1.In code that doesn’t use stringr, you’ll often see paste() and paste0(). What’s the difference between the two functions? What stringr function are they equivalent to? How do the functions differ in their handling of NA?

The difference can be seen by the example below.

paste("Hello","World")

## [1] "Hello World"

paste0("Hello","World")

## [1] "HelloWorld"

• By paste(): There is a space between each string;
• By paste0(): There is NO space between each string.

The equivalent stringr functions are shown beolow.

paste0("Hello","World")

## [1] "HelloWorld"

str_c("Hello","World")

## [1] "HelloWorld"

• paste0(): equivalent to str_c().

Now, let’s figure out how do these functions differ in their handling of NA.

paste("Hello","World",NA)

## [1] "Hello World NA"

paste0("Hello","World",NA)

## [1] "HelloWorldNA"

str_c("Hello","World",NA)

## [1] NA

• paste() and paste0(): see NA as a string and return all the strings incluing NA
• str_c(): only NA is returned regardless of the other strings

2.In your own words, describe the difference between the sep and collapse arguments to str_c().

str_c(c("2018","SEP","1"),"Sunny",sep="-")

## [1] "2018-Sunny" "SEP-Sunny"  "1-Sunny"

str_c(c("2018","SEP","1"),"Sunny",collapse="-")

## [1] "2018Sunny-SEPSunny-1Sunny"

• By sep: add the sep argument behind each string, the number of output is same to that of input;
• By collapse: add the collapse argument behind each string and return all the strings together as one whole string, only ONE output.

3.Use str_length() and str_sub() to extract the middle character from a string. What will you do if the string has an even number of characters?

For strings with odd number of characters: extract the middle character from the string.

x1 <- "abc" #odd number
n_x1 <- str_length(x1)
str_sub(x1,(n_x1+1)/2,(n_x1+1)/2)

## [1] "b"

For strings with even number of characters: extract the middle two characters from the string.

x2 <- "abcd" # even number
n_x2 <- str_length(x2) # calculate the length of the strings
str_sub(x2,(n_x2)/2,(n_x2)/2+1)

## [1] "bc"

Now, an obvious question is: Can we extract the middle character in the same way no matter the string has an odd or even characters?

The answer is YES! It can be done by ceiling().

x3 <- c("a", "abc", "abcd", "abcde")
n_x3 <- str_length(x3)
str_sub(x3, ceiling(n_x3/2), n_x3/2+1) #extract from the ceiling for the middle character

## [1] "a"  "b"  "bc" "c"

5. What does str_trim() do? What’s the opposite of str_trim()?

• str_trim() deletes the whitespace from a string.

str_trim(" a b c ",side = "left") # delete from left

## [1] "a b c "

str_trim(" a b c ",side = "right") # delete from right

## [1] " a b c"

str_trim(" a b c ",side = "both")# delete from both

## [1] "a b c"

• The opposite of str_trim() is str_pad(), which add whitespace(s) to a string.

str_pad("a b c",width=6, side = "left") # add left side,total width is 6

## [1] " a b c"

str_pad("a b c",width=7, side = "both") # add both side,total width is 7

## [1] " a b c "

6. Write a function that turns (e.g.) a vector c(“a”, “b”, “c”) into the string “a, b, and c”. Think carefully about what it should do if given a vector of length 0, 1, or 2.

First, I write a function as below to turn a vector into a string as required.

vector_to_string <- function(x){
   if(length(x) == 0){
    return(str_c("Please enter a vector")) # with vector of length 0, it's better to show a error information
  }
   if(length(x) ==1){
    return(x) # with vector of length 1, return itself
  }
   if(length(x)==2){
      str0=str_c(x[1],", and ",x[2])
      return(str0)
   }
   else{
      str1 <- str_c(x[1:length(x)-1],collapse=", ") # the first part of the string
      str2 <- str_c(str1,", and ", x[length(x)])
      return(str2)
   }
}

Now, let’s test it with vectors of length 0,1,2 or more.

str_0 <- c()
vector_to_string(str_0)

## [1] "Please enter a vector"

str_1 <- c("A")
vector_to_string(str_1)

## [1] "A"

str_2 <- c("A","B")
vector_to_string(str_2)

## [1] "A, and B"

str_3 <- c("A","B","C")
vector_to_string(str_3)

## [1] "A, B, and C"

1. Explain why each of these strings don’t match a : “",”\“,”\".

Regexps use the backslash, , to escape special behaviour.

String	Meaning
\	excape the next character
\	resolve to  in the regular expression
\\	the first two backslashes will resolve to a literal backslash in the regular expression, the third will escape the next character

2. How would you match the sequence "’?

x <- "\"\'\\"
str_view(x, "\\\"\\'\\\\") 

3. What patterns will the regular expression ...... match? How would you represent it as a string?

x<-"\\..\\..\\.."
writeLines(x)

## \..\..\..

str_c(writeLines(x))

## \..\..\..

## character(0)

1. How would you match the literal string “\(^\)”?

x <- "$^$"
str_view(x, "^\\$\\^\\$$")

2.Given the corpus of common words in stringr::words, create regular expressions that find all words that:

1. Start with “y”.

str_view(words, "^y", match =TRUE)

2. End with “x”

str_view(words, "x$", match =TRUE)

3. Are exactly three letters long. (Don’t cheat by using str_length()!)

str_view(words, "^...$", match = TRUE)

4. Have seven letters or more.

str_view(words, ".......", match = TRUE)

1. Create regular expressions to find all words that:

1. Start with a vowel.

str_view(words, "^[aeiou]", match = TRUE)

2. That only contain consonants. (Hint: thinking about matching “not”-vowels.)

str_view(words, "^[^aeiou]+$", match=TRUE)

3. End with ed, but not with eed.

str_view(words, "^ed$|[^e]ed$", match = TRUE)

4. End with ing or ise.

str_view(words, "i(ng|se)$", match = TRUE)

2. Empirically verify the rule “i before e except after c”.

str_view(words, "(cei|[^c]ie)", match = TRUE)

str_view(words, "(cie|[^c]ei)", match = TRUE)

3. Is “q” always followed by a “u”?

str_view(words, "q[^u]", match = TRUE)

We get no return, so the answer is NO.

1. Describe the equivalents of ?, +, * in {m,n} form.

expression	equivalents in {m,n} form
?	{0,1}
+	{1,}
*	{0,}

2. Describe in words what these regular expressions match: (read carefully to see if I’m using a regular expression or a string that defines a regular expression.)

1.^.*$ matches any string.

2. "\\{.+\\}" matches any string which contains {xx}, where xx can be any character(s).

3. \d{4}-\d{2}-\d{2} mastches xxxx-xx-xx, where x represents a digit.

4. "\\\\{4}" matches four backslashes.

3. Create regular expressions to find all words that:

1. Start with three consonants.

str_view(words, "^[^aeiou]{3}")

2. Have three or more vowels in a row.

str_view(words, "[aeiou]{3,}")

3. Have two or more vowel-consonant pairs in a row.

str_view(words, "([aeiou][^aeiou]){2,}")

1.For each of the following challenges, try solving it by using both a single regular expression, and a combination of multiple str_detect() calls.

1. Find all words that start or end with x.

A single regular expression:

str_subset(words,"(^x)|(x$)")

## [1] "box" "sex" "six" "tax"

A combination of multiple str_detect():

str_start <- str_detect(words, "^x")
str_end <- str_detect(words, "x$")
words[str_start|str_end]

## [1] "box" "sex" "six" "tax"

2. Find all words that start with a vowel and end with a consonant.

A single regular expression:

str_subset(words,"^[aeiou].*[^aeiou]$")%>%
   head() # just show the first six here to avoid large output

## [1] "about"   "accept"  "account" "across"  "act"     "actual"

A combination of multiple str_detect():

s1<-str_detect(words, "^[aeiou]")
s2<-str_detect(words,"[^aeiou]$")
head(words[s1&s2])

## [1] "about"   "accept"  "account" "across"  "act"     "actual"

3. Are there any words that contain at least one of each different vowel?

str_subset(words,"^[aeiou].*[aeiou]")%>%
   head() # just show the first six here to avoid large output

## [1] "able"     "about"    "absolute" "accept"   "account"  "achieve"

2. What word has the highest number of vowels? What word has the highest proportion of vowels? (Hint: what is the denominator?)

Find the word with the highest number of vowels:

n_max <- max(str_count(words,"[aeiou]")) # highest number of vowels
words[str_count(words,"[aeiou]")==n_max]

## [1] "appropriate" "associate"   "available"   "colleague"   "encourage"  
## [6] "experience"  "individual"  "television"

Find the word with the highest proportion of vowels:

n_vowels <- str_count(words,"[aeiou]") # count the number of vowels of each word
n_total <- str_count(words,".") # count the number of character of each word
max(n_vowels/n_total) # highest proportion of vowels

## [1] 1

words[n_vowels/n_total==max(n_vowels/n_total)]

## [1] "a"

1. In the previous example, you might have noticed that the regular expression matched “flickered”, which is not a colour. Modify the regex to fix the problem.

I match the boundary between words with \b.

colours <- c("red", "orange", "yellow", "green", "blue", "purple")
colour_match <- str_c(colours, collapse = "|")
b_colour_match <- str_c("\\b", colour_match, "\\b")
b_colour_match

## [1] "\\bred|orange|yellow|green|blue|purple\\b"

more <- sentences[str_count(sentences, b_colour_match) > 1]
str_view_all(more, colour_match)

2. From the Harvard sentences data, extract:

1. The first word from each sentence.

n1<-"^[^ ]+"
str_extract(sentences,n1) %>%
   head()

## [1] "The"   "Glue"  "It's"  "These" "Rice"  "The"

2. All words ending in ing.

n2<-"([^ ]+)ing"
s<-str_subset(sentences,n2)
str_extract(s,n2) %>%
   head()

## [1] "stocking" "spring"   "evening"  "morning"  "winding"  "living"

1.How would you find all strings containing  with regex() vs. with fixed()?

Let’s create test strings test_s first to find all strings containing .

test_s <- c("a\\b\\c","1\\2","hello")

Use regex():

str_view_all(test_s, regex(pattern = "\\\\"))

Use fixed():

str_detect(test_s, fixed(pattern = "\\"))

## [1]  TRUE  TRUE FALSE

2.What are the five most common words in sentences?

Install the library tidytext,then we can find the most common words

library(tidytext)
library(dplyr)
data_frame(text=sentences) %>% 
    unnest_tokens(word, text) %>%    # split words
    count(word, sort = TRUE) %>% # count occurrences
    head(5)

## # A tibble: 5 x 2
##   word      n
##   <chr> <int>
## 1 the     751
## 2 a       202
## 3 of      132
## 4 to      123
## 5 and     118

We can see that the five most common words are “a”, “an”, “the” etc. Futhermore, if we want take out these words, we can do it as below.

data_frame(text=sentences) %>% 
    unnest_tokens(word, text) %>%    # split words
    anti_join(stop_words) %>%    # take out "a", "an", "the", etc.
    count(word, sort = TRUE) %>% # count occurrences
    head(5)

## Joining, by = "word"

## # A tibble: 5 x 2
##   word       n
##   <chr>  <int>
## 1 red       11
## 2 fine      10
## 3 green     10
## 4 hot       10
## 5 strong    10

1. Find the stringi functions that:

1.Count the number of words.

stri_count_words("Today is the deadline of hw06")

## [1] 6

2. Find duplicated strings.

stri_duplicated(c("a","b","c",NA,"a","b","d","e",NA))

## [1] FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE  TRUE

stri_duplicated_any(c("a","b","c",NA,"a","b","d","e",NA))

## [1] 5

• stri_duplicated() shows whether this string is duplicated or not.
• stri_duplicated_any() shows the number of duplicated strings.

3. Generate random text. Random generation can be done by stri_rand_strings() and stri_rand_shuffle().

Generate random strings of specific length:

stri_rand_strings(5, 10) # 5 strings of length 10

## [1] "KonKuDMG8s" "IOHfygXKAo" "xNYl1qKwsw" "UXE61EmnZ8" "tHpUOlzUbw"

Generate random strings of random length:

stri_rand_strings(5, sample(1:10, 5, replace=TRUE)) # 5 strings of random lengths

## [1] "hmLPlNJ"   "E"         "KJKj"      "BqM"       "oMnJqtqAW"

Generate a string consisting of at least one digit, small and big ASCII letter, which is quite useful in creating a random password:

n <- 10 # number of strings
stri_rand_shuffle(stri_paste(
   stri_rand_strings(n, 1, '[0-9]'),
   stri_rand_strings(n, 1, '[a-z]'),
   stri_rand_strings(n, 1, '[A-Z]'),
   stri_rand_strings(n, sample(5:11, 5, replace=TRUE), '[a-zA-Z0-9]')
))

##  [1] "bRx7hwo3"       "Lt4q7J8o9yBB"   "YsupFizoRAJR4"  "b6rTTfTDHa"    
##  [5] "6MCQ1TnMe4vGuI" "w2qvJ0ZH"       "S3UBr7bOMmAc"   "1PjnJa93FydG3" 
##  [9] "T7yOpz4nln"     "L5K8tZVzdxN0ii"

2.How do you control the language that stri_sort() uses for sorting?

In order to control the language, we can set a locale to use when sorting by stri_sort(..., locale = ...).

stri_sort(c("hladny", "chladny"), locale="pl_PL")

## [1] "chladny" "hladny"

stri_sort(c("hladny", "chladny"), locale="sk_SK")

## [1] "hladny"  "chladny"