How to change a string according to an existing dictionary?

Question

Suppose, I have such a string 1gov2fg3job4qrt. I want to change the letters according to the following dictionary:

gov=a
fg=b
job=c
qrt=d

So, the output should be:

1a2b3c4d

How can I best do that?

Answer 1

As your post is only tagged with r, I assume you prefer a solution with base R only.

So, I have built a little function that should do the job! Please find below a reprex.

Reprex

• Code of the function

replacechar <- function(patterns, replacements, string) {
  .i <- 1
  while (.i <= length(patterns)) {
    string <- gsub(patterns[.i], replacements[.i], string)
    .i <- .i + 1
  }
  return(string)
}

• Using the function

mystring <- "1gov2fg3job4qrt"
entries <- c("gov", "fg", "job", "qrt")
outputs <- c("a", "b", "c", "d")

replacechar(entries, outputs, mystring)

• Output

#> [1] "1a2b3c4d"

^{Created on 2022-03-17 by the reprex package (v2.0.1)}

Answer 2

Here is another solution. Since you would like to apply multiple patterns and replacements to the same string, we just pass a named vector to pattern argument:

library(stringr)

str_replace_all(str, (c(gov = "a", fg = "b" , job = "c" , qrt = "d")))

[1] "1a2b3c4d"

Answer 3

Try the gsubfn package, though there are likely base R ways to do it:

#library("gsubfn")
gsubfn::gsubfn('gov|fg|job|qrt', list('gov' = 'a','fg' = 'b', 'job' = 'c', 'qrt' = 'd'), str)
# [1] "1a2b3c4d"

You can also use the stringi package:

stringi::stri_replace_all_regex(str,
                                pattern = c("gov", "fg","job","qrt"),
                                replacement = c("a","b","c","d"),
                                vectorize = FALSE)
# [1] "1a2b3c4d"

If your dictionary data are in a data frame or similar, you could use:

lib <- data.frame(pattern = c("gov", "fg","job","qrt"), sub = c("a","b","c","d"))

stringi::stri_replace_all_regex(str,
                                pattern = lib$pattern,
                                replacement = lib$sub,
                                vectorize = FALSE)
# [1] "1a2b3c4d"