library(stringr)    # more consistent than base r



  1. Using the 173 majors listed in fivethirtyeight.com’s College Majors dataset, provide code that identifies the majors that contain either “DATA” or “STATISTICS”



# note: github wraps data in formatting, to get raw data link, click the "Raw" button at the top-left of the data

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

str_view(majors$Major, "DATA|STATISTICS", match=TRUE)



  1. 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”)



The question asks to convert raw data into R code



vector_data<-c("bell pepper", "bilberry", "blackberry", "blood orange", "blueberry", "cantaloupe", "chili pepper", "cloudberry", "elderberry", "lime", "lychee", "mulberry", "olive", "salal berry")
dbl_quot<-'\"'
double_quote <- "\"" 
double_quote<-'"'
data_string<-str_c(double_quote,vector_data,double_quote,collapse = ", ")
# data_string<-str_c(vector_data, collapse = ", ") 
code_string<-str_c("c(",data_string,")")
writeLines(code_string)            # writeLines omits the escape characters
## c("bell pepper", "bilberry", "blackberry", "blood orange", "blueberry", "cantaloupe", "chili pepper", "cloudberry", "elderberry", "lime", "lychee", "mulberry", "olive", "salal berry")



  1. Describe, in words, what these expressions will match:



(.)\1\1



Whats inside the parenthesis is a “captured” match. In this case the dot means any character,
The back reference \1 then takes that match and duplicates a search of the next character only.
i.e. (.)\1\1 will macth any 3 consective characters. 

str<-c("BBAAAX", "bBaAAxx", "zzza", "aaaa","...","abaaaaaaax")

str_view(str,"(.)\\1\\1")



(.)(.)\2\1



\2 will duplicate a search of whats inside the second parenthesis,
thus \2\1 effectively will captured match #2 then captured match #1, 

str<-c("BAAX", "bBaAAxx", "zzza", "abba","...","aaaax")
str_view(str,"(.)(.)\\2\\1")



(..)\1

Any 2 characters followed the same 2 characters 

str<-c("BABAX", "bBaAAxx", "zzza", "abba","...","axaaaxa")
str_view(str,"(..)\1")



(.).\1.\1

Any 2 characters followed the first character followed by any character than the first character again. 

str<-c("BABAX", "bBaAAxx", "Z4Z5Z6za", "abba","...","XaXxXaaaxa")
str_view(str,"(.)(.)\\2\\1")



"(.)(.)(.).*\3\2\1" Any 3 characters followed by any number of characters we dont care about, followed by
the first 3 characters reversed. Asterix means 0 or more, so the “any number of”
characters in the middle could be zero." 

str<-c("BAZAZZA", "OBAZZABOO", "Z4Z5Z6za", "abbba","...","XaXxXaaaXaXxa")
str_view(str,"(.)(.)(.).*\\3\\2\\1")



  1. Construct regular expressions to match words that:

    Start and end with the same character. 

str<-c("BAZAZZA", "OBAZZABOO", "Z4Z5Z6za", "abbba","...","XaXxXaaaXaXxa")
str_view(str,"^(.).*\\1$")



Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.) 

str<-c("BBAAABBX", "bBaAAxx", "zzza", "aaaa","church")

str_view(str,"(..).*\\1")



Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)



str<-c("BANANA", "bBaAAxx", "zzza", "aaNOa","eleven","church")

str_view(str,"(.).*\\1.*\\1")