Instructions and the relevant questions can be found in the Instructions.txt file in the parent folder of this project.
There are only three majors in the 538 dataset of majors found in the majors-list.csv file that contain “DATA” or “STATISTICS”.
college_majors <- read_csv("https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv") |>
filter(grepl("DATA|STATISTICS", Major))
college_majors## # A tibble: 3 × 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 & MathematicsI found this question a little unclear, as the solution appears to be part of the question. Regardless, the code below creates a vector of the requested strings in the order requested.
fruits <- c("bell pepper", "bilberry", "blackberry", "blood orange", "blueberry",
"cantaloupe", "chili pepper", "cloudberry", "elderberry", "lime", "lychee",
"mulberry", "olive", "salal berry")
print(fruits)## [1] "bell pepper" "bilberry" "blackberry" "blood orange" "blueberry"
## [6] "cantaloupe" "chili pepper" "cloudberry" "elderberry" "lime"
## [11] "lychee" "mulberry" "olive" "salal berry"The first regex (.)\1\1 will match any sequence of three
matching characters since . matches any character and
\1 matches the same text as was most recently matched by
the first capture group, in this case (.). An example is
provided for clarity, we can see “aaa” matches, the substrings “bbb” and
“ccc” in “bbbccc” match, non-alphanumeric characters “!!!” match and the
repeating pattern of non-matching characters “abcabc” generates no
matches.
strings <- c("aaa", "bbbccc", "!!!", "abcabc")
str_view(strings, "(.)\\1\\1")## [1] │ <aaa>
## [2] │ <bbb><ccc>
## [3] │ <!!!>The second regex "(.)(.)\\2\\1" matches any four
character palindrome. The extra slash in \\2 escapes the
required slash since the regex is presented as a plain string.
\2 matches the most recent match in the second capture
group while \1 matches the most recent match in the first
capture group, and since both the first and second capture groups match
any character we end up with four character palindromes.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1")
str_view(strings, "(.)(.)\\2\\1")## [4] │ <xxxx>
## [5] │ <xyyx>
## [7] │ <1!!1>The third regex (..)\1 matches any four character string
where the first and second characters match the third and fourth
characters respectively. The capture group (..) matches any
two characters and \1 matches any match of the first
capture group.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1")
str_view(strings, "(..)\\1")## [4] │ <xxxx>
## [6] │ <xyxy>The fourth regex "(.).\\1.\\1" matches any five
character string where the first, third, and fifth characters match. The
second and third characters can be any non-newline character.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1", "abaca",
"!2!r!")
str_view(strings, "(.).\\1.\\1")## [8] │ <abaca>
## [9] │ <!2!r!>The fifth regex “(.)(.)(.).*\3\2\1” matches any string were the last three characters mirror the first three characters. The string can be any length six characters or greater that matches this pattern.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1", "abaca",
"!2!r!", "abccba", "xyz!231abzyx")
str_view(strings, "(.)(.)(.).*\\3\\2\\1")## [10] │ <abccba>
## [11] │ <xyz!231abzyx>Due to the ambiguity of the term “words” in the instructions, two solutions are provided for each question. The first assumes “word” to mean “any substring of the input string”. The second solution assumes a “word” is “a substring of the input string containing only consecutive english letters”. Hopefully both sets of solutions will sufficiently demonstrate a grasp of the topic.
The regex (.).*\1 will match any string that starts and
ends with the same character. This can include whitespace and
non-alphabet characters. If we follow the stricter definition of words
we can use ([a-zA-Z])\w*\1 instead.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1", "abaca",
"!2!r!", "abccba", "xyz!231abzyx", "my name is tom", "mom says hi")
str_view(strings, "(.).*\\1")## [1] │ <aaa>
## [2] │ <bbb><ccc>
## [3] │ <!!!>
## [4] │ <xxxx>
## [5] │ <xyyx>
## [6] │ <xyx>y
## [7] │ <1!!1>
## [8] │ <abaca>
## [9] │ <!2!r!>
## [10] │ <abccba>
## [11] │ <xyz!231abzyx>
## [12] │ <my name is tom>
## [13] │ <mom>< says >histr_view(strings, "([a-zA-Z])\\w*\\1")## [1] │ <aaa>
## [2] │ <bbb><ccc>
## [4] │ <xxxx>
## [5] │ <xyyx>
## [6] │ <xyx>y
## [8] │ <abaca>
## [10] │ <abccba>
## [13] │ <mom> <says> hiThe regex ([a-zA-Z][a-zA-Z]).*\1 will match any string
that contains a repeated pair of letters Alternatively, for the more
strict definition we can use ([a-zA-Z][a-zA-Z])\w*\1.
strings <- c("aaa", "bbbccc", "!!!!", "xxxx", "xyyx", "xyxy", "1!!1", "abaca",
"!2!r!", "abccba", "xyz!231abzyx", "my name is tom", "church", "he went to church",
"cheese chess")
str_view(strings, "([a-zA-Z][a-zA-Z]).*\\1")## [4] │ <xxxx>
## [6] │ <xyxy>
## [13] │ <church>
## [14] │ he went to <church>
## [15] │ <cheese ch>essstr_view(strings, "([a-zA-Z][a-zA-Z])\\w*\\1")## [4] │ <xxxx>
## [6] │ <xyxy>
## [13] │ <church>
## [14] │ he went to <church>The regex ([a-zA-Z]).*\1.*\1 will match any string with
an english letter that occurs at least three times. Once again, if we
want to restrict it to the stricter word definition we can instead use
([a-zA-Z])\w*\1\w*\1.
strings <- c("aaa", "bbbccc", "!!!", "xxxx", "xyyx", "xyxy", "1!!1", "abaca",
"!2!r!", "abccba", "xyz!231abzyx", "my name is tom", "church", "she shears sheep")
str_view(strings, "([a-zA-Z]).*\\1.*\\1")## [1] │ <aaa>
## [2] │ <bbb><ccc>
## [4] │ <xxxx>
## [8] │ <abaca>
## [12] │ <my name is tom>
## [14] │ <she shears s>heepstr_view(strings, "([a-zA-Z])\\w*\\1\\w*\\1")## [1] │ <aaa>
## [2] │ <bbb><ccc>
## [4] │ <xxxx>
## [8] │ <abaca>