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

问题

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

回答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.

回答2:

Not sure what you tried but this seems to work:

data$keep <- ifelse(grepl(paste(matches, collapse = "|"), data$animal), "Keep","Discard")

Edit:

In case you are doing this to later on subset the data.frame according to "Keep" and "Discard" entries, you could do this more directly using:

data[grepl(paste(matches, collapse = "|"), data$animal),]

This way, the results of grepl which are TRUE or FALSE are used for the subset.

回答3:

Try to avoid ifelse as much as possible. This, for example, works nicely

c("Discard", "Keep")[grepl("(dog|cat)", data$animal) + 1]

For a 123 seed you will get

##  [1] "Keep"    "Keep"    "Discard" "Keep"    "Keep"    "Keep"    "Discard" "Keep"   
##  [9] "Discard" "Discard" "Keep"    "Discard" "Keep"    "Discard" "Keep"    "Keep"   
## [17] "Keep"    "Keep"    "Keep"    "Keep"    "Keep"    "Keep"    "Keep"    "Keep"   
## [25] "Keep"    "Keep"    "Discard" "Discard" "Keep"    "Keep"    "Keep"    "Keep"   
## [33] "Keep"    "Keep"    "Keep"    "Discard" "Keep"    "Keep"    "Keep"    "Keep"   
## [41] "Keep"    "Discard" "Discard" "Keep"    "Keep"    "Keep"    "Keep"    "Discard"
## [49] "Keep"    "Keep"

来源：https://stackoverflow.com/questions/25391975/grepl-in-r-to-find-matches-to-any-of-a-list-of-character-strings

标签

regex

grep

grepl