Week 3 Assignment2

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”

There are 3 majors that contain either “DATA” or “STATISTICS”: “MANAGEMENT INFORMATION SYSTEMS AND STATISTICS”, “COMPUTER PROGRAMMING AND DATA PROCESSING”, and “STATISTICS AND DECISION SCIENCE”

df = read.csv(file = "https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv")
print(grep(('data|statistics'), df$Major,value=TRUE, ignore.case = TRUE))

## [1] "MANAGEMENT INFORMATION SYSTEMS AND STATISTICS"
## [2] "COMPUTER PROGRAMMING AND DATA PROCESSING"     
## [3] "STATISTICS AND DECISION SCIENCE"

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”)

#My attempt before I saw the message on slack. Converted it to a factor vector to resemble what the first format looked like and then converted it to a character vector.

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

fruits_factor <- factor(fruits)

factor_to_char <- function(x) {
  paste("c(", paste(paste("\"", as.character(x), "\""), collapse = ", "), ")", sep = "")
}

fruits_char <- factor_to_char(fruits_factor)
cat(fruits_char) #close enough

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

#My attempt after I saw the message on slack. Both have the same outcome.

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

convert_string_to_vector <- function(strStart) {
  fruits_raw <- gsub("\\[\\d+\\]", "", strStart)
  fruits_raw <- trimws(unlist(strsplit(fruits_raw, "\""))) 
  fruits_raw <- fruits_raw[fruits_raw != ""]
  fruits <- paste("\"", fruits_raw, "\"", collapse = ", ")
  cat(paste0("c(", fruits, ")"))
}

convert_string_to_vector(strStart)

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

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\1”: This expression does not yield a result but if it was (.)\1\1 then it would search for three of the same characters such as ‘aaa’ or ‘bbb’

exer_3 <- c('banana', 'daaa', 'bbb', 'abcd', 'abba', 'sbbsb', '444','32323', '1234321', 'abcddcba', 'church','eleven')

str_subset(exer_3, "(.)\1\1")

## character(0)

str_subset(exer_3, "(.)\\1\\1")

## [1] "daaa" "bbb"  "444"

2. “(.)(.)\2\1”: This would search for 4 characters. The last two characters are the reverse of the first two like a mirror between the first two letters. So “abba” or “1221”.

str_subset(exer_3, "(.)(.)\\2\\1")

## [1] "abba"     "sbbsb"    "abcddcba"

3. “(..)\1”: This does not yield any results but if it was (..)\1, it would search for two characters followed by the same two characters. For example: “coco” or “banana”

str_subset(exer_3, "(..)\\1")

## [1] "banana" "32323"

4. “(.).\1.\1”: This expression searches for five characters, where the first, third, and fifth characters are the same such as “banana” or “12121”. The second and fourth character can be anything.

str_subset(exer_3, "(.).\\1.\\1")

## [1] "banana" "32323"  "eleven"

5. “(.)(.)(.).*\3\2\1”: This expression searches for 7 or more characters. The last three characters are the reverse of the first three characters like “1234321” or “abcdcda”

str_subset(exer_3,"(.)(.)(.).*\\3\\2\\1")

## [1] "1234321"  "abcddcba"

4. Construct regular expressions to match words that:

1. Start and end with the same character: “^(.).*\1$”

str_subset(exer_3,"^(.).*\\1$")

## [1] "bbb"      "abba"     "444"      "32323"    "1234321"  "abcddcba"

2. Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.):“.([A-Za-z][A-Za-z]).\1.*”

str_subset(exer_3,".*([A-Za-z][A-Za-z]).*\\1.*")

## [1] "banana" "sbbsb"  "church"

3. Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.): “.([A-Za-z]).\1.\1.”

str_subset(exer_3,".*([A-Za-z]).*\\1.*\\1.*")

## [1] "banana" "daaa"   "bbb"    "sbbsb"  "eleven"