3a.

library(stringr)
raw.data <- "555-1239Moe Szyslak(636) 555-0113Burns, C. Montgomery555
-6542Rev. Timothy Lovejoy555 8904Ned Flanders636-555-3226Simpson, Homer5553642Dr. Julius Hibbert"
name <- unlist(str_extract_all(raw.data, "[[:alpha:]., ]{2,}"))
name
## [1] "Moe Szyslak"          "Burns, C. Montgomery" "Rev. Timothy Lovejoy"
## [4] "Ned Flanders"         "Simpson, Homer"       "Dr. Julius Hibbert"
location1 <- str_locate(name[2], "Burns")
location2 <- str_locate(name[2], "C.")
splitnames1 <- str_sub(name[2], location1[1,1], location1[1,2])
splitnames2 <- str_sub(name[2], location2[1,1], location2[1,2])
unitednames1 <- str_c(splitnames1, splitnames2, sep = " ")
location3 <- str_locate(name[5], "Simpson")
location4 <- str_locate(name[5], "Homer")
splitnames3 <- str_sub(name[5], location3[1,1], location3[1,2])
splitnames4 <- str_sub(name[5], location4[1,1], location4[1,2])
unitednames2 <- str_c(splitnames4, splitnames3, sep = " ")
location5 <- str_locate(name[3], "Timothy Lovejoy")
subnames1 <- str_sub(name[3], location5[1,1], location5[1,2])
location6 <- str_locate(name[6], "Julius Hibbert")
subnames2 <- str_sub(name[6], location6[1,1], location6[1,2])
newname <- str_c(c(name[1], unitednames1, subnames1, name[4], unitednames2, subnames2))

3b.

str_detect(name, "Rev.|Dr.")
## [1] FALSE FALSE  TRUE FALSE FALSE  TRUE

3c.

We have to count the number of characters in the vector that has two spaces. We have to exclude the titles.

withouttitle <- str_c(c(name[1], name[2], subnames1, name[4], name[5], subnames2))
str_count(withouttitle, " ")
## [1] 1 2 1 1 1 1

The reason why it is incorrect is because even though it specifies what the first character from the string should be, it will go all the way until the end. There is ambiguity in the command. The last character in the string is the same as the seventh.

news <- "<title>+++BREAKING NEWS+++</title>"
str_extract(news, "<.+>")
## [1] "<title>+++BREAKING NEWS+++</title>"

The following is the correct expression because the question mark indicates that the plus sign is optional and will be matched at most once.

str_extract(news, "<.+?>")
## [1] "<title>"
binomialthm <- "(5-3)^2=5^2-2*5*3+3^2 conforms to the binomial theorem."

The reason why the expression below is incorrect is because the caret symbol is included in the beginning of the character class. As a result, everything will be matched except the contents of the character class.

str_extract(binomialthm, "[^0-9=+*()]+")
## [1] "-"

A correction to the above command is the placement of the caret symbol at the end of the character class and the placement of a dash in the beginning.

str_extract(binomialthm, "[-0-9=+*()^]+")
## [1] "(5-3)^2=5^2-2*5*3+3^2"
msg = 'clcopCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0Tanwo
Uwisdij7Lj8kpf03AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1YwwojigO
d6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPa1otfb7wEm24k6t3sR9zqe5
fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr'
str_extract_all(msg, '[[:upper:]]|[[:punct:]]')
## [[1]]
##  [1] "C" "O" "N" "G" "R" "A" "T" "U" "L" "A" "T" "I" "O" "N" "S" "." "Y"
## [18] "O" "U" "." "A" "R" "E" "." "A" "." "S" "U" "P" "E" "R" "N" "E" "R"
## [35] "D" "!"