library(readr)
library(tidyverse)
library(RCurl)

Question 1

Using the 173 majors listed in fivethirtyeight.com’s College Majors dataset [https://fivethirtyeight.com/features/the-economic-guide-to-picking-a-college-major/], provide code that identifies the majors that contain either “DATA” or “STATISTICS”.

x <- getURL("https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv")
y <- read.csv(text = x)
str_subset(y$Major, "DATA") 
## [1] "COMPUTER PROGRAMMING AND DATA PROCESSING"
str_subset(y$Major, "STATISTICS")
## [1] "MANAGEMENT INFORMATION SYSTEMS AND STATISTICS"
## [2] "STATISTICS AND DECISION SCIENCE"

Question 2

Write code that transforms the data below: [1] “bell pepper” “bilberry” “blackberry” “blood orange” [5] “blueberry” “cantaloupe” “chili pepper” “cloudberry”
[9] “elderberry” “lime” “lychee” “mulberry”
[13] “olive” “salal berry”

Into a format like this: c(“bell pepper”, “bilberry”, “blackberry”, “blood orange”, “blueberry”, “cantaloupe”, “chili pepper”, “cloudberry”, “elderberry”, “lime”, “lychee”, “mulberry”, “olive”, “salal berry”)

strStart = '[1] "bell pepper" "bilberry"   "blackberry"  "blood orange"
[5] "blueberry"  "cantaloupe"  "chili pepper" "cloudberry"
[9] "elderberry"  "lime"     "lychee"    "mulberry"
[13] "olive"    "salal berry"'

temp <- str_replace_all(strStart, "[:digit:]", " ")
temp <- str_replace_all(temp, "\\[.*\\]", " ")
temp <- str_replace_all(temp, "\\n", " ")
temp <- str_trim(temp, side="both")
temp <- str_split_fixed(temp, '\\"' , n=Inf)
s <- c()
for (i in 1:length(temp)){
  if (str_detect(temp[i], "[A-Za-z]")){
    s <- append(s, temp[i])
  }
}
print(s) # As a vector
##  [1] "bell pepper"  "bilberry"     "blackberry"   "blood orange" "blueberry"   
##  [6] "cantaloupe"   "chili pepper" "cloudberry"   "elderberry"   "lime"        
## [11] "lychee"       "mulberry"     "olive"        "salal berry"
# From here I looked up how to make the vector print comma separated
# https://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector
cat("c(", paste(shQuote(s, type="cmd"), collapse=", "), ")") # Formatted
## c( "bell pepper", "bilberry", "blackberry", "blood orange", "blueberry", "cantaloupe", "chili pepper", "cloudberry", "elderberry", "lime", "lychee", "mulberry", "olive", "salal berry" )

Question 3

Describe, in words, what these expressions will match:

  1. (.)\1\1

The () indicates a group, the . is a meta-character where the . will match any character. So (.) is group 1 and will match any character. The \1 is a backreference, referencing group 1. So \1\1 indicates the same text matched by the first group should be matched again, twice, because there are two \1s. The full (.)\1\1 indicates the same character should appear three times in a row to be matched.

ts <- c("abc", "aabc", "aaabbc", "123", "12223")
str_subset(ts, "(.)\\1\\1")
## [1] "aaabbc" "12223"
  1. “(.)(.)\2\1”

The (.)(.) indicates two groups, both containing all characters. The \\2 is a backreference to group 2 and \\1 is a backreference to group 1. The full (.)(.)\\2\\1 indicates any character (say x) followed by another character (say y), followed by y followed by x should match. If x=y then any 4 of the same characters in a row will match.

ts <- c("aaaaa", "abba", "aaba", "abbbbb", "xyyx", "111222333")
str_subset(ts, "(.)(.)\\2\\1")
## [1] "aaaaa"  "abba"   "abbbbb" "xyyx"
  1. (..)\1

The (..) indicates group 1 is any two characters, the \1 indicates those same two characters should be repeated for a match. So for example abab would match.

ts <- c("aaaaa", "abab", "aaba", "abbbbb", "12213", "111222333")
str_subset(ts, "(..)\\1")
## [1] "aaaaa"  "abab"   "abbbbb"
  1. “(.).\1.\1”

The (.) indicates group 1 is any character, the . is any character, the \\1 refers to group 1. So in this case for a match the character in group 1, must be followed by a character, followed by group 1s character, followed by any character, followed by group 1s character again.

ts <- c("aaaaa", "abaca", "aaba", "12131", "12213", "111222333")
str_subset(ts, "(.).\\1.\\1")
## [1] "aaaaa" "abaca" "12131"
  1. “(.)(.)(.).*\3\2\1”

The three (.) indicates three groups of any character. The . indicates any character. The * is another meta character letting a pattern be optional or repeat any number of times. So the .* indicates a character repeated any number of times, including zero times is an option. The \\3\\2\\1 means that group 3, then group 2, then group 1 should be repeated to complete the match. So to get a match from (.)(.)(.).*\\3\\2\\1 the following must be true: character 1, then character 2, then character 3, then any character or no character, then character 3, then character 2, then character 1.

ts <- c("abczzzzzcba", "aaaaaa", "abcdefg", "123321", "12213", "111222333")
str_subset(ts, "(.)(.)(.).*\\3\\2\\1")
## [1] "abczzzzzcba" "aaaaaa"      "123321"

Question 4

Construct regular expressions to match words that:

  1. Start and end with the same character.
ts <- c("Hannah", "hello", "kayak", "aa", "121", "111222333")
str_subset(ts, regex("^(.).*\\1$", ignore_case = T))
## [1] "Hannah" "kayak"  "aa"     "121"
  1. Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.)
ts <- c("Church", "abab", "abc", "123321", "1212", "111222333")
str_subset(ts, regex("([A-Za-z][A-Za-z]).*\\1", ignore_case = T))
## [1] "Church" "abab"
  1. Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)
ts <- c("Eleven", "aaaaaa", "abcdefg", "12121", "12213", "111222333")
str_subset(ts, regex("([A-Za-z]).*\\1.*\\1", ignore_case = T))
## [1] "Eleven" "aaaaaa"