Strings

Jeff Stevens

2025-03-07

Introduction

Mental model of data analysis

The problem

What’s different between these data sets?

What is needed to create data2 from data1?

data1

# A tibble: 12 × 3
      id cond  resp 
   <int> <chr> <chr>
 1     1 cond1 yes  
 2     2 cond2 no   
 3     3 cond1 no   
 4     4 cond2 no   
 5     5 cond1 yes  
 6     6 cond2 no   
 7     7 cond1 no   
 8     8 cond2 no   
 9     9 cond1 yes  
10    10 cond2 no   
11    11 cond1 yes  
12    12 cond2 no

data2

# A tibble: 12 × 4
      id cond  resp  output                           
   <int> <chr> <chr> <glue>                           
 1     1 1     Yes   1 had condition 1 and said "Yes" 
 2     2 2     No    2 had condition 2 and said "No"  
 3     3 1     No    3 had condition 1 and said "No"  
 4     4 2     No    4 had condition 2 and said "No"  
 5     5 1     Yes   5 had condition 1 and said "Yes" 
 6     6 2     No    6 had condition 2 and said "No"  
 7     7 1     No    7 had condition 1 and said "No"  
 8     8 2     No    8 had condition 2 and said "No"  
 9     9 1     Yes   9 had condition 1 and said "Yes" 
10    10 2     No    10 had condition 2 and said "No" 
11    11 1     Yes   11 had condition 1 and said "Yes"
12    12 2     No    12 had condition 2 and said "No"

Set-up

library(tidyverse)
library(palmerpenguins)

Mental model

Character string basics

Useful character vectors

letters

 [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"

LETTERS

 [1] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K" "L" "M" "N" "O" "P" "Q" "R" "S"
[20] "T" "U" "V" "W" "X" "Y" "Z"

month.name

 [1] "January"   "February"  "March"     "April"     "May"       "June"     
 [7] "July"      "August"    "September" "October"   "November"  "December"

month.abb

 [1] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"

Quotes

Create strings with either single quotes or double quotes

(string1 <- "This is a string")

[1] "This is a string"

writeLines(string1)

This is a string

(string2 <- 'So is this.')

[1] "So is this."

writeLines(string2)

So is this.

Quotes

Including quotes in strings

(string3 <- 'To include a "double quote" inside a string, I use single quotes')

[1] "To include a \"double quote\" inside a string, I use single quotes"

writeLines(string3)

To include a "double quote" inside a string, I use single quotes

(string4 <- "And 'vice versa'")

[1] "And 'vice versa'"

writeLines(string4)

And 'vice versa'

Escaping quotes

Or use \ to “escape” it

double_quote <- "\""
writeLines(double_quote)

single_quote <- '\''
writeLines(single_quote)

Escaping quotes

Because \ escapes, you can’t just wrap it in quotes

If you want an actual backslash printed, you need two \\

backslash <- "\\"
writeLines(backslash)

Working with strings

Strings with {stringr}

library(stringr)

String length

Return number of characters in a string with stringr::str_length()

(r4ds_string <- c("a", "R for data science", NA))

[1] "a"                  "R for data science" NA

str_length(r4ds_string)  # nchar() in base R

[1]  1 18 NA

How does this differ from length()?

length(r4ds_string)

[1] 3

Extracting strings

Extract parts of a string based on position with stringr::str_sub()

x <- c("apple", "banana", "pear")
str_sub(x, 1, 3)  # substr() in base R

[1] "app" "ban" "pea"

# Negative numbers count backwards from end
str_sub(x, -3, -1)

[1] "ple" "ana" "ear"

Extracting strings

Useful when you don’t have delimiters. But use delimiters!

set.seed(1)
penguins |>
  slice_sample(n=8) |> 
  mutate(species = str_sub(species, 1, 1),
         island = str_sub(island, 1, 3),
         year = str_sub(year, -2, -1), .keep = "used")

# A tibble: 8 × 3
  species island year 
  <chr>   <chr>  <chr>
1 C       Dre    09   
2 G       Bis    07   
3 A       Tor    09   
4 C       Dre    07   
5 G       Bis    09   
6 G       Bis    08   
7 C       Dre    08   
8 A       Dre    08

Extracting strings

Also can substitute characters based on position

[1] "apple"  "banana" "pear"

str_sub(x, 1, 1) <- "#"  # replace first character
x

[1] "#pple"  "#anana" "#ear"

str_sub(x, -1, -1) <- "*"  # replace last character
x

[1] "#ppl*"  "#anan*" "#ea*"

str_sub(x, 0, 0) <- "~"  # add before first character
x

[1] "~#ppl*"  "~#anan*" "~#ea*"

Changing case

Control capitalization with stringr::str_to_lower() and stringr::str_to_upper()

(y <- "hello, World")

[1] "hello, World"

str_to_lower(y)  # tolower() in base R

[1] "hello, world"

str_to_upper(y)  # toupper() in base R

[1] "HELLO, WORLD"

Changing case

Plus super useful stringr::str_to_title() and stringr::str_to_sentence()

str_to_title(y)

[1] "Hello, World"

str_to_sentence(y)

[1] "Hello, world"

Changing case

Useful to change case of column entries

penguins |>
  mutate(sex_upper = str_to_sentence(sex), .keep = "used")

# A tibble: 344 × 2
   sex    sex_upper
   <fct>  <chr>    
 1 male   Male     
 2 female Female   
 3 female Female   
 4 <NA>   <NA>     
 5 female Female   
 6 male   Male     
 7 female Female   
 8 male   Male     
 9 <NA>   <NA>     
10 <NA>   <NA>     
# ℹ 334 more rows

But notice what happened to data type

Combining strings

Combine multiple strings into a single string with stringr::str_c():

c("x", "y", "z")

[1] "x" "y" "z"

str_c("x", "y", "z")

[1] "xyz"

str_c("x", "y", "z", sep = ", ")

[1] "x, y, z"

Collapsing strings

Collapse a vector of strings into a single string with collapse argument

str_c(c("x", "y", "z"), collapse = ", ")

[1] "x, y, z"

How is this different from using sep argument?

str_c("x", "y", "z", sep = ", ")

[1] "x, y, z"

Collapsing strings

When would this be useful?

str_c(month.name, collapse = ", ")

[1] "January, February, March, April, May, June, July, August, September, October, November, December"

unique(penguins$species)

[1] Adelie    Gentoo    Chinstrap
Levels: Adelie Chinstrap Gentoo

str_c(sort(unique(penguins$species)), collapse = ", ")

[1] "Adelie, Chinstrap, Gentoo"

Combining strings with output

Pasting character vectors with base R paste()

name <- "Fred"
age <- 50
paste("My name is", name, ", and my age next year is", age + 1, ".")

[1] "My name is Fred , and my age next year is 51 ."

paste0("My name is", name, ", and my age next year is", age + 1, ".")

[1] "My name isFred, and my age next year is51."

paste0("My name is ", name, ", and my age next year is ", age + 1, ".")

[1] "My name is Fred, and my age next year is 51."

paste("My name is ", name, ", and my age next year is ", age + 1, ".", sep = "")

[1] "My name is Fred, and my age next year is 51."

Combining strings with output

Gluing character vectors with stringr::str_glue()

str_glue("My name is {name}, and my age next year is {age + 1}.")

My name is Fred, and my age next year is 51.

str_glue("My name is {name}, and my age next year is {age + 1}.", 
         name = "Jane", age = 40)

My name is Jane, and my age next year is 41.

Combining strings with output

Apply to each row of a data frame

penguins |>
  mutate(full_island = str_glue("{island} Island")) |> 
  arrange(bill_length_mm) |> 
  select(species, island, full_island)

# A tibble: 344 × 3
   species island    full_island     
   <fct>   <fct>     <glue>          
 1 Adelie  Dream     Dream Island    
 2 Adelie  Dream     Dream Island    
 3 Adelie  Torgersen Torgersen Island
 4 Adelie  Dream     Dream Island    
 5 Adelie  Torgersen Torgersen Island
 6 Adelie  Torgersen Torgersen Island
 7 Adelie  Biscoe    Biscoe Island   
 8 Adelie  Torgersen Torgersen Island
 9 Adelie  Torgersen Torgersen Island
10 Adelie  Biscoe    Biscoe Island   
# ℹ 334 more rows

Cheatsheet

Solving the problem

library(tidyverse)
nrows <- 12
set.seed(20250307)
data1 <- tibble(id = 1:nrows, 
                cond = rep(c("cond1", "cond2"), times = 6), 
                resp = sample(c("yes", "no"), nrows, replace = TRUE))

Solving the problem

data1

# A tibble: 12 × 3
      id cond  resp 
   <int> <chr> <chr>
 1     1 cond1 yes  
 2     2 cond2 no   
 3     3 cond1 no   
 4     4 cond2 no   
 5     5 cond1 yes  
 6     6 cond2 no   
 7     7 cond1 no   
 8     8 cond2 no   
 9     9 cond1 yes  
10    10 cond2 no   
11    11 cond1 yes  
12    12 cond2 no

data2

# A tibble: 12 × 4
      id cond  resp  output                           
   <int> <chr> <chr> <glue>                           
 1     1 1     Yes   1 had condition 1 and said "Yes" 
 2     2 2     No    2 had condition 2 and said "No"  
 3     3 1     No    3 had condition 1 and said "No"  
 4     4 2     No    4 had condition 2 and said "No"  
 5     5 1     Yes   5 had condition 1 and said "Yes" 
 6     6 2     No    6 had condition 2 and said "No"  
 7     7 1     No    7 had condition 1 and said "No"  
 8     8 2     No    8 had condition 2 and said "No"  
 9     9 1     Yes   9 had condition 1 and said "Yes" 
10    10 2     No    10 had condition 2 and said "No" 
11    11 1     Yes   11 had condition 1 and said "Yes"
12    12 2     No    12 had condition 2 and said "No"

Let’s code!

Strings