something much faster if length(patterns) is large - I have a lot of data to process and I'm dissatisfied with the current results.

Exemplary toy data for benchmarking:

string <- readLines("http://www.gutenberg.org/files/31536/31536-0.txt", encoding="UTF-8")
patterns <- c("jak", "to", "do", "z", "na", "i", "w", "za", "tu", "gdy",
   "po", "jest", "Tadeusz", "lub", "razem", "nas", "przy", "oczy", "czy",
   "sam", "u", "tylko", "bez", "ich", "Telimena", "Wojski", "jeszcze")
replacements <- paste0(patterns, rev(patterns))

Answer 1

Using PCRE instead of fixed matching takes ~1/3 the time on my machine for your example.

xxx_replace_xxx_pcre <- function(string, patterns, replacements) {
   for (i in seq_along(patterns))
      string <- gsub(patterns[i], replacements[i], string, perl=TRUE)
   string
}
system.time(x <- xxx_replace_xxx(string, patterns, replacements))
#    user  system elapsed 
#   0.491   0.000   0.491 
system.time(p <- xxx_replace_xxx_pcre(string, patterns, replacements))
#    user  system elapsed 
#   0.162   0.000   0.162 
identical(x,p)
# [1] TRUE