Basics of Strings

• A general programming term for a unit of character data is a string

  • Strings are a sequence of characters

  • In R, "strings" and "character data" are mostly interchangeable.

  • Some languages have more precise distinctions, but we won't worry about that here!

• We can create strings by surrounding text, numbers, spaces, or symbols with quotes!

  • Examples: "Hello! My name is Michael" or "%*$#01234"

Basics of Strings

R can treat strings in funny ways!

"01" == "1"

## [1] FALSE

"01" == 1

## [1] FALSE

"1" == 1

## [1] TRUE

Reminder: We can check data types using the class() function!

c(class("1"),class(1))

## [1] "character" "numeric"

Data: King County Restaurant Inspections!

Today we'll study real data on food safety inspections in King County, collected from data.kingcounty.gov.

Note these data are fairly large. The following code can be used to download the data directly from my Github page:

load(url("https://pearce790.github.io/CSSS508/Lectures/Lecture8/restaurants.Rdata"))

Quick Examination of the Data

names(restaurants)

##  [1] "Name"                       "Program_Identifier"        
##  [3] "Inspection_Date"            "Description"               
##  [5] "Address"                    "City"                      
##  [7] "Zip_Code"                   "Phone"                     
##  [9] "Longitude"                  "Latitude"                  
## [11] "Inspection_Business_Name"   "Inspection_Type"           
## [13] "Inspection_Score"           "Inspection_Result"         
## [15] "Inspection_Closed_Business" "Violation_Type"            
## [17] "Violation_Description"      "Violation_Points"          
## [19] "Business_ID"                "Inspection_Serial_Num"     
## [21] "Violation_Record_ID"        "Grade"                     
## [23] "Date"

dim(restaurants)

## [1] 258630     23

Quick Examination of the Data

Good Questions to Ask:

• What does each row represent?
• Is the data in long or wide format?
• What are the key variables?
• How are the data stored? (data type)

nchar()

The nchar() function calculates the number of characters in a given string.

• length() doesn't work with strings!!
• Why not?

nchar("Mike Pearce")

## [1] 11

In our restaurants data, let's see how many characters are in each zip code:

length_zip <- nchar(restaurants$Zip_Code)
table(length_zip)

## length_zip
##      5     10 
## 258629      1

substr()

The substr() function allows us to extract characters from a string.

For example, we can extract the third through fifth elements of a string as follows:

substr("98126",3,5)

## [1] "126"

substr()

Let's extract the first five chararacters from each zip code in the restaurants data, and add it to our dataset.

library(dplyr)
restaurants$ZIP_5 <- substr(restaurants$Zip_Code,1,5)
restaurants %>% distinct(ZIP_5) %>% head()

## # A tibble: 6 × 1
##   ZIP_5
##   <chr>
## 1 98126
## 2 98109
## 3 98101
## 4 98032
## 5 98102
## 6 98004

paste()

We combine strings together using paste(). By default, it puts a space between different strings.

For example, we can combine "Michael" and "Pearce" as follows:

paste("Michael","Pearce")

## [1] "Michael Pearce"

More complex paste() commands

There are two additional common arguments to use with paste():

1. sep= controls what separates vectors, entry-wise
2. collapse= controls if/how multiple outputs are collapsed into a single string.

paste("CSSS","508",sep= "_")

## [1] "CSSS_508"

paste(c("CSSS","STAT"),"508",sep= "_")

## [1] "CSSS_508" "STAT_508"

paste(c("CSSS","STAT"),"508",sep= "_",collapse=" , ")

## [1] "CSSS_508 , STAT_508"

When do we get one string as output vs. two?

paste()

Let's use paste() to create complete mailing addresses for each restaurant:

restaurants$mailing_address <- 
  paste(restaurants$Address,", ",
        restaurants$City,", WA ",restaurants$ZIP_5,
        sep = "")
restaurants %>% distinct(mailing_address) %>% head()

## # A tibble: 6 × 1
##   mailing_address                                  
##   <chr>                                            
## 1 2920 SW AVALON WAY, Seattle, WA 98126            
## 2 10 MERCER ST, Seattle, WA 98109                  
## 3 1001 FAIRVIEW AVE N Unit 1700A, SEATTLE, WA 98109
## 4 1225 1ST AVE, SEATTLE, WA 98101                  
## 5 18114 E VALLEY HWY, KENT, WA 98032               
## 6 121 11TH AVE E, SEATTLE, WA 98102

stringr

stringr is yet another R package from the Tidyverse (like ggplot2, dplyr, tidyr, lubridate, readr).

It provides TONS of functions for working with strings:

• Some are equivalent/better versions of Base R functions
• Some can do fancier tricks with strings

Most stringr functions begin with "str_" to make RStudio auto-complete more useful.

We'll cover the basics today, but know there's much more out there!

library(stringr)

Equivalencies: str_length()

str_length() is equivalent to nchar():

nchar("weasels")

## [1] 7

str_length("weasels")

## [1] 7

Equivalencies: str_sub()

str_sub() is like substr():

str_sub("Washington", 2,4)

## [1] "ash"

str_sub() also lets you put in negative values to count backwards from the end (-1 is the end, -3 is third from end):

str_sub("Washington", 4, -3)

## [1] "hingt"

Equivalencies: str_c()

str_c() ("string combine") is just like paste() but where the default is sep = "" (no space!)

str_c(c("CSSS","STAT"),508)

## [1] "CSSS508" "STAT508"

str_c(c("CSSS","STAT"),508,sep=" ")

## [1] "CSSS 508" "STAT 508"

str_c(c("CSSS","STAT"),508,sep = " ",collapse = ", ")

## [1] "CSSS 508, STAT 508"

Changing Cases

str_to_upper(), str_to_lower(), str_to_title() convert cases, which is often a good idea to do before searching for values:

unique_cities <- unique(restaurants$City)
unique_cities %>% head()

## [1] "Seattle"  "SEATTLE"  "KENT"     "BELLEVUE" "KENMORE"  "Issaquah"

str_to_upper(unique_cities) %>% head()

## [1] "SEATTLE"  "SEATTLE"  "KENT"     "BELLEVUE" "KENMORE"  "ISSAQUAH"

str_to_lower(unique_cities) %>% head()

## [1] "seattle"  "seattle"  "kent"     "bellevue" "kenmore"  "issaquah"

str_to_title(unique_cities) %>% head()

## [1] "Seattle"  "Seattle"  "Kent"     "Bellevue" "Kenmore"  "Issaquah"

Whitespace: str_trim()

Extra leading or trailing whitespace is common in text data:

unique_names <- unique(restaurants$Name)
unique_names %>% head(3)

## [1] "@ THE SHACK, LLC "    "10 MERCER RESTAURANT"
## [3] "100 LB CLAM"

We can remove the whitespace using str_trim():

str_trim(unique_names) %>% head(3)

## [1] "@ THE SHACK, LLC"     "10 MERCER RESTAURANT"
## [3] "100 LB CLAM"

Patterns!

It's common to want to see if a string satisfies a certain pattern.

We did this with numeric values earlier in this course!

cars %>% filter(speed < 5 | speed > 24)

##   speed dist
## 1     4    2
## 2     4   10
## 3    25   85

cars %>% filter(dist > 2 & dist <= 10)

##   speed dist
## 1     4   10
## 2     7    4
## 3     9   10

Patterns: str_detect()

We can do similar pattern-checking using str_detect():

str_detect(string,pattern)

• string is the character string (or vector of strings) we want to examine
• pattern is the pattern that we're checking for inside string
• Output: TRUE/FALSE vector indicating if pattern was found

str_detect(string = c("Hello","my name","is Michael"),
           pattern = "m")

## [1] FALSE  TRUE FALSE

str_detect(string = c("Hello","my name","is Michael"),
           pattern = "M")

## [1] FALSE FALSE  TRUE

Results are case-sensitive!!

Patterns: str_detect()

Let's see which phone numbers are in the 206 area code:

unique_phones <- unique(restaurants$Phone)
unique_phones %>% tail(4)

## [1] "(360) 698-0417" "(206) 525-7747" "(206) 390-9205"
## [4] "(425) 557-4474"

str_detect(unique_phones,"206") %>% tail(4)

## [1] FALSE  TRUE  TRUE FALSE

Replacement: str_replace()

What about if you want to replace a string with something else? Use str_replace()!

This function works very similarly to str_detect(), but with one extra argument:

str_replace(string, pattern, replacement)

• replacement is what pattern is substituted for.

str_replace(string="Hi, I'm Michael",
            pattern="Hi",replacement="Hello")

## [1] "Hello, I'm Michael"

Replacement: str_replace()

In the Date variable, let's replace each dash ("-") with an underscore ("_")

dates <- restaurants$Date
dates %>% tail(3)

## [1] "2017-03-21" "2017-03-21" "2016-10-10"

str_replace(dates,"-","_") %>% tail(3)

## [1] "2017_03-21" "2017_03-21" "2016_10-10"

Wait, what?

Replacement: str_replace_all()

str_replace() only changes the first instance of a pattern in each string!

If we want to replace all patterns, use str_replace_all()

dates <- restaurants$Date
dates %>% tail(3)

## [1] "2017-03-21" "2017-03-21" "2016-10-10"

str_replace_all(dates,"-","_") %>% tail(3)

## [1] "2017_03_21" "2017_03_21" "2016_10_10"

Activity: My Answers

The variable Inspection_Date is in the format "MM/DD/YYYY". In this question, we'll change the format using functions for strings.

1.How long is each character string in this variable?

table(nchar(restaurants$Inspection_Date))

## 
##     10 
## 258000

2.Use substr() to extract the month of each entry and save it to an object called "months"

3.Use substr() to extract the year of each entry and save it to an object called "years"

months <- substr(restaurants$Inspection_Date,1,2)
years <- substr(restaurants$Inspection_Date,7,10)

Activity: My Answers

4.Use paste() to combine each month and year, separated by an underscore (_). Save this as a new variable in the data called "Inspection_Date_Formatted"

restaurants$Inspection_Date_Formatted <- 
  paste(months,years,sep="_")
restaurants %>% 
  select(Name,Inspection_Date,Inspection_Date_Formatted) %>%
  head(5)

## # A tibble: 5 × 3
##   Name                   Inspection_Date Inspection_Date_Formatted
##   <chr>                  <chr>           <chr>                    
## 1 "@ THE SHACK, LLC "    <NA>            NA_NA                    
## 2 "10 MERCER RESTAURANT" 01/24/2017      01_2017                  
## 3 "10 MERCER RESTAURANT" 01/24/2017      01_2017                  
## 4 "10 MERCER RESTAURANT" 01/24/2017      01_2017                  
## 5 "10 MERCER RESTAURANT" 10/10/2016      10_2016

Activity: My Solutions

1. Filter your data to only include rows in which the Name includes the word "coffee" (in any case!)

coffee <- restaurants
coffee$Name <- str_to_lower(coffee$Name)
coffee <- coffee %>% filter(str_detect(Name,"coffee"))

Activity: My Solutions

2.Create a new variable in your data which includes the length of the business name, after removing beginning/trailing whitespace.

3. Create a new variable in your data for the inspection year.

coffee$NameLength <- str_length(str_trim(coffee$Name))
coffee$Year <- str_sub(coffee$Inspection_Date,-4,-1)

Activity: My Solutions

4. Create side-by-side boxplots for the length of business name vs. year.

library(ggplot2)
ggplot(coffee,aes(Year,NameLength))+geom_boxplot()

Activity: My Solutions

5. Calculate the maximum Inspection_Score by business and year.

coffee_summary <- coffee %>% group_by(Name,Year) %>% 
  summarize(MaxScore=max(Inspection_Score))

## `summarise()` has grouped output by 'Name'. You can override using
## the `.groups` argument.

Activity: My Solutions

6. Create a line plot of maximum score ("MaxScore") over time ("Year"), by business ("Name"). That is, you should have a single line for each business. (Don't try to label them, as there are far too many!)

ggplot(coffee_summary,aes(Year,MaxScore,group=Name))+
  geom_line(alpha=.2)