The purpose of this assignment was to practice using regular expressions in R. These questions are from the end of Chapter 8 in the textbook Automated Data Collection with R.

3a: Use the tools of this chapter to rearrange the vector so that all elements conform to the standard first_name last_name.

(i) To start, I loaded the raw data and extracted the names.

namesraw <- c("555-1239Moe Szyslak(636) 555-0113Burns, C. Montgomery555-6542Rev. Timothy Lovejoy555 8904Ned Flanders636-555-3226Simpson, Homer5553642Dr. Julius Hibbert")

names <- unlist(str_extract_all(namesraw, "[[:alpha:]., ]{2,}"))

names
## [1] "Moe Szyslak"          "Burns, C. Montgomery" "Rev. Timothy Lovejoy"
## [4] "Ned Flanders"         "Simpson, Homer"       "Dr. Julius Hibbert"

(ii) Then, I removed the titles from the names, and trimmed the white space left over. I identified the titles by looking for a pattern of 2 to 3 characters plus a period.

trimmed <- str_replace_all(names, "[[:alpha:]]{2,3}\\.", "")

trimmed <- str_trim(trimmed)

trimmed
## [1] "Moe Szyslak"          "Burns, C. Montgomery" "Timothy Lovejoy"     
## [4] "Ned Flanders"         "Simpson, Homer"       "Julius Hibbert"

(iii) To fix the names, my goal was to create clean lists of first names and last names that I could easily concatenate back together in the right order. I started by creating a clean list of last names. I identified last names as:

  • the pattern of all printable characters before a comma; OR
  • the specific pattern of 1 character + 1 blank space + any number of characters.

Then, I removed the 1 character + 1 blank space pattern, and also removed the commas. This left me with a clean list of last names.

last_names <- str_extract_all(trimmed,"[[:print:]]+[\\,]|[[:alpha:]][[:blank:]]{1}[[:alpha:]]+")

last_names <- str_replace_all(last_names,"[[:alpha:]]{1}[[:blank:]]{1}", "")

last_names <- str_replace_all(last_names, "\\,","")

last_names
## [1] "Szyslak"  "Burns"    "Lovejoy"  "Flanders" "Simpson"  "Hibbert"

(iv) The clean list of first names was easier. I identified first names as:

  • the pattern of all printable characters after a comma, OR
  • the pattern of characters + blank spaces.

Then, I removed the commas and trimmed the blank spaces, and got a clean list of first names.

first_names <- str_extract_all(trimmed,"[\\,][[:print:]]+|[[:alpha:]]+[[:blank:]]+")

first_names <- str_replace_all(first_names,"[\\,]+[[:blank:]]{1,2}", "")

first_names <- str_trim(first_names)

first_names
## [1] "Moe"           "C. Montgomery" "Timothy"       "Ned"          
## [5] "Homer"         "Julius"

(v) Finally, I concatenated both lists in the right order and separated the names with a space.

final <- str_c(first_names, last_names, sep=" ")

final
## [1] "Moe Szyslak"         "C. Montgomery Burns" "Timothy Lovejoy"    
## [4] "Ned Flanders"        "Homer Simpson"       "Julius Hibbert"

3b: Construct a logical vector indicating whether the character has a title.

A title is a pattern of two or three characters followed by a period. Two of the names fit this pattern.

title <- grepl("[[:alpha:]]{2,3}\\.", names)

title
## [1] FALSE FALSE  TRUE FALSE FALSE  TRUE

3c: Construct a logical vector indicating whether the character has a second name.

I used my cleaned list of first and last names to answer this question, since it only contains names (without titles). I identified a second name as characters followed by a period. Only one name fit this pattern.

secondname <- grepl("[[:alpha:]]+\\.", final)

secondname
## [1] FALSE  TRUE FALSE FALSE FALSE FALSE

Describe the types of strings that conform to the following regular expressions and construct an example that is matched by the regular expression.

4a: [0-9]+\\$

This pattern matches a string that has a number from 0 to 9 (at least once) followed by a dollar sign.

Example: “4$”

a1 <- c("4$", "4", "10$")

a2 <- grepl("[0-9]+\\$", a1)

a2
## [1]  TRUE FALSE  TRUE

4b: \\b[a-z]{1,4}\\b

This pattern describes a string that has 1-4 letters from edge to edge.

Example: “dddd”

b1 <- c("hodc", "ff", "dddd", "5gs", "ggggg")

b2 <- grepl("\\b[a-z]{1,4}\\b", b1)

b2
## [1]  TRUE  TRUE  TRUE FALSE FALSE

4c: .*?\\.txt$

This pattern describes a string of any length (found at least once) that ends in “.txt”.

Example: “assignment.txt”

c1 <- c("assignment.txt", "m.txt", ".txtm", "r.txt.y.txt")

c2 <- grepl(".*?\\.txt$", c1)

c2
## [1]  TRUE  TRUE FALSE  TRUE

4d: \\d{2}/\\d{2}/\\d{4}

This pattern describes a string with 2 digits, a forward slash, two digits, another forward slash, and 4 digits. It is constructed like a date.

Example: “11/22/2018”

d1 <- c("11222018", "11/22/2018")

d2 <- grepl("\\d{2}/\\d{2}/\\d{4}", d1)

d2
## [1] FALSE  TRUE

4e: <(.+?)>.+?</\\1>

This pattern matches an HTML tag with at least one character between the tags. It also has a backreference \1.

Example: “<b>bold</b>”

e1 <- c("<b>bold</b>", "<i>italics</i>")

e2 <- grepl("<(.+?)>.+?</\\1>", e1)

e2
## [1] TRUE TRUE

9: The following code hides a secret message. Crack it with R and regular expressions.

secret <- "clcopCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0TanwoUwisdij7Lj8kpf03AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1YwwojigOd6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPa1otfb7wEm24k6t3sR9zqe5fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr"

secret <- str_replace_all(secret, "[[:digit:]]", "")

secret <- str_replace_all(secret, "[[:lower:]]", "")

secret <- str_replace_all(secret, "\\.", " ")

secret
## [1] "CONGRATULATIONS YOU ARE A SUPERNERD!"