In this lab activity, we’ll discuss Data Wrangling/Transformation via the dplyr package. We’ll explore ways to choose subsets of data, aggregate data to create summaries, make new variables, and sort your data frames. It is recommended you also explore the RStudio Cheatsheet on Data Transformation as we discuss this content.

Back to `gapminder`

Here is a look at the gapminder data frame in the gapminder package.

library(gapminder)
rmarkdown::paged_table(gapminder)

# Has been set as the default way to print data frames
# via df_print: paged in the YAML at the top

Say we wanted mean life expectancy across all years for Asia

# Base R
asia <- gapminder[gapminder$continent == "Asia", ]
mean(asia$lifeExp)

[1] 60.0649

library(dplyr)
gapminder %>% 
  filter(continent == "Asia") %>%
  summarize(mean_exp = mean(lifeExp))

# A tibble: 1 × 1
  mean_exp
     <dbl>
1     60.1

The pipe `%>%`

A way to chain together commands
It is essentially the dplyr equivalent to the + in ggplot2

The Five Main Verbs (5MV) of data wrangling

`filter()`
`summarize()`
`group_by()`
`mutate()`
`arrange()`

`filter()`

Select a subset of the rows of a data frame.
The arguments are the “filters” that you’d like to apply.

library(gapminder); library(dplyr)
gap_2007 <- gapminder %>% filter(year == 2007)
gap_2007

# A tibble: 142 × 6
   country     continent  year lifeExp       pop gdpPercap
   <fct>       <fct>     <int>   <dbl>     <int>     <dbl>
 1 Afghanistan Asia       2007    43.8  31889923      975.
 2 Albania     Europe     2007    76.4   3600523     5937.
 3 Algeria     Africa     2007    72.3  33333216     6223.
 4 Angola      Africa     2007    42.7  12420476     4797.
 5 Argentina   Americas   2007    75.3  40301927    12779.
 6 Australia   Oceania    2007    81.2  20434176    34435.
 7 Austria     Europe     2007    79.8   8199783    36126.
 8 Bahrain     Asia       2007    75.6    708573    29796.
 9 Bangladesh  Asia       2007    64.1 150448339     1391.
10 Belgium     Europe     2007    79.4  10392226    33693.
# ℹ 132 more rows

Use == to compare a variable to a value

Logical operators

Use | to check for any in multiple filters being true:

gapminder %>% 
  filter(year == 2002 | continent == "Europe")

# A tibble: 472 × 6
   country     continent  year lifeExp      pop gdpPercap
   <fct>       <fct>     <int>   <dbl>    <int>     <dbl>
 1 Afghanistan Asia       2002    42.1 25268405      727.
 2 Albania     Europe     1952    55.2  1282697     1601.
 3 Albania     Europe     1957    59.3  1476505     1942.
 4 Albania     Europe     1962    64.8  1728137     2313.
 5 Albania     Europe     1967    66.2  1984060     2760.
 6 Albania     Europe     1972    67.7  2263554     3313.
 7 Albania     Europe     1977    68.9  2509048     3533.
 8 Albania     Europe     1982    70.4  2780097     3631.
 9 Albania     Europe     1987    72    3075321     3739.
10 Albania     Europe     1992    71.6  3326498     2497.
# ℹ 462 more rows

Use & or , to check for all of multiple filters being true:

gapminder %>% 
  filter(year == 2002, continent == "Europe")

# A tibble: 30 × 6
   country                continent  year lifeExp      pop gdpPercap
   <fct>                  <fct>     <int>   <dbl>    <int>     <dbl>
 1 Albania                Europe     2002    75.7  3508512     4604.
 2 Austria                Europe     2002    79.0  8148312    32418.
 3 Belgium                Europe     2002    78.3 10311970    30486.
 4 Bosnia and Herzegovina Europe     2002    74.1  4165416     6019.
 5 Bulgaria               Europe     2002    72.1  7661799     7697.
 6 Croatia                Europe     2002    74.9  4481020    11628.
 7 Czech Republic         Europe     2002    75.5 10256295    17596.
 8 Denmark                Europe     2002    77.2  5374693    32167.
 9 Finland                Europe     2002    78.4  5193039    28205.
10 France                 Europe     2002    79.6 59925035    28926.
# ℹ 20 more rows

Use %in% to check for any being true (shortcut to using | repeatedly with ==)

gapminder %>% 
  filter(country %in% c("Argentina", "Belgium", "Mexico"),
         year %in% c(1987, 1992))

# A tibble: 6 × 6
  country   continent  year lifeExp      pop gdpPercap
  <fct>     <fct>     <int>   <dbl>    <int>     <dbl>
1 Argentina Americas   1987    70.8 31620918     9140.
2 Argentina Americas   1992    71.9 33958947     9308.
3 Belgium   Europe     1987    75.4  9870200    22526.
4 Belgium   Europe     1992    76.5 10045622    25576.
5 Mexico    Americas   1987    69.5 80122492     8688.
6 Mexico    Americas   1992    71.5 88111030     9472.

`summarize()`

Any numerical summary that you want to apply to a column of a data frame is specified within summarize().

max_exp_1997 <- gapminder %>% 
  filter(year == 1997) %>% 
  summarize(max_exp = max(lifeExp))
max_exp_1997

# A tibble: 1 × 1
  max_exp
    <dbl>
1    80.7

Combining `summarize()` with `group_by()`

When you’d like to determine a numerical summary for all levels of a different categorical variable

max_exp_1997_by_cont <- gapminder %>% 
  filter(year == 1997) %>% 
  group_by(continent) %>%
  summarize(max_exp = max(lifeExp))
max_exp_1997_by_cont

# A tibble: 5 × 2
  continent max_exp
  <fct>       <dbl>
1 Africa       74.8
2 Americas     78.6
3 Asia         80.7
4 Europe       79.4
5 Oceania      78.8

Without the `%>%`

It’s hard to appreciate the %>% without seeing what the code would look like without it:

max_exp_1997_by_cont <- 
  summarize(
    group_by(
      filter(
        gapminder, 
          year == 1997), 
      continent),
    max_exp = max(lifeExp))
max_exp_1997_by_cont

# A tibble: 5 × 2
  continent max_exp
  <fct>       <dbl>
1 Africa       74.8
2 Americas     78.6
3 Asia         80.7
4 Europe       79.4
5 Oceania      78.8

`mutate()`

Allows you to
1. create a new variable based on other variables OR
2. change the contents of an existing variable

create a new variable based on other variables

gap_w_gdp <- gapminder %>% mutate(gdp = pop * gdpPercap)
gap_w_gdp

# A tibble: 1,704 × 7
   country     continent  year lifeExp      pop gdpPercap          gdp
   <fct>       <fct>     <int>   <dbl>    <int>     <dbl>        <dbl>
 1 Afghanistan Asia       1952    28.8  8425333      779.  6567086330.
 2 Afghanistan Asia       1957    30.3  9240934      821.  7585448670.
 3 Afghanistan Asia       1962    32.0 10267083      853.  8758855797.
 4 Afghanistan Asia       1967    34.0 11537966      836.  9648014150.
 5 Afghanistan Asia       1972    36.1 13079460      740.  9678553274.
 6 Afghanistan Asia       1977    38.4 14880372      786. 11697659231.
 7 Afghanistan Asia       1982    39.9 12881816      978. 12598563401.
 8 Afghanistan Asia       1987    40.8 13867957      852. 11820990309.
 9 Afghanistan Asia       1992    41.7 16317921      649. 10595901589.
10 Afghanistan Asia       1997    41.8 22227415      635. 14121995875.
# ℹ 1,694 more rows

`mutate()`

change the contents of an existing variable

gap_weird <- gapminder %>% mutate(pop = pop + 1000)
gap_weird

# A tibble: 1,704 × 6
   country     continent  year lifeExp      pop gdpPercap
   <fct>       <fct>     <int>   <dbl>    <dbl>     <dbl>
 1 Afghanistan Asia       1952    28.8  8426333      779.
 2 Afghanistan Asia       1957    30.3  9241934      821.
 3 Afghanistan Asia       1962    32.0 10268083      853.
 4 Afghanistan Asia       1967    34.0 11538966      836.
 5 Afghanistan Asia       1972    36.1 13080460      740.
 6 Afghanistan Asia       1977    38.4 14881372      786.
 7 Afghanistan Asia       1982    39.9 12882816      978.
 8 Afghanistan Asia       1987    40.8 13868957      852.
 9 Afghanistan Asia       1992    41.7 16318921      649.
10 Afghanistan Asia       1997    41.8 22228415      635.
# ℹ 1,694 more rows

`arrange()`

Reorders the rows in a data frame based on the values of one or more variables

gapminder %>%
  arrange(year, country)

# A tibble: 1,704 × 6
   country     continent  year lifeExp      pop gdpPercap
   <fct>       <fct>     <int>   <dbl>    <int>     <dbl>
 1 Afghanistan Asia       1952    28.8  8425333      779.
 2 Albania     Europe     1952    55.2  1282697     1601.
 3 Algeria     Africa     1952    43.1  9279525     2449.
 4 Angola      Africa     1952    30.0  4232095     3521.
 5 Argentina   Americas   1952    62.5 17876956     5911.
 6 Australia   Oceania    1952    69.1  8691212    10040.
 7 Austria     Europe     1952    66.8  6927772     6137.
 8 Bahrain     Asia       1952    50.9   120447     9867.
 9 Bangladesh  Asia       1952    37.5 46886859      684.
10 Belgium     Europe     1952    68    8730405     8343.
# ℹ 1,694 more rows

Can also put into descending order

gapminder %>%
  filter(year > 2000) %>%
  arrange(desc(lifeExp))

# A tibble: 284 × 6
   country          continent  year lifeExp       pop gdpPercap
   <fct>            <fct>     <int>   <dbl>     <int>     <dbl>
 1 Japan            Asia       2007    82.6 127467972    31656.
 2 Hong Kong, China Asia       2007    82.2   6980412    39725.
 3 Japan            Asia       2002    82   127065841    28605.
 4 Iceland          Europe     2007    81.8    301931    36181.
 5 Switzerland      Europe     2007    81.7   7554661    37506.
 6 Hong Kong, China Asia       2002    81.5   6762476    30209.
 7 Australia        Oceania    2007    81.2  20434176    34435.
 8 Spain            Europe     2007    80.9  40448191    28821.
 9 Sweden           Europe     2007    80.9   9031088    33860.
10 Israel           Asia       2007    80.7   6426679    25523.
# ℹ 274 more rows

Other useful `dplyr` verbs

select
top_n
sample_n
slice
glimpse
rename

Your Task

Determine which African country had the highest GDP per capita in 1982 using the gapminder data in the gapminder package.

gapminder %>%
  filter(continent == "Africa", year == 1982) %>%
  arrange(desc(gdpPercap)) %>%
  slice(1)

# A tibble: 1 × 6
  country continent  year lifeExp     pop gdpPercap
  <fct>   <fct>     <int>   <dbl>   <int>     <dbl>
1 Libya   Africa     1982    62.2 3344074    17364.

filter(continent == “Africa”, year == 1982) %>%-This selects only African countries in the year 1982 (arrange(desc(gdpPercap))- sorts countries from highest to lowest GDP per capita slice (1)- Extracts the single country at the top of the list — the highest GDP per capita.

Challenge

For both of these problems below, use the bechdel data frame in the fivethirtyeight package:

Use the count function in the dplyr package to determine how many movies in 2013 fell into each of the different categories for clean_test

bechdel %>%
  filter(year == 2013) %>%
  count(clean_test)

# A tibble: 5 × 2
  clean_test     n
  <ord>      <int>
1 nowomen        6
2 notalk        23
3 men           15
4 dubious        9
5 ok            46

filter(year == 2013) - Restricts the dataset to movies released in 2013. count(clean_test) -Counts how many movies fall into each clean_test category

Determine the percentage of movies that received the value of "ok" across all years

Your Task

Determine the top five movies in terms of domestic return on investment for 2013 scaled data using the bechdel data frame in the fivethirtyeight package.

library(fivethirtyeight)
bechdel %>%
  filter(year == 2013) %>%
  mutate(domestic_roi = (domgross - budget) / budget) %>%
  arrange(desc(domestic_roi)) %>%
  select(title, year, budget, domgross, domestic_roi) %>%
  slice_head(n = 5)

# A tibble: 5 × 5
  title                 year   budget  domgross domestic_roi
  <chr>                <int>    <int>     <dbl>        <dbl>
1 The Purge             2013  3000000  64473115        20.5 
2 Insidious: Chapter 2  2013  5000000  83586447        15.7 
3 The Conjuring         2013 20000000 137400141         5.87
4 Despicable Me 2       2013 76000000 368065385         3.84
5 Mama                  2013 15000000  71628180         3.78

filter(year == 2013) -Selects only movies released in 2013. mutate(domestic_roi = (domgross - budget) / budget) -Creates a new variable which is the domestic return on investment (ROI). arrange(desc(domestic_roi)) -Sorts movies from highest to lowest ROI. select(…)- Keeps only the most relevant columns slice_head(n = 5) - Returns the top five movies based on highest ROI.

Determine the percentage of movies that received the value of "ok" across all years

bechdel %>%
  summarize(
    total_movies = n(),
    ok_movies = sum(clean_test == "ok", na.rm = TRUE),
    percent_ok = (ok_movies / total_movies) * 100
  )

# A tibble: 1 × 3
  total_movies ok_movies percent_ok
         <int>     <int>      <dbl>
1         1794       803       44.8

summarize()- Creates summary statistics for the entire dataset. total_movies = n() -Counts the total number of movies. ok_movies = sum(clean_test == “ok”) - Counts only the movies where clean_test equals ok percent_ok = (ok_movies / total_movies) * 100 - the percentage of movies rated “ok”.

Data Wrangling Lab

D’Mya Tucker

2025-11-18

Back to `gapminder`

The pipe `%>%`

The Five Main Verbs (5MV) of data wrangling

`filter()`
`summarize()`
`group_by()`
`mutate()`
`arrange()`

`filter()`

Logical operators

`summarize()`

Combining `summarize()` with `group_by()`

Without the `%>%`

`mutate()`

`mutate()`

`arrange()`

Other useful `dplyr` verbs

Your Task

Your Task

Data Wrangling Lab

D’Mya Tucker

2025-11-18

Back to gapminder

The pipe %>%

The Five Main Verbs (5MV) of data wrangling

filter() summarize() group_by() mutate() arrange()

filter()

Logical operators

summarize()

Combining summarize() with group_by()

Without the %>%

mutate()

mutate()

arrange()

Other useful dplyr verbs

Your Task

Your Task

Back to `gapminder`

The pipe `%>%`

`filter()`
`summarize()`
`group_by()`
`mutate()`
`arrange()`

`filter()`

`summarize()`

Combining `summarize()` with `group_by()`

Without the `%>%`

`mutate()`

`mutate()`

`arrange()`

Other useful `dplyr` verbs