regex

Ch. 15

exercises with babynames and regular expressions

Code

library(babynames)
library(tidyverse)
library(scales)
babynames |> 
       mutate(
            first = str_sub(name, 1, 1),
             last = str_sub(name, -1, -1)
         ) |> 
  count(year, first, wt = n, sort = TRUE) |>
  filter(n > 2e+03) |>
      ggplot(aes(x = year, y = n, 
                 color = first, 
                 group = first)) + geom_line()   + 
  scale_y_continuous(labels = label_comma())  + 
  labs(title="Babynames: Distribution of First Letters of First")

Code

babynames |> 
  mutate(
    first = str_sub(name, 1, 1),
    last = str_sub(name, -1, -1)
  ) |> group_by(last, wt = n) |>
  mutate(r = row_number()) |> 
  slice_head(n = 2) |> 
  ggplot(aes(x = year, y = last)) + geom_hex() + 
  labs(title="Babynames: Distribution of First Letters of Last Name Over Time")

What baby name has the most vowels?

Code

babynames |> 
  mutate(
    vowels = str_count(name, "[^aeiouAEIOU]"), 
    .keep = "used") |> 
  arrange(-vowels)

# A tibble: 1,924,665 × 2
   name            vowels
   <chr>            <int>
 1 Christophermich     11
 2 Johnchristopher     11
 3 Christophermich     11
 4 Christopherjohn     11
 5 Christopherjohn     11
 6 Christopherjohn     11
 7 Christophermich     11
 8 Christopherjohn     11
 9 Johnchristopher     11
10 Johnchristopher     11
# ℹ 1,924,655 more rows

What name has the highest proportion of vowels? (Hint: what is the denominator?)

Code

babynames |> 
  mutate(
    vowels = str_count(name, "[^aeiouAEIOU]"), 
    length = str_length(name),
    prop = vowels/length,
    .keep = "used") |> 
  arrange(-vowels, -length, prop)

# A tibble: 1,924,665 × 4
   name             prop vowels length
   <chr>           <dbl>  <int>  <int>
 1 Christophermich 0.733     11     15
 2 Johnchristopher 0.733     11     15
 3 Christophermich 0.733     11     15
 4 Christopherjohn 0.733     11     15
 5 Christopherjohn 0.733     11     15
 6 Christopherjohn 0.733     11     15
 7 Christophermich 0.733     11     15
 8 Christopherjohn 0.733     11     15
 9 Johnchristopher 0.733     11     15
10 Johnchristopher 0.733     11     15
# ℹ 1,924,655 more rows

Replace all forward slashes in “a/b/c/d/e” with backslashes.

Code

s <- str_replace_all("a/b/c/d/e","/","\\\\")
print(s)

[1] "a\\b\\c\\d\\e"

Code

str_view(s)

[1] │ a\b\c\d\e

Puzzle

What happens if you attempt to undo the transformation by replacing all backslashes with forward slashes? (We’ll discuss the problem very soon.)

Code

str_replace_all(s,"\\\\","/")

[1] "a/b/c/d/e"

Implement a simple version of str_to_lower() using str_replace_all().

Code

str_replace_all("JAY ENJOYS ISLANDS OF UBER","[A-Z]",tolower)

[1] "jay enjoys islands of uber"

Create a regular expression that will match telephone numbers.

Code

a <- "423-687-4291"
str_view(a,"[0-9]?+-\\d+-\\d+-\\d+")

How would you match the literal string “’? How about”\(^\)“?

Explain why each of these patterns don’t match a : “",”\“,”\".

Given the corpus of common words in stringr::words, create regular expressions that find all words that:

1. Start with “y”.
2. Don’t start with “y”.
3. End with “x”.
4. Are exactly three letters long. (Don’t cheat by using str_length()!)
5. Have seven letters or more.
6. Contain a vowel-consonant pair.
7. Contain at least two vowel-consonant pairs in a row.
8. Only consist of repeated vowel-consonant pairs.
9. Write a function that mixes up words into a bad poem.

Code

# so silly
w <- tibble(words)

a <- {}
l <- function(.x, .y){
  for (i in c(1:.y)){
  a[i] <- slice_sample(w,n=.x) |> pull() |> str_flatten_comma() |>
    str_replace_all(","," ") 
  a[i] <- str_c(a[i],"  \\n")
  }
  return(str_view(a))
}