r- grepl to find multiple strings exists

问题

grepl("instance|percentage", labelTest$Text)

will return true if any one of instance or percentage is present.

How will i get true only when both the terms are present.

回答1:

Text <- c("instance", "percentage", "n", 
          "instance percentage", "percentage instance")

grepl("instance|percentage", Text)
# TRUE  TRUE FALSE  TRUE  TRUE

grepl("instance.*percentage|percentage.*instance", Text)
# FALSE FALSE FALSE TRUE  TRUE

The latter one works by looking for:

('instance')(any character sequence)('percentage')  
OR  
('percentage')(any character sequence)('instance')

Naturally if you need to find any combination of more than two words, this will get pretty complicated. Then the solution mentioned in the comments would be easier to implement and read.

Another alternative that might be relevant when matching many words is to use positive look-ahead (can be thought of as a 'non-consuming' match). For this you have to activate perl regex.

# create a vector of word combinations
set.seed(1)
words <- c("instance", "percentage", "element",
           "character", "n", "o", "p")
Text2 <- replicate(10, paste(sample(words, 5), collapse=" "))

# grepl with multiple positive look-ahead
longperl <- grepl("(?=.*instance)(?=.*percentage)(?=.*element)(?=.*character)",
  Text2, perl=TRUE)

# this is equivalent to the solution proposed in the comments
longstrd <- grepl("instance", Text2) & 
          grepl("percentage", Text2) & 
             grepl("element", Text2) & 
           grepl("character", Text2)

# they produce identical results
identical(longperl, longstrd)

Furthermore, if you have the patterns stored in a vector you can condense the expressions significantly, giving you

pat <- c("instance", "percentage", "element", "character")

longperl <- grepl(paste0("(?=.*", pat, ")", collapse=""), Text2, perl=TRUE)
longstrd <- rowSums(sapply(pat, grepl, Text2) - 1L) == 0L

As asked for in the comments, if you want to match on exact words, i.e. not match on substrings, we can specify word boundaries using \\b. E.g:

tx <- c("cent element", "percentage element", "element cent", "element centimetre")

grepl("(?=.*\\bcent\\b)(?=.*element)", tx, perl=TRUE)
# TRUE FALSE  TRUE FALSE
grepl("element", tx) & grepl("\\bcent\\b", tx)
# TRUE FALSE  TRUE FALSE

回答2:

Use intersect and feed it a grep for each word

library(data.table) #used for subsetting text vector below

vector_of_text[ intersect( grep(vector_of_text , pattern = "pattern1") , grep(vector_of_text , pattern = "pattern2") ) ]

来源：https://stackoverflow.com/questions/44152970/r-grepl-to-find-multiple-strings-exists

标签

grepl