data_607-week_03

week 3 assignment

Please deliver links to an R Markdown file (in GitHub and rpubs.com) with solutions to problems 3 and 4 from chapter 8 of Automated Data Collection in R. Problem 9 is extra credit.

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

# Extract information
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"

phone <- unlist(str_extract_all(raw.data, "\\(?(\\d{3})?\\)?(-| )?\\d{3}(-| )?\\d{4}"))
phone

## [1] "555-1239"       "(636) 555-0113" "555-6542"       "555 8904"      
## [5] "636-555-3226"   "5553642"

data.frame(name = name, phone = phone)

##                   name          phone
## 1          Moe Szyslak       555-1239
## 2 Burns, C. Montgomery (636) 555-0113
## 3 Rev. Timothy Lovejoy       555-6542
## 4         Ned Flanders       555 8904
## 5       Simpson, Homer   636-555-3226
## 6   Dr. Julius Hibbert        5553642

3.1

Use the tools of this chapter to rearrange the vector so that all elements conform to the standard “first_name last_name”.

Answer:

# use grep to find strings with commas to identify names with "last, first" convention
first_lasts <- grep(",", name, value=TRUE)
first_lasts

## [1] "Burns, C. Montgomery" "Simpson, Homer"

# split each string by comma & remove leading whitespace
name_a <- str_trim(unlist(strsplit(first_lasts[1], ",")), side = "left")
name_1 <- paste(name_a[2], name_a[1])
name_b <- str_trim(unlist(strsplit(first_lasts[2], ",")), side = "left")
name_2 <- paste(name_b[2], name_b[1])
first_last <- c(name_1, name_2)

# replace names in original vector
name[str_detect(name,"[[:alpha:]],")] <- first_last
name

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

3.2.

Construct a logical vector indicating whether a character has a title (i.e., Rev. and Dr.).

Answer:

# look for strings with a period following the second or third character
title <- str_detect(name, "[[:alpha:]]{2,3}\\.")
title

## [1] FALSE FALSE  TRUE FALSE FALSE  TRUE

# show matching names
name[title]

## [1] "Rev. Timothy Lovejoy" "Dr. Julius Hibbert"

3.3.

Construct a logical vector indicating whether a character has a second name.

Answer:

# assumption: "second name" == "middle name"
# remove titles and trailing spaces
no_titles <- str_trim(str_remove(name, "[[:alpha:]]{2,3}\\."))
no_titles

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

# find names with strings with leading and trailing spaces to indicate presence of a middle name
grep("[[:space:]]+[[:alpha:]]+[[:space:]]", no_titles, value=TRUE)

## [1] "C. Montgomery Burns"

4.1

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

Answer:

#1 [0-9]+\\$ 
# match any digit one or more times, followed by a '$'
grep(pattern = "[0-9]+\\$", "2756$", value = TRUE)

## [1] "2756$"

#2 \\b[a-z]{1,4}\\b
# match 'b', followed by any lower case ASCII letter at least 1 time, but not more than 4 times, followed by a 'b' 
grep(pattern = "\\b[a-z]{1,4}\\b", "blab", value = TRUE )

## [1] "blab"

#3 .*?\\.txt$
# match an optional '.' zero or one time, then a '.txt' at the end 
grep(pattern = ".*?\\.txt$", ".foo.txt", value = TRUE )

## [1] ".foo.txt"

#4 \\d{2}/\\d{2}/\\d{4} 
# match digits exactly twice, forward slash, digits exactly twice, forward slash, then digits four times
grep(pattern = "\\d{2}/\\d{2}/\\d{4}", "09/15/2019", value = TRUE )

## [1] "09/15/2019"

#5 <(.+?)>.+?</\\1> 
# match any optional character immediately preceded by any character wrapped by less-than and greater-than, followed by any optional character immediately preceded by any character, followed by a forward-slash that is followed by the first parenthesis-delimited expression    
grep(pattern = "<(.+?)>.+?</\\1>", "It is possible to parse HTML markup <strong>formatted</strong> <em>text</em>.", value = TRUE )

## [1] "It is possible to parse HTML markup <strong>formatted</strong> <em>text</em>."

9.1

The following code hides a secret message. Crack it with R and regular expressions. Hint: Some of the characters are more revealing than others! The code snippet is also available in the materials www.r-datacollection.com.

Answer:

code <- "clcopCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0TanwoUwisdij7Lj8kpf03AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1YwwojigOd6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPa1otfb7wEm24k6t3sR9zqe5fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr"
str_replace_all(code, "[[:lower:]]|[[:digit:]]", "")

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