grepl in R to find matches to any of a list of character strings

Question

Is it possible to use a grepl argument when referring to a list of values, maybe using the %in% operator? I want to take the data below and if the animal name has "dog" or "cat" in it, I want to return a certain value, say, "keep"; if it doesn't have "dog" or "cat", I want to return "discard".

data <- data.frame(animal = sample(c("cat","dog","bird", 'doggy','kittycat'), 50, replace = T))

Now, if I were just to do this by strictly matching values, say, "cat" and "dog', I could use the following approach:

matches <- c("cat","dog")

data$keep <- ifelse(data$animal %in% matches, "Keep", "Discard")

But using grep or grepl only refers to the first argument in the list:

data$keep <- ifelse(grepl(matches, data$animal), "Keep","Discard")

returns

Warning message:
In grepl(matches, data$animal) :
  argument 'pattern' has length > 1 and only the first element will be used

Note, I saw this thread in my search, but this doesn't appear to work: grep using a character vector with multiple patterns

Answer 1

You can use an "or" (|) statement inside the regular expression of grepl.

ifelse(grepl("dog|cat", data$animal), "keep", "discard")
# [1] "keep"    "keep"    "discard" "keep"    "keep"    "keep"    "keep"    "discard"
# [9] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "discard" "keep"   
#[17] "discard" "keep"    "keep"    "discard" "keep"    "keep"    "discard" "keep"   
#[25] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"   
#[33] "keep"    "discard" "keep"    "discard" "keep"    "discard" "keep"    "keep"   
#[41] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"   
#[49] "keep"    "discard"

The regular expression dog|cat tells the regular expression engine to look for either "dog" or "cat", and return the matches for both.