Recode if string (with punctuation) contains certain text

January 10, 2022

How can I search through a character vector and, if the string at a given index contains a pattern, replace that index’s value?

I tried this:

List <- c(1:8)
  Types<-as.character(c(
    "ABC, the (stuff).\n\n\n fun", "meaningful", "relevant", "rewarding", 
    "unpleasant", "enjoyable", "engaging", "disinteresting"))
  for (i in List) {
    if (grepl(Types[i], "fun", fixed = TRUE))
    {Types[i]="1"
    } else if (grepl(Types[i], "meaningful", fixed = TRUE))
    {Types[i]="2"}}

The code works for "meaningful", but doesn’t when there’s punctuation or other things in the string, as with "fun".

>Solution :

The first argument to grepl is the pattern, not the string.

This would be a literal fix of your code:

for (i in seq_along(Types)) {
  if (grepl("fun", Types[i], fixed = TRUE)) {
    Types[i] = "1"
  } else if (grepl("meaningful", Types[i], fixed = TRUE)) {
    Types[i] = "2"
  }
}
Types
# [1] "1"              "2"              "relevant"       "rewarding"      "unpleasant"    
# [6] "enjoyable"      "engaging"       "disinteresting"

BTW, the use of List works, but it’s a little extra: when you have separate variables like that, it is possible that one might go out of sync with the other. For instance, if you update Types and forget to update List, then it will break (or fail). For this, I used seq_along(Types) instead.

BTW: here’s a slightly different version that leaves Types untouched and returns a new vector, and is introducing you to the power of vectorization:

Types[grepl("fun", Types, fixed = TRUE)] <- "1"
Types[grepl("meaningful", Types, fixed = TRUE)] <- "2"
Types
# [1] "1"              "2"              "relevant"       "rewarding"      "unpleasant"    
# [6] "enjoyable"      "engaging"       "disinteresting"

The next level (perhaps over-complicating?) would be to store the patterns and recoding replacements in a frame (always a 1-to-1, you’ll never accidentally update one without the other, can be stored in CSV if needed) and Reduce on it:

ptns <- data.frame(ptn = c("fun", "meaningful"), repl = c("1", "2"))
Reduce(function(txt, i) {
  txt[grepl(ptns$ptn[i], txt, fixed = TRUE)] <- ptns$repl[i]
  txt
}, seq_len(nrow(ptns)), init = Types)
# [1] "1"              "2"              "relevant"       "rewarding"      "unpleasant"    
# [6] "enjoyable"      "engaging"       "disinteresting"