Data 607 Assignment 3

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Question 3

Copy the introductory example. The vector name stores the extracted names.

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

library(stringr)
library(stringi)

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

# extract data to get the names
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"

#extract data to get the phone numbers
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"

#combine both using a different method than the book and change column names
combined.data <- list(name,phone)
combined_df<- as.data.frame(combined.data)
colnames(combined_df) [1:2]<- c("user name","phone number") 

combined_df

##              user name   phone number
## 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

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

# write a for/if loop to find names with ","; split them apart and reverse the order

for(i in 1:length(name)) {
  if(str_detect(name[i],",")==TRUE) {
    temp<-unlist(str_split(name[i],","))
    name[i]<-paste(temp[2], temp[1], sep=" ")
  }
}
name

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

# keep only numbers for the phone column
phone<- as.numeric(gsub("\\D", "", phone))
phone

## [1]    5551239 6365550113    5556542    5558904 6365553226    5553642

#write a ifelse statement focused on number of characters
#if the string is 7 characters long, split the first 3 characters, add "-"
#if the string is not 7 characters, split first 3 characters 
#          add "-", the second 3 characters add "-" 

phone<- ifelse(stri_length(phone)== 7, 
       (gsub("(^\\d{3})(\\d{4}$)","\\1-\\2", phone[nchar(phone) == 7])),
       (gsub("(^\\d{3})(\\d{3})(\\d{4}$)", "\\1-\\2-\\3", phone[nchar(phone) == 10]))
      )       
phone

## [1] "555-1239"     "636-555-3226" "555-8904"     "555-3642"    
## [5] "636-555-0113" "555-6542"

# recreate the dataframe with the updated columns
combined_df<- data.frame(name, phone)
combined_df

##                   name        phone
## 1          Moe Szyslak     555-1239
## 2  C. Montgomery Burns 636-555-3226
## 3 Rev. Timothy Lovejoy     555-8904
## 4         Ned Flanders     555-3642
## 5        Homer Simpson 636-555-0113
## 6   Dr. Julius Hibbert     555-6542

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

#Detects if Rev. or Dr. is in the name
title_present<- str_detect(name, "Rev.|Dr.")
title_present

## [1] FALSE FALSE  TRUE FALSE FALSE  TRUE

#Construct a new list identifing titles
title <- c()
for(t in 1:length(title_present)) {
  if(title_present[t]==FALSE){
    title[t]<-'NA'
  } else if(str_detect(name[t], "Rev.")=='TRUE') {
    title[t]<-"Rev."
  } else {
    title[t]<-"Dr."
  }
}
title

## [1] "NA"   "NA"   "Rev." "NA"   "NA"   "Dr."

#clean up the Name column to remove titles
name<- gsub("Rev.", "", name)
name<- gsub("Dr.", "", name)

#reconstruct the combined dataframe with new colum
combined_df<- data.frame(title, name, phone)
combined_df

##   title                 name        phone
## 1    NA          Moe Szyslak     555-1239
## 2    NA  C. Montgomery Burns 636-555-3226
## 3  Rev.      Timothy Lovejoy     555-8904
## 4    NA         Ned Flanders     555-3642
## 5    NA        Homer Simpson 636-555-0113
## 6   Dr.       Julius Hibbert     555-6542

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

#creating a regular expression to find the literal "."
dot <- "\\."

#finding any case where DOT is present, since I cleaned up all titles, 
#this will find middle names.

middle <- str_extract(name, dot)

#replace all found instances of dot with Yes
middle <- ifelse(middle == dot, "n/a", "yes")

#updating the dataframe to add new column
combined_df<- data.frame(title, name, phone, middle)
combined_df

##   title                 name        phone middle
## 1    NA          Moe Szyslak     555-1239   <NA>
## 2    NA  C. Montgomery Burns 636-555-3226    yes
## 3  Rev.      Timothy Lovejoy     555-8904   <NA>
## 4    NA         Ned Flanders     555-3642   <NA>
## 5    NA        Homer Simpson 636-555-0113   <NA>
## 6   Dr.       Julius Hibbert     555-6542   <NA>

Question 4

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

q4<- c("12$3","RTg","29/49/24","1$45","578$","drt","Lenovo.txt","02/24/1954","487$", "34G$","few", "Rate", "data", "phone8.txt","01/08/2019", "<html> table </html>","<table> Science </load>")

(a) [0-9]+\$

• any string of numbers with a $ at the end of the string

unlist(str_extract_all(q4,"[0-9]+\\$"))

## [1] "12$"  "1$"   "578$" "487$"

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

• a character groupings of lower case letters between 1 to 4 characters

unlist(str_extract_all(q4,"\\b[a-z]{1,4}\\b"))

## [1] "drt"  "txt"  "few"  "data" "txt"  "html" "html" "load"

(c) .?\.txt$
Any string which ends with .txt

unlist(str_extract_all(q4,".*?\\.txt$"))

## [1] "Lenovo.txt" "phone8.txt"

(d) \d{2}/\d{2}/\d{4}

• Any string which matches “2 digits / 2 digits / 4 digits” (think a date)

unlist(str_extract_all(q4,"\\d{2}/\\d{2}/\\d{4}"))

## [1] "02/24/1954" "01/08/2019"

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

• Any string which matches any text
  
• Where text1 is the same characters and second set of Text1 begins with backslash

unlist(str_extract_all(q4,"<(.+?)>.+?</\\1>"))

## [1] "<html> table </html>"

Secret Message

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 at www.r-datacollection.com.

message_raw<-"clcopCow1zmstc0d87wnkig7OvdicpNuggvhryn92Gjuwczi8hqrfpRxs5Aj5dwpn0TanwoUwisdij7Lj8kpf03AT5Idr3coc0bt7yczjatOaootj55t3Nj3ne6c4Sfek.r1w1YwwojigOd6vrfUrbz2.2bkAnbhzgv4R9i05zEcrop.wAgnb.SqoU65fPa1otfb7wEm24k6t3sR9zqe5fy89n6Nd5t9kc4fE905gmc4Rgxo5nhDk!gr"
message<- unlist (str_extract_all(message_raw, "[[:upper:].]{1,}"))
message<- str_replace_all(paste(message, collapse = ''), "[.]", " ")
message

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