UC Business Analytics R Programming Guide

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Regular Expression Functions in Base R

R contains a set of functions in the base package that we can use to find pattern matches. Alternatively, the R package stringr also provides several functions for regex operations. This section covers the base R functions that provide pattern finding, pattern replacement, and string splitting capabilities.


Pattern Finding Functions

There are five functions that provide pattern matching capabilities. The three functions that I provide examples for are ones that are most common. The two other functions which I do not illustrate are gregexpr() and regexec() which provide similar capabilities as regexpr() but with the output in list form.

grep( )

To find a pattern in a character vector and to have the element values or indices as the output use grep():

# use the built in data set `state.division`
head(as.character(state.division))
## [1] "East South Central" "Pacific" "Mountain"          
## [4] "West South Central" "Pacific" "Mountain"


# find the elements which match the patter
grep("North", state.division)
##  [1] 13 14 15 16 22 23 25 27 34 35 41 49


# use 'value = TRUE' to show the element value
grep("North", state.division, value = TRUE)
##  [1] "East North Central" "East North Central" "West North Central"
##  [4] "West North Central" "East North Central" "West North Central"
##  [7] "West North Central" "West North Central" "West North Central"
## [10] "East North Central" "West North Central" "East North Central"


# can use the 'invert' argument to show the non-matching elements
grep("North | South", state.division, invert = TRUE)
##  [1]  2  3  5  6  7  8  9 10 11 12 19 20 21 26 28 29 30 31 32 33 37 38 39
## [24] 40 44 45 46 47 48 50


grepl( )

To find a pattern in a character vector and to have logical (TRUE/FALSE) outputs use grep():

grepl("North | South", state.division)
##  [1]  TRUE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [12] FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE  TRUE
## [23]  TRUE  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE
## [34]  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE
## [45] FALSE FALSE FALSE FALSE  TRUE FALSE

# wrap in sum() to get the count of matches
sum(grepl("North | South", state.division))
## [1] 20


regexpr( )

To find exactly where the pattern exists in a string use regexpr():

x <- c("v.111", "0v.11", "00v.1", "000v.", "00000")

regexpr("v.", x)
## [1]  1  2  3  4 -1
## attr(,"match.length")
## [1]  2  2  2  2 -1
## attr(,"useBytes")
## [1] TRUE

The output of regexpr() can be interepreted as follows. The first element provides the starting position of the match in each element. Note that the value -1 means there is no match. The second element (attribute “match length”) provides the length of the match. The third element (attribute “useBytes”) has a value TRUE meaning matching was done byte-by-byte rather than character-by-character.


Pattern Replacement Functions

In addition to finding patterns in character vectors, its also common to want to replace a pattern in a string with a new patter. There are two options for this:

sub( )

To replace the first matching occurrence of a pattern use sub():

new <- c("New York", "new new York", "New New New York")
new
## [1] "New York"         "new new York"     "New New New York"

# Default is case sensitive
sub("New", replacement = "Old", new)
## [1] "Old York"         "new new York"     "Old New New York"

# use 'ignore.case = TRUE' to perform the obvious
sub("New", replacement = "Old", new, ignore.case = TRUE)
## [1] "Old York"         "Old new York"     "Old New New York"


gsub( )

To replace all matching occurrences of a pattern use gsub():

# Default is case sensitive
gsub("New", replacement = "Old", new)
## [1] "Old York"         "new new York"     "Old Old Old York"

# use 'ignore.case = TRUE' to perform the obvious
gsub("New", replacement = "Old", new, ignore.case = TRUE)
## [1] "Old York"         "Old Old York"     "Old Old Old York"


Splitting Character Vectors

To split the elements of a character string use strsplit():

x <- paste(state.name[1:10], collapse = " ")

# output will be a list
strsplit(x, " ")
## [[1]]
##  [1] "Alabama"     "Alaska"      "Arizona"     "Arkansas"    "California" 
##  [6] "Colorado"    "Connecticut" "Delaware"    "Florida"     "Georgia"

# output as a vector rather than a list
unlist(strsplit(x, " "))
##  [1] "Alabama"     "Alaska"      "Arizona"     "Arkansas"    "California" 
##  [6] "Colorado"    "Connecticut" "Delaware"    "Florida"     "Georgia"