#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”
library(tidyverse)## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.1 ──## ✓ ggplot2 3.3.5 ✓ purrr 0.3.4
## ✓ tibble 3.1.6 ✓ dplyr 1.0.7
## ✓ tidyr 1.1.4 ✓ stringr 1.4.0
## ✓ readr 2.1.1 ✓ forcats 0.5.1## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()# read in csv file
majors <- read.csv("https://raw.githubusercontent.com/justinm0rgan/data607/main/Assignments/3/majors-list.csv")
# enter search query and return results
str_view(majors$Major,"DATA|STATISTICS", match = T)#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”)
# would not take a second [] without producing an error, so created sep str and
# concatenated them.
fruits_org_1 <- '[1] "bell pepper" "bilberry" "blackberry" "blood orange"'
fruits_org_2 <- '[5] "blueberry" "cantaloupe" "chili pepper" "cloudberry"'
fruits_org_3 <- '[9] "elderberry" "lime" "lychee" "mulberry"'
fruits_org_4 <- '[13] "olive" "salal berry"'
# full string needing to be transformed
fruits_org <- str_c(fruits_org_1,fruits_org_2,fruits_org_3,fruits_org_4)
fruits_org## [1] "[1] \"bell pepper\" \"bilberry\" \"blackberry\" \"blood orange\"[5] \"blueberry\" \"cantaloupe\" \"chili pepper\" \"cloudberry\"[9] \"elderberry\" \"lime\" \"lychee\" \"mulberry\"[13] \"olive\" \"salal berry\""fruits_org %>%
str_extract_all("\"([a-z]+.[a-z]+)\"") %>%
unlist() %>%
str_remove_all("\"") %>%
as_tibble() %>%
paste0(sep = "", collapse = ",") %>%
writeLines()## 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 (.) equals any character, then \1 equals a repetition of that character in the (), a second \1 equals another repetition of the character in the (). However, if this is a string, then it needs to have double backslash to escape the backslash, otherwise it will not match. Essentially it will match any three like characters in a row, i.e. “aaa” or 111.
“(.)(.)\2\1” (.) again matches any character, with this twice, any two characters. This example is a string so the \2 is actually regex \2 which means it matches a repetition of the the group right before it. This is followed by a \1 which denotes a repetition of the first group. Therefore this will match any two characters then their reverse i.e. “abba” or 2332.
(..)\1 (..) matches any two characters, then \1 denotes a repetition of those characters. However, once again, if this is a string then it needs a second backslash to be a literal backslash, otherwise it will not match. For example this would match “abab” or 1212
str_view(c("1212"), "(..)\1")“(.).\1.\1” (.) will match any character, the parenthesis denote a group. This is followed by any character (not in a group). This pattern is a string, so the \1 will then match the first character that was enclosed in the parenthesis. This is followed by a . for any character.Then once more the \1 repeats the first character enclosed by parenthesis. For example, this pattern would match any 5 character term where the first, third and last are the same i.e.”abaca” or 12131
“(.)(.)(.).\3\2\1” (.) create groups where each can match any character. The . without parenthesis denotes any character WITHOUT a group, this is followed by an , which denotes 0 or more of any character, so the . after the (.) could be repeated many times or it may not be. This pattern is string once more, so the following \3\2\1 equate to repetitions of the each group in a mirror image of its previous rendition. For example this pattern would match any three characters followed by zero or more of a fourth character, then a mirror image of the first three characters i.e. “abccba” or “abc99789cba” or “123lkj34321”. As long as the first three and the last three are mirror images of each other, the pattern would match.
#4 Construct regular expressions to match words that: Start and end with the same character.
str_view(stringr::words, "^(.).*\\1$", match = T)Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.)
str_subset(stringr::words, "([a-z][a-z]).*\\1")## [1] "appropriate" "church" "condition" "decide" "environment"
## [6] "london" "paragraph" "particular" "photograph" "prepare"
## [11] "pressure" "remember" "represent" "require" "sense"
## [16] "therefore" "understand" "whether"Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)
str_view(stringr::words, ".*([a-z]).*\\1.*\\1.*", match = T)