R grep: is there an AND operator?

Question

Suppose I have the following data frame:

User.Id    Tags
34234      imageUploaded,people.jpg,more,comma,separated,stuff
34234      imageUploaded
12345      p         


        

Answer 1

I love @Chase's answer, and it makes good sense to me, but it can be a bit dangerous to use constructs that one doesn't totally understand.

This answer is meant to reassure anyone who'd like to use @thelatemail's more straightforward approach that it works just as well and is completely competitive speedwise. It's certainly what I'd use in this case. (It's also reassuring that the more sophisticated Perl-compatible-regex pays no performance cost for its power and easy extensibility.)

library(rbenchmark)
x <- paste0(sample(letters, 1e6, replace=T), ## A longer vector of
            sample(letters, 1e6, replace=T)) ## possible matches

## Both methods give identical results
tlm <- grepl("a", x, fixed=TRUE) & grepl("b", x, fixed=TRUE)
pat <- "(?=.*a)(?=.*b)"
Chase <- grepl(pat, x, perl=TRUE)
identical(tlm, Chase)
# [1] TRUE    

## Both methods are similarly fast
benchmark(
    tlm = grepl("a", x, fixed=TRUE) & grepl("b", x, fixed=TRUE),
    Chase = grepl(pat, x, perl=TRUE))
#          test replications elapsed relative user.self sys.self
# 2       Chase          100    9.89    1.105      9.80     0.10
# 1 thelatemail          100    8.95    1.000      8.47     0.48

Answer 2

Below is an alternative to grep using hadley's stringr::str_detect(). This avoids the use of perl=true @jan-stanstrup. Additionally, the dplyr::filter() will return the rows within the dataframe itself so you never need to leave the df.

library(stringr)
libary(dplyr)
 x <- data.frame(User.Id =c(34234,34234,12345), 
                 Tags=c("imageUploaded,people.jpg,more,comma,separated,stuff",
                        "imageUploaded",
                        "people.jpg"))

 data.people <- x %>% filter(str_detect(Tags,"(?=.*imageUploaded)(?=.*people\\.jpg)"))
 data.people

# returns
#  User.Id                                                Tags
# 1   34234 imageUploaded,people.jpg,more,comma,separated,stuff

This is simpler and works if "people.jpg" always follows "imageUploaded"

str_extract(x,"imageUploaded.*people\\.jpg")

Answer 3

For readability's sake, you could just do:

x <- c(
       "imageUploaded,people.jpg,more,comma,separated,stuff",
       "imageUploaded",
       "people.jpg"
       )

xmatches <- intersect(
                      grep("imageUploaded",x,fixed=TRUE),
                      grep("people.jpg",x,fixed=TRUE)
                     )
x[xmatches]
[1] "imageUploaded,people.jpg,more,comma,separated,stuff"

Answer 4

Thanks to this answer, this regex seems to work. You want to use grepl() which returns a logical to index into your data object. I won't claim to fully understand the inner workings of the regex, but regardless:

x <- c("imageUploaded,people.jpg,more,comma,separated,stuff", "imageUploaded", "people.jpg")

grepl("(?=.*imageUploaded)(?=.*people\\.jpg)", x, perl = TRUE)
#-----
[1]  TRUE FALSE FALSE