Assignment_3

String Manipulation

1. Using the 173 majors listed in fivethirtyeight.com’s College Majors dataset [https://fivethirtyeight.com/features/the-economic-guide-to-picking-a-college-major/], provide code that identifies the majors that contain either “DATA” or “STATISTICS”

Regular expression was implemented for this problem. str_detect from regular expression is used to identify “DATA” and “STATISTICS” majors.

setwd("C:/Users/malia/OneDrive/Desktop/MSDS DATA 607")
assigment3_dat<-read.csv("https://raw.githubusercontent.com/fivethirtyeight/data/master/college-majors/majors-list.csv")
assigment3_dat1<-c(assigment3_dat$Major)

head(assigment3_dat1)

## [1] "GENERAL AGRICULTURE"                  
## [2] "AGRICULTURE PRODUCTION AND MANAGEMENT"
## [3] "AGRICULTURAL ECONOMICS"               
## [4] "ANIMAL SCIENCES"                      
## [5] "FOOD SCIENCE"                         
## [6] "PLANT SCIENCE AND AGRONOMY"

assigment3_dat1[str_detect(assigment3_dat1,
   pattern = "DATA"
)]

## [1] "COMPUTER PROGRAMMING AND DATA PROCESSING"

assigment3_dat1[str_detect(assigment3_dat1,
   pattern = "STATISTICS")]

## [1] "MANAGEMENT INFORMATION SYSTEMS AND STATISTICS"
## [2] "STATISTICS AND DECISION SCIENCE"

2 Write code that transforms the data below:

[1] “bell pepper” “bilberry” “blackberry” “blood orange”

[5] “blueberry” “cantaloupe” “chili pepper” “cloudberry”

[9] “elderberry” “lime” “lychee” “mulberry”

[13] “olive” “salal berry”

Into a format like this:

c(“bell pepper”, “bilberry”, “blackberry”, “blood orange”, “blueberry”, “cantaloupe”, “chili pepper”, “cloudberry”, “elderberry”, “lime”, “lychee”, “mulberry”, “olive”, “salal berry”)

In order to solve this problem, glue_collapse was used. After that c() is used to compress the data sets in a single vector.For further manipulation st_replace_all is used to clean data by removing unwanted space, brackets etc.

FRUITS1 <- '[1] "bell pepper"  "bilberry"     "blackberry"   "blood orange"

[5] "blueberry"    "cantaloupe"   "chili pepper" "cloudberry"  

[9] "elderberry"   "lime"         "lychee"       "mulberry"    

[13] "olive"        "salal berry"'

FRUITS1 <- str_extract_all(FRUITS1, "(.+?)+")
FRUITS1 <- unlist(FRUITS1)

FRUITS1

## [1] "[1] \"bell pepper\"  \"bilberry\"     \"blackberry\"   \"blood orange\""
## [2] "[5] \"blueberry\"    \"cantaloupe\"   \"chili pepper\" \"cloudberry\"  "
## [3] "[9] \"elderberry\"   \"lime\"         \"lychee\"       \"mulberry\"    "
## [4] "[13] \"olive\"        \"salal berry\""

Fruitlist1<-glue_collapse(FRUITS1,sep = ", ")
print(Fruitlist1)

## [1] "bell pepper"  "bilberry"     "blackberry"   "blood orange", [5] "blueberry"    "cantaloupe"   "chili pepper" "cloudberry"  , [9] "elderberry"   "lime"         "lychee"       "mulberry"    , [13] "olive"        "salal berry"

c(Fruitlist1)

## [1] "[1] \"bell pepper\"  \"bilberry\"     \"blackberry\"   \"blood orange\", [5] \"blueberry\"    \"cantaloupe\"   \"chili pepper\" \"cloudberry\"  , [9] \"elderberry\"   \"lime\"         \"lychee\"       \"mulberry\"    , [13] \"olive\"        \"salal berry\""

Fruitlist1<- str_replace_all(Fruitlist1, "[\\[\\]]", "") # for removing square brackets
Fruitlist1 <- str_replace_all(Fruitlist1, "[!^[:digit:]]", "") #removing digits
Fruitlist1 <- str_replace_all(Fruitlist1, "\\\n", "") #removing "\n"
Fruitlist1 <- str_replace_all(Fruitlist1, '[\"]', "'") #replacing '\"' with "'"
Fruitlist1<- trimws(Fruitlist1) #removing leading / trailing whitespace
Fruitlist1 <- str_replace_all(Fruitlist1, "\\s+", " ") #compresseng whitespace
Fruitlist1 <- str_replace_all(Fruitlist1, "' '", "','") 
print(Fruitlist1)

## [1] "'bell pepper','bilberry','blackberry','blood orange', 'blueberry','cantaloupe','chili pepper','cloudberry' , 'elderberry','lime','lychee','mulberry' , 'olive','salal berry'"

3 Describe, in words, what these expressions will match:

(.)\1\1

This expression will not match anything because it is incorrectly done. If it was done like this : (.)\1\1 it would repeat a charecter in first group then repeat it again.

“(.)(.)\2\1”

This expression matches a first letter and a second letter in the following order: 1st letter - 2nd letter - 2nd letter - 1st letter. For instance“Beeb” “pepper”.

(..)\1

This expression will match nothing because the correct statement looks like this:(..)\1.if it was done correctly it will match charecters repeated twice.For instance: “LELELI”.

“(.).\1.\1”

This expression will match charecter shows up 3 times with one charecter between each.For instance: Eleven. “e” shows up three times in this word.

“(.)(.)(.).\3\2\1" ### This expression will match three charecters in one order with charecters between, then three charecters with three letters in the reverse order of what it started.For instance”lizpizil“,”bantannab"

4 Construct regular expressions to match words that:

###Start and end with the same character.

word_list <- c("Apple","LIL","Orange","Blue","ELLE")  
same_char <- "^(.).*\\1$"
word_list %>% 
  str_subset(same_char)

## [1] "LIL"  "ELLE"

Contain one letter repeated in at least three places (e.g. “eleven” contains three “e”s.)

repeated_thrice <- c("eleven","committee","Emimtmt")  
slicing <- "(.).\\1.\\1"
repeated_thrice %>% 
  str_subset(slicing)

## [1] "eleven"  "Emimtmt"

Contain a repeated pair of letters (e.g. “church” contains “ch” repeated twice.)

repeated_twice <- c("church","papa","apple")  
slicing1 <- "(.)(.).*\\1"
repeated_twice %>% 
  str_subset(slicing1)

## [1] "church" "papa"