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

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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

Jun 22, 2022 in Data Science by avinash
• 1,840 points
2,784 views

1 answer to this question.

0 votes

Inside of a grepl regular expression, you can utilise a "or" (|) statement.

Ifelse(grepl("dog|cat", data$animal), "keep," "discard," # [1] "keep," "keep," "discard," # [9] "keep," "keep," "discard," "keep," "keep," "keep," "keep," "discard," "keep," and "keep" #[17] Discard" Keep" Discard" Keep" Discard" Keep" Discard" Keep" #[25] "keep," "keep," "keep," "keep," "keep," "keep," and "keep" #[33] "keep," "discard," "keep," "discard," "keep," "keep," and "discard" #[41] "keep," "keep," "keep," "keep," "keep," "keep," "keep," "keep," and "discard" are all forms of "keep."
Dog|Cat instructs the regular expression engine to search for either "dog" or "cat," returning matches for both.

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answered Jun 24, 2022 by Sohail
• 3,040 points

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