Chapter 14 - Strings
Load the libraries needed for these exercises.
library(tidyverse)14.2 - String basics
Problem 1
In code that doesn’t use stringr, you’ll often see paste() and paste0(). What’s the difference between the two functions? What stringr function are they equivalent to? How do the functions differ in their handling of NA?
paste() automatically includes a space between each character string it combines. paste0() does not include a space. They are ~equivalent to str_c() from library(stringr). paste() and paste0() include NA as text. str_c() returns an NA for the entire string if the string contains an NA.
Problem 2
In your own words, describe the difference between the sep and collapse arguments to str_c().
sep is a character string to insert between input vectors. Its input vector and output vector always have the same length.
length(str_c("Letter", letters, sep = ": "))## [1] 26collapse is a character string to insert between input vectors and to turn the vector into a single string. collapse always returns a vector with length one.
length(str_c("Letter", letters, collapse = ": "))## [1] 1Problem 3
Use str_length() and str_sub() to extract the middle character from a string. What will you do if the string has an even number of characters.
string_middle <- function(string) {
string_length <- str_length(string)
if (string_length %% 2 == 1) {
str_sub(string, floor((string_length + 1) / 2), ceiling((string_length) / 2))
}
else if (string_length %% 2 == 0) {
NULL
}
else {"Error!"}
}
string_middle("abc")## [1] "b"string_middle("abcd")## NULLIt returned a NULL if string_length() is even.
Problem 4
What does str_wrap() do? When might you want to use it?
It implements the Knuth-Plass paragraph wrapping algorithm. It “breaks text paragraphs into lines, of total width - if it is possible - of at most given width.
graph <- "It implements the Knuth-Plass paragraph wrapping algorithm. It breaks text paragraphs into lines, of total width - if it is possible - of at most given width."
str_wrap(graph, width = 20)## [1] "It implements\nthe Knuth-Plass\nparagraph wrapping\nalgorithm. It breaks\ntext paragraphs\ninto lines, of total\nwidth - if it is\npossible - of at\nmost given width."This could be useful for formatting in html and rmarkdown. Especially for graphics and sidebars. Custom width is useful - especially in reproducible documents.
Problem 5
What does str_trim() do? What’s the opposite of str_trim()?
It trims whitespace from the left, right, or both sides of a character string. It is the string version of trimws().
str_pad() is the opposite of str_trim(). It adds whitespace to the left, right, or both sides of a character string.
Problem 6
Write a function that turns (e.g.) a vector c("a", "b", "c") into a string a, b, and c. Think carefully about what it should do if given a vector of length 0, 1, or 2.
list_maker <- function(string) {
if (length(string) > 1) {
stringa <- string[1:length(string)-1]
stringb <- string[length(string)]
stringa <- str_c(stringa, collapse = ", ")
str_c(stringa, ", and ", stringb, collapse = "")
} else {
string
}
}
string <- c("a", "b", "c", "d", "e")
list_maker(string)## [1] "a, b, c, d, and e"14.3 - Matching patterns with regular expressions {-}
Problem 1
Explain why each of these strings don’t match a : “",”\“,”\".
- “" escapes the quotation mark and isn’t a valid character string in R.
- “\” returns a character string with two backslashes which doesn’t match one backslash.
- “\" escapes the quotation mark and isn’t a valid character string.
Problem 2
How would you match the sequence “’?
Problem 3
What patterns will the regular expression ...... match? How would you represent it as a string?
It will match a string of three periods separated by characters. \\..\\..\\...
str_view(".a.b.c", "\\..\\..\\..")14.3.2 - Anchors
Problem 1
How would you match the literal string “\(^\)”?
x <- "$^$"
str_view(x, "\\$\\^\\$")Problem 2
Given the corpus of common words in stringr::words, create regular expressions that will find all words that:
- Start with “y”
str_view(words, "^y", match = TRUE)- End with “x”
table(str_detect(words, "x$"))##
## FALSE TRUE
## 976 4- Are exactly three letters long. (Don’t cheat by using
str_length()!)
table(str_detect(words, "^...$"))##
## FALSE TRUE
## 870 110- Have seven letters or more.
table(str_detect(words, "^......."))##
## FALSE TRUE
## 761 21914.3.2 - Character classes and alternatives
Problem 1
Create regular expressions to find all words that:
- Start with a vowel
str_view(words[1:10], "^[aeiou]", match = TRUE)- Only contain consonants. (Hint: think about match “not”-vowels.)
str_view(words, "^[^aeiou]+$", match = TRUE)I’m not sure if this can be done with + which is introduced on page 204 after the exercises.
- End with ed, but not with eed.
str_view(words, "[^e]ed$", match = TRUE)- End with ing or ize.
str_view(words, "ing$|ize$", match = TRUE)Problem 2
Empirically verify the rule “i before e except after c.”
Let’s try this with proof by contradiction. We need to look for two conditions:
- ie after c
- ei
str_view(words, "ei|[c]ie", match = TRUE)Six words violate the rules. “i before e except after c” is and always will be rubbish.
Problem 3
Is “q” always followed by a “u”?
Proof by contradiction: look for a “q” not followed by a “u”.
str_view(words, "q^[u]", match = TRUE)Yes, “q” is always followed by a “u” in this data set.
Problem 4
Write a regular expression that matches a word if it’s probably written in British English, not American English.
str_view(words, "our|ise|ogue", match = TRUE)Problem 5
Create a regular expression that will match telephone numbers as commonly written in your country.
phone <- c("212-555-7891", "(212)-555-7891")
str_view(phone, "\\d\\d\\d-\\d\\d\\d-\\d\\d\\d\\d|\\(\\d\\d\\d\\)-\\d\\d\\d-\\d\\d\\d\\d", match = TRUE)14.3.4 - Repetition
Problem 1
Describe the equivalents of ?, +, and * in {n, m} form.
? == {1}+ == {1,}* == {0,}
Problem 2
Describe in words what these regular expressions match (read carefully to if I’m using a regular expression or a string that defines a regular expressions):
^.*$matches an entire string.^matches the start of a string..is any character which is repeated 0 or more times with*.$matches the end of a string."\\{.+\\}"3.\d{4}-\d{2}-\d{2}matches exactly 4 digits followed by a dash followed by exactly two digits followed by a dash followed by exactly two digits. This is the same as the ISO8601 date international standard.\\\\{4}matches exactly four backslashes.
Problem 3
Create regular expressions to find all words that:
- Start with three consonants.
string <- c("scratch", "apple")
str_view(string, "^[^aeiou]{3}")- Have three or more vowels in a row.
string <- c("scratch", "aaapple")
str_view(string, "^[aeiou]{3,}")- Have two or more vowel consonant pairs in a row.
string <- c("banana", "coconut")
str_view(string, "([aeiou][^aeiou]){2,}")Problem 4
Solve the beginner regexp crosswords at http://regexcrossword.com/challenges/beginner
14.3.5 - Grouping and backreferences
Problem 1
Describe in words what these expressions will match:
(.)\1\1will match any string of three repeated letters or symbols."(.)(.)\\2\\1"will match a four letter palindrome (spelled the same forwards and backwards).(..)\1will match a four letter string where the second half is a reptition of the first half."(.).\\1.\\1"will match and repetition of the same character three times where each character is spearated by a character (ex. “ababa” and “&&&&&”)."(.)(.)(.)*\\3\\2\\1"will match a string of characters where the first three characters are repeted in reverse and the middle character can be repeated multiple times (ex. “abccba” and “abcccccba”).
Problem 2
0.0.0.0.1 2. Construct regular expressions to match words that:
- Start and end with the same character.
"^(.).*\\1$" - Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice)
".*(..).*\\1.*" - Contain one letter repeated in at least three places (e.g., “eleven” contains three “e”s).
".*(.).*\\1.*\\1.*"
14.4 - Tools
Problem 1
For each of the following challenges, try solving it by using both a singular regular expression, and a combination of multiple str_detect() calls: 1. Find all words that start of end with x. str_detect(words, "^x.*x$") & str_detect(str_detect(words, "^x"), "x$") 2. Find all words that start with a vowel and end with a consonant. str_detect(words, "^[aeiou].*[^aeiou]$") & str_detect(str_detect(words, "^[aeiou]"), "[^aeiou]$") 3. Are there any words that contain at least one of each different vowel? TODO(aaron): hmm? 4. What word has the highest number of vowels? What word has the highest proportion of vowels? (Hint: what is the denominator)
as_tibble(words) %>%
mutate(vowels = str_count(value, "[aeiou]")) %>%
filter(vowels == max(vowels))## # A tibble: 8 x 2
## value vowels
## <chr> <int>
## 1 appropriate 5
## 2 associate 5
## 3 available 5
## 4 colleague 5
## 5 encourage 5
## 6 experience 5
## 7 individual 5
## 8 television 5as_tibble(words) %>%
mutate(letters = str_count(value),
vowels = str_count(value, "[aeiou]"),
proportion = vowels / letters) %>%
filter(proportion == max(proportion))## # A tibble: 1 x 4
## value letters vowels proportion
## <chr> <int> <int> <dbl>
## 1 a 1 1 114.4.3 - Extract Matches
Problem 1
In the previous example, you might have noticed that the regular expression matched “flickered”, which is not a color. Modify the regex to fix the problem.
colors <- "\\b(red|orange|yellow|green|blue|purple)\\b"
more <- sentences[str_count(sentences, colors) > 1]
str_view_all(more, colors)Problem 2
From the Harvard sentences data, extract: 1. The first word from each sentence. str_extract(sentences, "[^\\s]*") 2. All words ending in ing. str_extract_all(sentences, "\\b[^\\s]*ing\\b") 3. All plurals. TODO(aaron): hmm?
14.4.4 - Grouped matches
Problem 1
Find all words that come after a “number” like “one”, “two”, “three”, etc. Pull out the number and the word.
numbers <- "([Oo]ne|[Tt]wo|[Tt]hree|[Ff]our|[Ff]ive|[Ss]ix|[Ss]even|[Ee]ight|[Nn]ine|[Tt]en) ([^ ]+)"
tibble(sentence = sentences) %>%
extract(
sentence, c("number", "word"), numbers,
remove = FALSE
) %>%
filter(!is.na(number))## # A tibble: 46 x 3
## sentence number word
## <chr> <chr> <chr>
## 1 Rice is often served in round bowls. ten served
## 2 Four hours of steady work faced us. Four hours
## 3 Two blue fish swam in the tank. Two blue
## 4 Lift the square stone over the fence. one over
## 5 The rope will bind the seven books at once. seven books
## 6 The two met while playing on the sand. two met
## 7 There are more than two factors here. two factors
## 8 He lay prone and hardly moved a limb. one and
## 9 Ten pins were set in order. Ten pins
## 10 Type out three lists of orders. three lists
## # ... with 36 more rowsProblem 2
Find all contractions. Separate out the pieces before and after the apostrophe.
"[^ ]*'[^ ]*" could be used, but it returns possessive nouns. The following string of regular expressions gets around this problem.
contractions <- "[^ ]*'m|[^ ]*n't|[^ ]*'ve|[^ ]*'d|[^ ]*'re|[^ ]*'ll|[Ll]et's|[Ss]he's|[Hh]e's"
tibble(sentence = sentences) %>%
mutate(contraction = str_extract(sentences, contractions)) %>%
filter(!is.na(contraction)) %>%
extract(contraction, c("before", "apostrophe", "after"), "(.*)(')(.*)")## # A tibble: 4 x 4
## sentence before apostrophe after
## <chr> <chr> <chr> <chr>
## 1 Open the crate but don't break the glass. don ' t
## 2 Let's all join as we sing the last chorus. Let ' s
## 3 We don't get much money but we have fun. don ' t
## 4 We don't like to admit our small faults. don ' t14.4.5 - Replacing matches
Problem 1
Replace all forward slashes in a string with backslashes. str_replace_all("a/b/c", "/", "\\\\")
Problem 2
Implement a simple version of str_to_lower() using str_replace_all(). str_replace_all("AbC", "[A-Z]", tolower)
Problem 3
Switch the first and last letters in words. Which of those strings are still words?
new.words <- str_replace(words, "(^.)(.*)(.$)", "\\3\\2\\1")
words[new.words %in% words]## [1] "a" "america" "area" "dad" "dead"
## [6] "deal" "dear" "depend" "dog" "educate"
## [11] "else" "encourage" "engine" "europe" "evidence"
## [16] "example" "excuse" "exercise" "expense" "experience"
## [21] "eye" "god" "health" "high" "knock"
## [26] "lead" "level" "local" "nation" "no"
## [31] "non" "on" "rather" "read" "refer"
## [36] "remember" "serious" "stairs" "test" "tonight"
## [41] "transport" "treat" "trust" "window" "yesterday"14.4.6 - Splitting
Problem 1
Split up a string like “apples, pears, and bananas” into individual components.
str_split("apples, pears, and bananas", boundary("word"))
Problem 2
Why is it better to split up by boundary(“word”) than " “?
" " captures non-words like the space after the period while boundary("word") only captures words.
Problem 3
What does splitting with an empty string (“”) do? Experiment and read the documentation.
“An empty pattern,”“, is equivalent to boundary(”character“).”
14.5 - Other types of pattern
Problem 1
How would you find all strings containing “" with regex() versus fixed. regex("\\\\") & fixed("\")
Problem 2
What are the five most common words in setences?
The five most common words are “the”, “a”, “of”, “to”, and “and”.
str_split(sentences, boundary("word")) %>%
flatten_chr() %>%
str_to_lower() %>%
as_tibble() %>%
group_by(value) %>%
count() %>%
arrange(-n) %>%
ungroup() %>%
top_n(5)## Selecting by n## # A tibble: 5 x 2
## value n
## <chr> <int>
## 1 the 751
## 2 a 202
## 3 of 132
## 4 to 123
## 5 and 11814.6 - Other uses of regular expressions
Problem 1
Find the stringi function that: 1. Count the number of words stri_count_words 2. Find duplicated strings. stri_duplicated() 3. Generate random text. stri_rand_strings()
Problem 2
How do you control the language that str_sort() uses for sorting? With the locale = argument in the opts_collator argument.