Chapter 15

Strings

Author

Aditya Dahiya

Published

September 10, 2023

Important uses of quoting in R: –

Code Purpose
⁠\n⁠ newline (aka ‘line feed’)
\t⁠ tab
\b⁠ backspace
⁠\v⁠ vertical tab
\\⁠ backslash ‘⁠\⁠’
\'⁠ ASCII apostrophe ‘⁠'⁠’
⁠\"⁠ ASCII quotation mark ‘⁠"⁠’
⁠\nnn⁠ character with given octal code (1, 2 or 3 digits)
⁠\xnn⁠ character with given hex code (1 or 2 hex digits)

15.2.4 Exercises

Question 1

Create strings that contain the following values:

  1. He said "That's amazing!"

    x = "He said \"That's amazing!\""
str_view(x)
    [1] │ He said "That's amazing!"

  2. \a\b\c\d

    x = "\\a\\b\\c\\d"
str_view(x)
    [1] │ \a\b\c\d

  3. \\\\\\

    x = "\\\\\\\\\\\\"
str_view(x)
    [1] │ \\\\\\

Question 2

Create the string in your R session and print it. What happens to the special “\u00a0”? How does str_view() display it? Can you do a little googling to figure out what this special character is?

x <- "This\u00a0is\u00a0tricky"

The "\u00a0" represents a white space. By google, I find out that this represents No-Break Space (NBSP). But, str_view() displays it in form of a greenish-blue font {\u00a0}.

"\u00a0" # This represents a white space
[1] " "
str_view("\u00a0")
[1] │ {\u00a0}
x <- "This\u00a0is\u00a0tricky"
print(x)
[1] "This is tricky"
str_view(x)
[1] │ This{\u00a0}is{\u00a0}tricky

The "\u00a0" represents a non-breaking space character in Unicode encoding. Unicode is a standardized character encoding system that assigns a unique numerical code to almost every character from every writing system in the world, including various symbols, letters, and special characters.

In Unicode, “\u” is used to indicate that the following four characters represent a Unicode code point in hexadecimal notation. In this case, "\u00a0" represents the code point for the non-breaking space character.

A non-breaking space is a type of space character that is used in typography and word processing to prevent a line break or word wrap from occurring at that particular space.

It is similar to a regular space character (ASCII code 32), but it has the special property of keeping adjacent words or characters together on the same line when text is justified or formatted.

15.3.4 Exercises

Question 1

Compare and contrast the results of paste0() with str_c() for the following inputs:

str_c("hi ", NA) 
str_c(letters[1:2], letters[1:3])

As we can see below, paste0 converts NA into a string "NA" and simply joins it with another string. However, str_c() behaves more sensibly - it generates NA if any of the strings being joined is NA.

str_c("hi ", NA)
[1] NA
paste0("hi ", NA)
[1] "hi NA"

Further, we see below that we are joining two string vectors of unequal length, i.e., letters[1:2] is "a" "b" and letters[1:3] is "a" "b" "c" , both str_c() and paste0() behave differently.

  • str_c() throws an error and informs us that the string vectors being joined are of unequal length.

  • paste0 simple recycles the shorter string vector silently.

# str_c(letters[1:2], letters[1:3])
paste0(letters[1:2], letters[1:3])
[1] "aa" "bb" "ac"

Question 2

What’s the difference between paste() and paste0()? How can you recreate the equivalent of paste() with str_c()?

In R, both paste() and paste0() functions are used to concatenate strings together. However, they differ in how they handle separating the concatenated elements.

paste() concatenates its arguments with a space character as the default separator. We can specify a different separator using the sep argument.

paste0() is similar to paste(), but it does not add any separator between the concatenated elements. It simply combines them as-is.

Here is an example: –

vec1 <- c("Hello", "Hi")
vec2 <- c("Amy", "Tom", "Neal")
paste(vec1, vec2)
[1] "Hello Amy"  "Hi Tom"     "Hello Neal"
paste(vec1, vec2, sep = ", ")
[1] "Hello, Amy"  "Hi, Tom"     "Hello, Neal"
paste0(vec1, vec2)
[1] "HelloAmy"  "HiTom"     "HelloNeal"

We can recreate the equivalent of paste() using the str_c() function from the stringr package in R. To do this, we can specify the separator using the sep argument in str_c() as follows: –

vec1 <- c(vec1, "Hallo")
paste(vec1, vec2)
[1] "Hello Amy"  "Hi Tom"     "Hallo Neal"
str_c(vec1, vec2, sep = " ")
[1] "Hello Amy"  "Hi Tom"     "Hallo Neal"

Note: We had to add a string to vec1 so that both vec1 and vec2 are of length 3. Else, str_c will throw up an error.

Question 3

Convert the following expressions from str_c() to str_glue() or vice versa:

  1. str_c("The price of ", food, " is ", price)

    str_glue("The price of {food} is {price}")

  2. str_glue("I'm {age} years old and live in {country}")

    str_c("I'm ", age, " years old and live in ", country)

  3. str_c("\\section{", title, "}")

    str_glue("\\\\section{{{title}}}")

asd

find the distribution of lengths of US babynames and then with filter() to look at the longest names, which happen to have 15 letters

data("babynames")
babynames |>
  mutate(name_lgth = str_length(name)) |>
  count(name_lgth, wt = n)
# A tibble: 14 × 2
   name_lgth        n
       <int>    <int>
 1         2   338150
 2         3  8589596
 3         4 48506739
 4         5 87011607
 5         6 90749404
 6         7 72120767
 7         8 25404066
 8         9 11926551
 9        10  1306159
10        11  2135827
11        12    16295
12        13    10845
13        14     3681
14        15      830
babynames |>
  filter(str_length(name) == 15) |>
  count(name, wt = n, sort = TRUE) |>
  slice_head(n = 5) |>
  select(name) |>
  as_vector() |>
  unname() |>
  str_sub(start = -3, end = -1)
[1] "ier" "ohn" "her" "ame" "ich"

15.5.3 Exercises

Question 1

When computing the distribution of the length of babynames, why did we use wt = n?

The babynames data-set (Table 1) displays the column n to reflect the frequency, i.e., number of observations of that name in that year. Thus, when we are computing the distribution of the length of baby names (Table 2), we need to weigh the observations by n otherwise each row will be treated as 1 (Table 2 column 3), instead of the actual number reflected in n leading to erroneous results.

Code 
babynames |>
  slice_head(n = 5) |>
  gt() |>
  fmt_number(prop, decimals = 4)
Table 1:

The babynames data-set
year sex name n prop
1880 F Mary 7065 0.0724
1880 F Anna 2604 0.0267
1880 F Emma 2003 0.0205
1880 F Elizabeth 1939 0.0199
1880 F Minnie 1746 0.0179
Code 
df1 = babynames |>
  mutate(name_length = str_length(name)) |>
  count(name_length, wt = n) |>
  rename(correct_frequency = n)

df2 = babynames |>
  mutate(name_length = str_length(name)) |>
  count(name_length) |>
  rename(wrong_frequency_without_weights = n)

inner_join(df1, df2, by = "name_length") |>
  gt() |>
  fmt_number(-name_length , decimals = 0) |>
  cols_label_with(
    fn = ~ janitor::make_clean_names(., case = "title")
  ) |>
  gt_theme_538()
Table 2:

The distribution of the length of babynames
Name Length Correct Frequency Wrong Frequency without Weights
2 338,150 4,660
3 8,589,596 41,274
4 48,506,739 177,838
5 87,011,607 404,291
6 90,749,404 546,519
7 72,120,767 424,360
8 25,404,066 213,803
9 11,926,551 78,946
10 1,306,159 23,437
11 2,135,827 6,461
12 16,295 1,610
13 10,845 946
14 3,681 390
15 830 130

Question 2

Use str_length() and str_sub() to extract the middle letter from each baby name. What will you do if the string has an even number of characters?

The code displayed below extracts the middle letter from each baby name, and the results for first 10 names are displayed in Table 3 . If the string has an even number of characters, we can pick the middle two characters.

df3 = babynames |>
  mutate(
    name_length = str_length(name),
    middle_letter_start = if_else(name_length %% 2 == 0,
                             name_length/2,
                             (name_length/2) + 0.5),
    middle_letter_end = if_else(name_length %% 2 == 0,
                                (name_length/2) + 1,
                                (name_length/2) + 0.5),
    middle_letter = str_sub(name, 
                            start = middle_letter_start, 
                            end = middle_letter_end)
  ) |>
  select(-c(year, sex, n, prop)) |>
  slice_head(n = 10)

df3 |>
  gt() |>
  cols_label_with(fn = ~ janitor::make_clean_names(., case = "title")) |>
  cols_align(align = "center",
             columns = -name) |>
  gt_theme_538()
Table 3:

Middle letters of names
Name Name Length Middle Letter Start Middle Letter End Middle Letter
Mary 4 2 3 ar
Anna 4 2 3 nn
Emma 4 2 3 mm
Elizabeth 9 5 5 a
Minnie 6 3 4 nn
Margaret 8 4 5 ga
Ida 3 2 2 d
Alice 5 3 3 i
Bertha 6 3 4 rt
Sarah 5 3 3 r

Question 3

Are there any major trends in the length of baby names over time? What about the popularity of first and last letters?

The Figure 1, Figure 2 and Figure 3 show the trends over time.

Code 
df4 = babynames |>
  mutate(
    name_length = str_length(name),
    name_start = str_sub(name, 1, 1),
    name_end = str_sub(name, -1, -1)
  )
y_coord = c(5.4, 6.3)
df4 |>
  group_by(year) |>
  count(name_length, wt = n) |>
  summarise(mean_length = weighted.mean(name_length, w = n)) |>
  ggplot(aes(x = year, y = mean_length)) +
  theme_classic() +
  labs(y = "Average name length (for each year)",
       x = "Year", 
       title = "Baby names have become longer over the past 12 decades",
       subtitle = "Between 1890-1920, and 1960-1990 baby names became longer\nBut, since 1990 the names are becoming shorter again") +
  scale_x_continuous(breaks = seq(1880, 2000, 20)) +
  geom_rect(mapping = aes(xmin = 1890, xmax = 1920,
                          ymin = y_coord[1], ymax = y_coord[2]),
            alpha = 0.01, fill = "grey") +
  geom_rect(mapping = aes(xmin = 1960, xmax = 1990,
                          ymin = y_coord[1], ymax = y_coord[2]),
            alpha = 0.01, fill = "grey") +
  geom_line(lwd = 1) +
  coord_cartesian(ylim = y_coord) +
  theme(plot.title.position = "plot")

Figure 1: Length of babynames over time
Code 
ns_vec = df4 |>
  count(name_start, wt = n, sort = TRUE) |>
  slice_head(n = 5) |>
  select(name_start) |>
  as_vector() |>
  unname()

df4 |>
  filter(name_start %in% ns_vec) |>
  group_by(year) |>
  count(name_start, wt = n) |>
  mutate(prop = 100*n/sum(n)) |>
  mutate(lbl = if_else(year == 2017, 
                       name_start, 
                       NA)) |>
  ggplot(aes(x = year, y = prop, 
             col = name_start, label = lbl)) +
  geom_line(lwd = 1) +
  ggrepel::geom_label_repel(nudge_x = 1) +
  labs(x = "Year",
       y = "Percentage of names starting with character",
       title = "People's preferences for baby names' starting letter change over time",
       subtitle = "Names starting with A are most popular now\nNames starting with J were popular in the 1940s\nIn 1950s, names starting with D became popular, while those starting with A lost popularity") +
  theme_classic() +
  theme(legend.position = "none",
        plot.title.position = "plot") +
  scale_x_continuous(breaks = seq(1880, 2020, 20))

Figure 2: Trends on the starting letter of babynames over time
Code 
ns_vec = df4 |>
  count(name_end, wt = n, sort = TRUE) |>
  slice_head(n = 5) |>
  select(name_end) |>
  as_vector() |>
  unname()

df4 |>
  filter(name_end %in% ns_vec) |>
  group_by(year) |>
  count(name_end, wt = n) |>
  mutate(prop = 100*n/sum(n)) |>
  mutate(lbl = if_else(year == 2017, 
                       name_end, 
                       NA)) |>
  ggplot(aes(x = year, y = prop, 
             col = name_end, label = lbl)) +
  geom_line(lwd = 1) +
  ggrepel::geom_label_repel(nudge_x = 1) +
  labs(x = "Year",
       y = "Percentage of names ending with character",
       title = "People's preferences for baby names' ending letter change over time",
       subtitle = "Names ending in N have risen in popularity over the decades.\nNames ending with E have become less popular over time") +
  theme_classic() +
  theme(legend.position = "none",
        plot.title.position = "plot") +
  scale_x_continuous(breaks = seq(1880, 2020, 20))

Figure 3: Trends on the ending letter of babynames over time