#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”
#load in data into a dataframe and parse columns
urlfile <- 'https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv'
majors <- read_csv(url(urlfile))## Rows: 174 Columns: 3
## -- Column specification --------------------------------------------------------
## Delimiter: ","
## chr (3): FOD1P, Major, Major_Category
##
## i Use `spec()` to retrieve the full column specification for this data.
## i Specify the column types or set `show_col_types = FALSE` to quiet this message.#Filter dataset
df1 <- majors |>
filter(str_detect(majors$Major, regex("DATA|STATISTICS")))
print(df1)## # A tibble: 3 x 3
## FOD1P Major Major_Category
## <chr> <chr> <chr>
## 1 6212 MANAGEMENT INFORMATION SYSTEMS AND STATISTICS Business
## 2 2101 COMPUTER PROGRAMMING AND DATA PROCESSING Computers & Mathematics
## 3 3702 STATISTICS AND DECISION SCIENCE 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”)
#load in data into a data frame and parse columns
var_str <- '[1] "bell pepper" "bilberry" "blackberry" "blood orange"
[5] "blueberry" "cantaloupe" "chili pepper" "cloudberry"
[9] "elderberry" "lime" "lychee" "mulberry"
[13] "olive" "salal berry"'
#Remove all white spaces
var_str <- str_replace_all(var_str,'((?![a-z])\\s+(?![a-z]))', '') #remove all whitespace
#Remove all brackets and all digits inside
var_str <- str_replace_all(var_str,'(\\[\\d*\\])', '') #replace all brackets along w/ all digits inside
#Remove all "\" after a fruit name
var_str <- str_replace_all(var_str, '[a-z]+(?<=(.))\\)', "") #remove all backslashes
print(var_str)## [1] "\"bell pepper\"\"bilberry\"\"blackberry\"\"blood orange\"\"blueberry\"\"cantaloupe\"\"chili pepper\"\"cloudberry\"\"elderberry\"\"lime\"\"lychee\"\"mulberry\"\"olive\"\"salal berry\""#3 Describe, in words, what these expressions will match:
(.)\1 =====================> Looks for any character followed by two of the same.Any character that repeats 3 times. Cannot be new line.
“(.)(.)\2\1”==============> Looks for something like this: ‘“ab\2\1”’ where a and b can be any character excluding new line.
(..)\1 =====================> Looks for any 2 characters except new line. Continues to look for 2 more within the capture. Ex: abab and grgr.
“(.).\1.\1” ==============> Looks for something like this: “je\1\1” where j and e can be any character excluding new line. The tail needs to be \1(anycharacter)1. string must be enclosed in quotations.
“(.)(.)(.).*\3\2\1” =====> Looks for something like this: “degaaaaaa\3\2\1” where the first four characters can be anything but new line. The fifth character matches the 4th character an unlimited number of times. followed by literal \3\2\1. String must be enclosed in quotations.
#4 Construct regular expressions to match words that:
#Start and end with the same character.
#^([a-z])(([a-z]+\\1$)|\\1?$)
words <- c("car", "nun", "room", "bob", "picture", "racecar", 'whenever', "mama","church", "mississippi", "eleven")
rgx1 ="^([a-z])(([a-z]+\\1$)|\\1?$)"
ss1 <- str_subset(words, rgx1)
ss1## [1] "nun" "bob" "racecar"#Contain a repeated pair of letters (e.g. "church" contains "ch" repeated twice.)
rgx2 ="([A-Za-z][A-Za-z]).*\\1"
ss2 <- str_subset(words, rgx2)
ss2## [1] "mama" "church" "mississippi"#Contain one letter repeated in at least three places (e.g. "eleven" contains three "e"s.)
rgx3 ="([A-Za-z]).*\\1.*\\1"
ss3 <- str_subset(words, rgx3)
print(ss3)## [1] "whenever" "mississippi" "eleven"