name stores the extracted names.First, load the ‘stringr’ package:
install.packages("stringr",repos='http://mirrors.nics.utk.edu/cran/')
library(stringr)Then, re-create the introductory example:
# Load data into 'raw.data' character vector
raw.data <- "555-1239Moe Szyslak(636) 555-0113Burns, C. Montgomery555-6542Rev. Timothy Lovejoy555 8904Ned Flanders636-555-3226Simpson, Homer5543642Dr. Julius Hibbert"
# Use code provided in book to create character vector 'name'
# which contains the names of the Simpson's characters
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"Problems
(a) Use the tools of this chapter to rearrange the vector so that all elements conform to the standard first_name last_name.
# First name for strings with commas
fc <- str_trim(str_sub(name, start = str_locate(name, ",")[,1] + 1, end = str_length(name)))
# Last name for strings with commas
lc <- str_trim(str_sub(name, start = 1, end = str_locate(name, ",")[,1] - 1))
# First name for strings without commas and a single space
fs <- str_sub(name, start = 1, end = str_locate(name, " ")[,1] - 1)
# Last name for strings without commas and a single space
ls <- str_sub(name, start = str_locate(name, " ")[,1] + 1, end = str_length(name))
# Everything after the first space for all strings
as <- str_sub(name, start = str_locate(name, " ")[,1] + 1, end = str_length(name))
# First name
firstname <- ifelse(str_detect(name, ","), fc, ifelse(str_count(name, " ") == 2, ifelse(str_detect(as, " "), str_sub(as, start = 1, end = str_locate(as, " ")[,1] - 1), as), fs))
# Last name
lastname <- ifelse(str_detect(name, ","), lc, ifelse(str_count(name, " ") == 2, ifelse(str_detect(as, " "), str_sub(as, start = str_locate(as, " ")[,1] + 1, end = str_length(as)), as), ls))
# Formatted name
fullname <- str_c(firstname, " ", lastname)
fullname## [1] "Moe Szyslak" "C. Montgomery Burns" "Timothy Lovejoy"
## [4] "Ned Flanders" "Homer Simpson" "Julius Hibbert"(b) Construct a logical vector indicating whether a character has a title (i.e., Rev. and Dr.)
# Look for the presence of any of a number
# of name prefixes/titles/honorifics
str_detect(name, "Ms.|Miss|Mrs.|Mr.|Mister|Rev.|Reverend|Dr.|Doctor|Prof.|Professor|Father|Hon.|Honorable|Pres.|President|Gov.|Governer|Msgr.|Monsignor|Sen.|Senator")## [1] FALSE FALSE TRUE FALSE FALSE TRUE(c) Construct a logical vector indicating whether a character has a second name.
# Copy name into new vector prior to replacing it
name_count <- name
# I took 'second name' to mean the individual
# has two names prior to their last name.
# Therefore, I first replaced any name prefixes/titles/honorifics
# with a null string, then I trimmed the results, and then created
# a boolean condition based on whether the count of spaces was
# greater than one, which would be true for anyone with a 2nd name.
str_count(str_trim(str_replace_all(name_count, "Ms.|Miss|Mrs.|Mr.|Mister|Rev.|Reverend|Dr.|Doctor|Prof.|Professor|Father|Hon.|Honorable|Pres.|President|Gov.|Governer|Msgr.|Monsignor|Sen.|Senator", "")), " ") > 1## [1] FALSE TRUE FALSE FALSE FALSE FALSEProblems
(a) [0-9]+\\$
The [0-9]+ part looks for any string of digits 0 thru 9 that is 1 or more characters long. The two backslahes tells us to regard the $ as a character to be matched, not a metacharacter. Hence, any string of digits followed by a dollar sign would be matched by this regular expression.
# Create example string
t <- "This is an odd string: 234$. Why the trailing dollar sign?"
# Test regular expression to see if explanation provided is correct
unlist(str_extract_all(t, "[0-9]+\\$"))## [1] "234$"(b) \\b[a-z]{1,4}\\b
This regular expression looks at each word edge and matches lower case letters at least once, but not more than four times, and then requires there to be a word edge at the end of the string. Therefore, it will only match lower case words that are four characters or less in length.
# Create example string
t <- "Any lower case word in this sentence with four or less characters will be matched."
# Test regular expression to see if explanation provided is correct
unlist(str_extract_all(t, "\\b[a-z]{1,4}\\b"))## [1] "case" "word" "in" "this" "with" "four" "or" "less" "will" "be"(c) .*?\\.txt$
The dot represents any character. It is followed by an asterisk, which means that the character can be matched zero or more times. The question mark tells us that the preceding item is optional, which means we don’t have to have any characters at all. The two backslashes tell us to treat the second dot literally (as a character instead of a metacharacter), which means we’re trying to match “.txt” within the string. The dollar sign tells us the “.txt” should be at the end of the string. This regular expression should match “.txt” or any string of characters followed by “.txt”.
# Create example strings
t <- c(".txt","a.txt","test.txt", "big_data.txt", "2345.txt")
# Test regular expression to see if explanation provided is correct
unlist(str_extract_all(t, ".*?\\.txt$"))## [1] ".txt" "a.txt" "test.txt" "big_data.txt"
## [5] "2345.txt"(d) \\d{2}/\\d{2}/\\d{4}
The two backslashes and the ‘d’ looks for numerical digits, and the {x} tells us how many numerical digits to look for. In between three sets of numerical digits the expression looks for the forward slash character. Thus, this regular expression would match any date in a mm/dd/yyyy or dd/mm/yyyy format, or even any string in that format even it was not a valid date (i.e., “34/99/0002”). It would not match any dates that did not use a two-digit day or month, or a year which was not four digits.
# Create example strings
t <- c("2/15/2016","04/12/2015","26/03/1968","1/1/2011","34/99/0002","2/3/978")
# Test regular expression to see if explanation provided is correct
unlist(str_extract_all(t, "\\d{2}/\\d{2}/\\d{4}"))## [1] "04/12/2015" "26/03/1968" "34/99/0002"(e) <(.+?)>.+?</\\1>
This regular expression matches any string that starts with ‘<’, followed by one or more characters. Note that the one or more characters part (dot - plus - question mark) is in parentheses. After this, the ‘>’ character is matched, then one or more characters again, and then ‘</’. After this, it matches the same string which was matched earlier using the code inside the aforementioned parentheses (this is what the \1 does). Then, it looks for ‘>’. This looks like it is meant to match any HTML tag set that has one or more characters between the beginning tag and the closing tag. For example, ‘<h1>My First Heading</h1>’ would match, but ‘<h1></h1>’ would not. Nor would any HTML codings which don’t have an opening tag and closing tag, such as img src.
# Create example strings
t <- c("<h1>My First Heading</h1>","<h1></h1>","<p>P. Sherman<br>42 Wallaby Way<br>Sydney</p>","<img src='smiley.gif' alt='Smiley face' height='42' width='42'>")
# Test regular expression to see if explanation provided is correct
unlist(str_extract_all(t, "<(.+?)>.+?</\\1>"))## [1] "<h1>My First Heading</h1>"
## [2] "<p>P. Sherman<br>42 Wallaby Way<br>Sydney</p>"c1copCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0Ta woUwisdij7Lj8kpf0w3AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1Yw
wojigOd6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPalotfb7wEm24k6
t3sR9zqe5fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr
# Put code into character vector
t <- "c1copCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0TanwoUwisdij7Lj8kpf0w3AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1YwwojigOd6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPalotfb7wEm24k6t3sR9zqe5fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr"
# Extract capital letters and punctuation characters.
ec <- unlist(str_extract_all(t, "[[:upper:]]|[[:punct:]]"))
# Combine all elements of extracted vector into one string
os <- paste(ec, collapse = '')
# Replace the periods with a space
secret_message <- str_replace_all(os, "\\.", " ")
secret_message## [1] "CONGRATULATIONS YOU ARE A SUPERNERD!"And proud of it, mind you.