2.1 String definition

A string can be defined within double " or simple ' quote

string1 <- "This is a string"
string2 <- 'If I want to include a "quote"
inside a string, I use single quotes'

If you forget to close a quote, you’ll see +, the continuation character:

> "This is a string without a closing quote
+ 
+ 
+ HELP I'M STUCK

If this happens to you, press Escape and try again!

To include a literal single or double quote in a string you can use \ to escape it:

double_quote <- "\"" # or '"'
single_quote <- '\'' # or "'"

If you want to include a literal backslash, you’ll need to double it up: "\\".

2.2 String representation

The printed representation of a string is not the same as string itself

x <- c("\"", "\\")
x

[1] "\"" "\\"

writeLines(x)

"
\

Some characters have a special representation, they are called special characters. The most common are "\n", newline, and "\t", tabulation, but you can see the complete list by requesting help on ": ?'"'

2.3 String operation

You can perform basic operation on strings like

• String length

str_length(c("a", "R for data science", NA))

[1]  1 18 NA

• Combining strings

str_c("x", "y", "z")

[1] "xyz"

• Subsetting strings

x <- c("Apple", "Banana", "Pear")
str_sub(x, 1, 3)

[1] "App" "Ban" "Pea"

• Subsetting strings negative numbers count backwards from the end

str_sub(x, -3, -1)

[1] "ple" "ana" "ear"

• Lower case transform

str_to_lower(x)

[1] "apple"  "banana" "pear"  

• ordering

str_sort(x)

[1] "Apple"  "Banana" "Pear"  

3.1 Exercises

• Explain why each of these strings doesn’t match a : “\”, “\\”, “\\\”.
• How would you match the sequence "'\?
• What patterns will the regular expression \..\..\.. match? How would you represent it as a string?

3.2 Anchors

Until now we searched for patterns anywhere in the target string. But we can use anchors to be more precise.

• ^ Match the start of the string.
• $ Match the end of the string.

x <- c("apple", "banana", "pear")
str_view(x, "^a")

str_view(x, "a$")

x <- c("apple pie", "apple", "apple cake")
str_view(x, "^apple$")

3.3 Exercices

• How would you match the literal string "$^$"?
• Given the corpus of common words in stringr::words, create regular expressions that find all words that: -Start with “y”.
  • End with “x”
  • Are exactly three letters long. (Don’t cheat by using str_length()!)
  • Have seven letters or more.

Since this list is long, you might want to use the match argument to str_view() to show only the matching or non-matching words.

3.4 Character classes and alternatives

In regular expression we have special character and patterns that match groups of characters.

• \d: matches any digit.
• \s: matches any whitespace (e.g. space, tab, newline).
• [abc]: matches a, b, or c.
• [^abc]: matches anything except a, b, or c.

str_view(c("abc", "a.c", "a*c", "a c"), "a[.]c")

str_view(c("abc", "a.c", "a*c", "a c"), ".[*]c")

str_view(c("abc", "a.c", "a*c", "a c"), "a[ ]")

You can use alternations to pick between one or more alternative patterns. For example, abc|d..f will match either abc, or deaf. Note that the precedent for | is low, so that abc|xyz matches abc or xyz not abcyz or abxyz. Like with mathematical expressions, if presidents ever get confusing, use parentheses to make it clear what you want:

str_view(c("grey", "gray"), "gr(e|a)y")

3.5 Exercices

Create regular expressions to find all words that:

• Start with a vowel.
• That only contains consonants. (Hint: thinking about matching “not”-vowels.)
• End with ed, but not with eed.
• End with ing or ise.

3.6 Repetition

Now that you know how to search for groups of characters you can define the number of times you want to see them.

• ?: 0 or 1
• +: 1 or more
• *: 0 or more

x <- "1888 is the longest year in Roman numerals: MDCCCLXXXVIII"
str_view(x, "CC?")

str_view(x, "CC+")

str_view(x, 'C[LX]+')

You can also specify the number of matches precisely:

• {n}: exactly n
• {n,}: n or more
• {,m}: at most m
• {n,m}: between n and m

str_view(x, "C{2}")

str_view(x, "C{2,}")

str_view(x, "C{2,3}")

3.7 Exercices

• Describe in words what these regular expressions match: (read carefully to see if I’m using a regular expression or a string that defines a regular expression.)
  • ^.*$
  • "\\{.+\\}"
  • \d{4}-\d{2}-\d{2}
  • "\\\\{4}"
• Create regular expressions to find all words that:
  • Start with three consonants.
  • Have three or more vowels in a row.
  • Have two or more vowel-consonant pairs in a row.

3.8 Grouping

You learned about parentheses as a way to disambiguate complex expressions. Parentheses also create a numbered capturing group (number 1, 2 etc.). A capturing group stores the part of the string matched by the part of the regular expression inside the parentheses. You can refer to the same text as previously matched by a capturing group with back references, like \1, \2 etc.

str_view(fruit, "(..)\\1", match = TRUE)

3.9 Exercices

• Describe, in words, what these expressions will match:
  • "(.)\1\1"
  • "(.)(.)\\2\\1"
  • "(..)\1"
  • "(.).\\1.\\1"
  • "(.)(.)(.).*\\3\\2\\1"
• Construct regular expressions to match words that:
  • Start and end with the same character.
  • Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.)
  • Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)

3.10 Detect matches

x <- c("apple", "banana", "pear")
str_detect(x, "e")

[1]  TRUE FALSE  TRUE

How many common words start with t?

sum(str_detect(words, "^t"))

[1] 65

What proportion of common words ends with a vowel?

mean(str_detect(words, "[aeiou]$"))

[1] 0.2765306

3.11 Combining detection

Find all words containing at least one vowel, and negate

no_vowels_1 <- !str_detect(words, "[aeiou]")

Find all words consisting only of consonants (non-vowels)

no_vowels_2 <- str_detect(words, "^[^aeiou]+$")
identical(no_vowels_1, no_vowels_2)

[1] TRUE

3.12 With tibble

df <- tibble(
  word = words, 
  i = seq_along(word)
)
df %>% 
  filter(str_detect(word, "x$"))

# A tibble: 4 × 2
  word      i
  <chr> <int>
1 box     108
2 sex     747
3 six     772
4 tax     841

3.13 Extract matches

head(sentences)

[1] "The birch canoe slid on the smooth planks." 
[2] "Glue the sheet to the dark blue background."
[3] "It's easy to tell the depth of a well."     
[4] "These days a chicken leg is a rare dish."   
[5] "Rice is often served in round bowls."       
[6] "The juice of lemons makes fine punch."      

We want to find all sentences that contain a colour. We first create a vector of colour names, and then turn it into a single regular expression:

colours <- c("red", "orange", "yellow", "green", "blue", "purple")
colour_match <- str_c(colours, collapse = "|")
colour_match

[1] "red|orange|yellow|green|blue|purple"

3.14 Extract matches

We can select the sentences that contain a colour, and then extract the colour to figure out which one it is:

has_colour <- str_subset(sentences, colour_match)
matches <- str_extract(has_colour, colour_match)
head(matches)

[1] "blue" "blue" "red"  "red"  "red"  "blue"

3.15 Grouped matches

Imagine we want to extract nouns from the sentences. As a heuristic, we’ll look for any word that comes after “a” or “the”.

noun <- "(a|the) ([^ ]+)"
has_noun <- sentences %>%
  str_subset(noun) %>%
  head(10)
has_noun %>% 
  str_extract(noun)

 [1] "the smooth" "the sheet"  "the depth"  "a chicken"  "the parked"
 [6] "the sun"    "the huge"   "the ball"   "the woman"  "a helps"   

str_extract() gives us the complete match; str_match() gives each individual component.

has_noun %>% 
  str_match(noun)

      [,1]         [,2]  [,3]     
 [1,] "the smooth" "the" "smooth" 
 [2,] "the sheet"  "the" "sheet"  
 [3,] "the depth"  "the" "depth"  
 [4,] "a chicken"  "a"   "chicken"
 [5,] "the parked" "the" "parked" 
 [6,] "the sun"    "the" "sun"    
 [7,] "the huge"   "the" "huge"   
 [8,] "the ball"   "the" "ball"   
 [9,] "the woman"  "the" "woman"  
[10,] "a helps"    "a"   "helps"  

3.16 Exercises

• Find all words that come after a number like one, two, three etc. Pull out both the number and the word.

3.17 Replacing matches

Instead of replacing with a fixed string, you can use back references to insert components of the match. In the following code, I flip the order of the second and third words.

sentences %>% 
  str_replace("([^ ]+) ([^ ]+) ([^ ]+)", "\\1 \\3 \\2") %>%
  head(5)

[1] "The canoe birch slid on the smooth planks." 
[2] "Glue sheet the to the dark blue background."
[3] "It's to easy tell the depth of a well."     
[4] "These a days chicken leg is a rare dish."   
[5] "Rice often is served in round bowls."       

3.18 Exercices

• Replace all forward slashes in a string with backslashes.
• Implement a simple version of str_to_lower() using replace_all().
• Switch the first and last letters in words. Which of those strings are still words?

3.19 Splitting

sentences %>%
  head(5) %>% 
  str_split("\\s")

[[1]]
[1] "The"     "birch"   "canoe"   "slid"    "on"      "the"     "smooth" 
[8] "planks."

[[2]]
[1] "Glue"        "the"         "sheet"       "to"          "the"        
[6] "dark"        "blue"        "background."

[[3]]
[1] "It's"  "easy"  "to"    "tell"  "the"   "depth" "of"    "a"     "well."

[[4]]
[1] "These"   "days"    "a"       "chicken" "leg"     "is"      "a"      
[8] "rare"    "dish."  

[[5]]
[1] "Rice"   "is"     "often"  "served" "in"     "round"  "bowls."

3.20 See you in R.8: Factors