CUNY 607_Week#4

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The following document is solution to exercises from Text: Automated Data Collection with ‘R’. Exercise 3 page 217: In this exercise we are using string manipulation techniques to reformat names and create logical vector to carry some information about content of name strings.

# Invoque String Package
library(stringr)

# Set Name vector

raw.data <- "555-1239Moe Szyslak(636) 555-0113Burns, C. Montgomery555-6542Rev. Timothy Lovejoy555 8904Ned Flanders636-555-3226Simpson, Homer5543642Dr. Julius Hibbert"

# Extract name from raw.data

name <- unlist(str_extract_all(raw.data, "[[:alpha:]., ]{2,}"))

# Dispay name

name

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

1. Reformatting name to display first_name Last_name format

## Unfortunately, I am sill trying to get the function to work properly in all cases...



## Assume that name will contain at least first and last name

format_name <- function(raw_string){
  
  # reset name variables
  first_name <- " "
  last_name <- " " 
  title_name <- " "
  middle_name <- " "
  
  name_str <- unlist(str_split(raw_string, "[[:blank:]]"))
  
  for (i in 1:length(name_str) ) {
    name_tst <- name_str[i]
  
    name_tst <- str_trim(name_tst)
    name_alpha <- str_extract(name_tst, "[[:alpha:]]+")
     
    # check for ending ',' or '.'  
    if (str_detect(name_tst, ",")){
        last_name <-name_alpha
      }
    }else if(str_detect(name_tst, "\\.")){
      if (i == 1) { 
        title_name <- name_tst
      }else (i<length(name_tst)){
        middle_name <-name_tst
      }
    }else{ # no punctuation character detected
      if ( i== 1){
        first_name <- name_alpha
      }else if (i<length(name_tst)){
        middle_name <- name_alpha
      }else{
        last_name <- name_alpha
      }
    } # end of else
  } # end of for loop
  
  out_string <- str_c(first_name, last_name, sep = " ")
  return(out_string)
}

format_name(name[1])
format_name(name[2])
format_name(name[3])
format_name(name[4])
format_name(name[5])
format_name(name[6])

2. Construct a logical vector indicating whether a character has a title

v_rev <- str_detect(name, "Rev.")
v_dr <- str_detect(name, "Dr.")
v_title <- v_rev | v_dr
# display logical vector
v_title

## [1] FALSE FALSE  TRUE FALSE FALSE  TRUE

3. Construct a logical vector indicating whether a character has a middle name

v_middle <- str_detect(name, "[[:upper:]]\\.")
# Display logical vector
v_middle

## [1] FALSE  TRUE FALSE FALSE FALSE FALSE

Exercise # 7 page 218

Consider the string

, we would like to extract first HTML tag, we write expression: <.+> Explain why it is not working and fix it. First let us see what we get:

test_str <- "<title>+++BREAKING NEWS+++</title>"

str_extract(test_str, "<.+>")

## [1] "<title>+++BREAKING NEWS+++</title>"

When we execute this command, we get the all string. This is due to R applying “greedy quantification”. That is R will extract the greatest possible sequence of any characters before “<….>” to modify this behavior we have to use the ? to indicate that we only want shortest possible expression. Hence by modifying the sequence by adding ?

test_str <- "<title>+++BREAKING NEWS+++</title>"

str_extract(test_str, "<.+?>")

## [1] "<title>"

Exercise # 8 page 218 Consider the string: (5-3)²⁼⁵2-253+3^2, we would like to extract the formula to the string by writing following regular expression “[^0-9=+*()]+”. This does not lead to the desire result. Explain why and fix it. Again, we will try this expression and consider the results:

test_str2 <- "(5-3)^2=5^2-2*5*3+3^2"

str_extract(test_str2, "[^0-9=+*()]+")

## [1] "-"

The result is ‘-’. This is due to the Metacharacters and their meaning… ^ indicates “not in” and the - is interpreted as a range between digit. Most Metacharacter are interpreted literally when included within bracket in expression. However, this is not the case for ^ at beginning of expression and - between digits. Hence we need to make the following modifications to the expression… Move the ^ within expression not at beginning and add - in expression to account for - sign in expression, we still want to interpret 0-9 as range of digits.

str_extract(test_str2, "[0-9=+*^()-]+")

## [1] "(5-3)^2=5^2-2*5*3+3^2"

Exercise # 9 page 218

I have not craked the code. I am still trying to figure it out…