Assignment 3 - R Character Manipulation

Please deliver links to an R Markdown file (in GitHub and rpubs.com) with solutions to the problems below. You may work in a small group, but please submit separately with names of all group participants in your submission.

#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”

majors_df <- read.csv("https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv")

filt_majors <- majors_df %>% filter(grepl("DATA | STATISTICS", Major))

filt_majors

##   FOD1P                                         Major          Major_Category
## 1  6212 MANAGEMENT INFORMATION SYSTEMS AND STATISTICS                Business
## 2  2101      COMPUTER PROGRAMMING AND DATA PROCESSING Computers & Mathematics

#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”)

goal_vector <- c("bell pepper", "bilberry", "blackberry", "blood orange", "blueberry", "cantaloupe", "chili pepper", "cloudberry", "elderberry", "lime", "lychee", "mulberry", "olive", "salal berry")

sample_string <- '[1] "bell pepper"  "bilberry"     "blackberry"   "blood orange" 

[5] "blueberry"    "cantaloupe"   "chili pepper"  "cloudberry"  

[9] "elderberry"   "lime"         "lychee"       "mulberry"    

[13] "olive"        "salal berry"'

sample_string

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

setdiff(sample_string, goal_vector)

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

# read in everything into a list of 1

sample_string <- sample_string %>% str_split("\\\\")

# replace all non characters with blanks 
only_chars <- '([^a-z" ])'

sample_string <- sample_string %>% str_replace_all(only_chars, "")

sample_string 

## [1] " \"bell pepper\"  \"bilberry\"     \"blackberry\"   \"blood orange\"  \"blueberry\"    \"cantaloupe\"   \"chili pepper\"  \"cloudberry\"   \"elderberry\"   \"lime\"         \"lychee\"       \"mulberry\"     \"olive\"        \"salal berry\""

# trim whitespace

sample_string <- sample_string %>% str_trim()

# remove escaped quotes
sample_string <- sample_string %>% str_replace_all('(["])', "")

# expand string with multiple spaces
sample_string <- sample_string %>% str_split("  ")

#change from list into character vector
sample_string <- sample_string[[1]]

#remove empty characters
sample_string <- sample_string[sample_string != ""]
#trim again
sample_string <- sample_string %>% str_trim()
sample_string

##  [1] "bell pepper"  "bilberry"     "blackberry"   "blood orange" "blueberry"   
##  [6] "cantaloupe"   "chili pepper" "cloudberry"   "elderberry"   "lime"        
## [11] "lychee"       "mulberry"     "olive"        "salal berry"

setdiff(sample_string,goal_vector)

## character(0)

The two exercises below are taken from R for Data Science, 14.3.5.1 in the on-line version:

#3 Describe, in words, what these expressions will match:

• (.)\1\1
  • While this would not be valid regex in R - in theory this would capture any character except line breaks (.) and then \1 is a back-reference that would fit the first capture group and since it is repeated, it would be match the first capture group twice. So something like aaa would match. However, since the backslashes are not escaped (“\\1” and then “\\1”), it would not match anything.
• “(.)(.)\\2\\1”
  • This regular expression has two capture groups that match any character and then back-references that refer to the second capture group and then the first capture group. Thus it would match anything that follows a pattern of abba where the second character repeats and the first character ends the four character phrase
• (..)\1
  • This regex has a capture group that matches any two character and then a back-reference to the capture group of the same set. So this would match the pattern abab. However, like the first example, this is not properly formed regex in R and it would fail to actually match the expected pattern.
• “(.).\\1.\\1”
  • There is a capture group matching any character followed by any character, a back-reference to the capture group, any character, and then another back-reference to the first capture group. This means the regex would match a pattern like azaxa or anything where the first character follows any single character consecutively bybib or ;z;w;
• "(.)(.)(.).*\\3\\2\\1"
  • First, you start off capturing a single character in a row. Then you match on any character 0 or more times until it ends with the third capture group then the second capture group and lastly the first captured character. A simple example would be abc anythinginhere cba

#4 Construct regular expressions to match words that:

• Start and end with the same character.
  • "\b([A-Za-z])[A-Za-z]*\1\b"
  • Start with a \\b word boundary and then capture any a-z or A-Z character. Then you can have 0 to any characters from a-z or A-Z until it ends with the first character that was captured

sentence <- "This sentence has all the types of words that we are looking for like elevate or church or eleven"
pattern <- "\\b([A-Za-z])[A-Za-z]*\\1\\b"

str_extract_all(string = sentence, pattern = pattern)

## [[1]]
## [1] "that"    "elevate"

• Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.)
  • "\b(..)[A-Za-z]*\1\b"
  • \b is a word boundary to ensure you match on a word, then a two character capture group followed by any number of characters a-z case insensitive and then repeated by the two character capture group

sentence <- "This sentence has all the types of words that we are looking for like elevate or church or eleven"
pattern <- "\\b(..)[A-Za-z]*\\1\\b"

str_extract_all(string = sentence, pattern = pattern)

## [[1]]
## [1] "church"

• Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)
  • “\w(\w)\w\1\w\1\w”
  • Start with \\w* to look for any word character as many times and then capture a word character. Afterwords, allow any word character and look for the captured word 2 other times then match if the criteria is met.

sentence <- "This sentence has all the types of words that we are looking for like elevate or church or eleven"
pattern <- "\\w*(\\w)\\w*\\1\\w*\\1\\w*"

str_extract_all(string = sentence, pattern = pattern)

## [[1]]
## [1] "sentence" "elevate"  "eleven"