R-DataCamp-Data Manipulation with dplyr in R
1. Transforming Data with dplyr
1.1 The countries dataset (video)
1.2 Understanding you data
1.3 Selecting columns
Select the following four columns from the counties variable:
statecountypopulationpoverty
You don’t need to save the result to a variable.
Instruction:
Select the columns listed from the counties variable.

# Select the columns 
counties %>%
select(state, county, population, poverty)

1.4 The filter and arrange verbs (video)
1.5 Arranging observations
Here you see the counties_selected dataset with a few interesting variables selected. These variables: private_work, public_work, self_employed describe whether people work for the government, for private companies, or for themselves.
In these exercises, you’ll sort these observations to find the most interesting cases.
Instruction:
Add a verb to sort the observations of the public_work variable in descending order.

counties_selected <- counties %>%
select(state, county, population, private_work, public_work, self_employed)

# Add a verb to sort in descending order of public_work
counties_selected %>%
arrange(desc(public_work))

1.6 Filtering for conditions
You use the filter() verb to get only observations that match a particular condition, or match multiple conditions.
Instruction 1:
Find only the counties that have a population above one million (1000000).

counties_selected <- counties %>%
select(state, county, population)

# Filter for counties with a population above 1000000
counties_selected %>%
filter(population > 1000000)

Instruction 2:
Find only the counties in the state of California that also have a population above one million (1000000).

counties_selected <- counties %>%
select(state, county, population)

# Filter for counties in the state of California that have a population above 1000000
counties_selected %>%
filter(state == "California" & population > 1000000)

1.7 Filtering and arranging
We’re often interested in both filtering and sorting a dataset, to focus on observations of particular interest to you. Here, you’ll find counties that are extreme examples of what fraction of the population works in the private sector.
Instruction:
Filter for counties in the state of Texas that have more than ten thousand people (10000), and sort them in descending order of the percentage of people employed in private work.

counties_selected <- counties %>%
select(state, county, population, private_work, public_work, self_employed)

# Filter for Texas and more than 10000 people; sort in descending order of private_work
counties_selected %>%
filter(state == 'Texas', population > 10000)%>%
arrange(desc(private_work))

1.8 Mutate (video)
1.9 Calculating the number of government employees
In the video, you used the unemployment variable, which is a percentage, to calculate the number of unemployed people in each county. In this exercise, you’ll do the same with another percentage variable: public_work.
The code provided already selects the state, county, population, and public_work columns.
Instruction 1:
Use mutate() to add a column called public_workers to the dataset, with the number of people employed in public (government) work.

counties_selected <- counties %>%
select(state, county, population, public_work)

# Add a new column public_workers with the number of people employed in public work
counties_selected %>%
mutate(public_workers = population * public_work/100)

Instruction 2:
Sort the new column in descending order.

counties_selected <- counties %>%
select(state, county, population, public_work)

# Sort in descending order of the public_workers column
counties_selected %>%
mutate(public_workers = public_work * population/100) %>%
arrange(desc(public_workers))

1.10 Calculating the percentage of women in a country
The dataset includes columns for the total number (not percentage) of men and women in each county. You could use this, along with the population variable, to compute the fraction of men (or women) within each county.
In this exercise, you’ll select the relevant columns yourself.
Instruction:
Select the columns state, county, population, men, and women. Add a new variable called proportion_women with the fraction of the county’s population made up of women.

# Select the columns state, county, population, men, and women
counties_selected <- counties %>%
select(state, county, population, men,women)
  
# Calculate proportion_women as the fraction of the population made up of women
counties_selected %>%
mutate(proportion_women = women/population)

1.11 Select, mutate, filter, and arrange
In this exercise, you’ll put together everything you’ve learned in this chapter (select(), mutate(), filter() and arrange()), to find the counties with the highest proportion of men.
Instruction:
Select only the columns state, county, population, men, and women. Add a variable proportion_men with the fraction of the county’s population made up of men. Filter for counties with a population of at least ten thousand (10000). Arrange counties in descending order of their proportion of men.

counties %>%
# Select the five columns 
select(state, county, population, men, women)%>%
# Add the proportion_men variable
mutate(proportion_men = men/population)%>%
# Filter for population of at least 10,000
filter(population >= 10000)%>%
# Arrange proportion of men in descending order 
arrange(desc(proportion_men))

2. Aggregating Data
2.1The count verb (video)
2.2 Counting by region
The counties dataset contains columns for region, state, population, and the number of citizens, which we selected and saved as the counties_selected table. In this exercise, you’ll focus on the region column.

counties_selected <- counties %>%
select(region, state, population, citizens)

Instruction:
Use count() to find the number of counties in each region, using a second argument to sort in descending order.

# Use count to find the number of counties in each region
counties_selected %>%
count(region, sort = TRUE)

2.3 Counting citizens by state
You can weigh your count by particular variables rather than finding the number of counties. In this case, you’ll find the number of citizens in each state.

counties_selected <- counties %>%
select(region, state, population, citizens)

Instruction:
Count the number of counties in each state, weighted based on the citizens column, and sorted in descending order.

# Find number of counties per state, weighted by citizens
counties_selected %>%
count(state, wt = citizens, sort = TRUE) 

2.4 Mutating and counting
You can combine multiple verbs together to answer increasingly complicated questions of your data. For example: “What are the US states where the most people walk to work?”
You’ll use the walk column, which offers a percentage of people in each county that walk to work, to add a new column and count based on it.

counties_selected <- counties %>%
select(region, state, population, walk)

Instruction:
Use mutate() to calculate and add a column called population_walk, containing the total number of people who walk to work in a county. Use a (weighted and sorted) count() to find the total number of people who walk to work in each state.

counties_selected %>%
# Add population_walk containing the total number of people who walk to work 
mutate(population_walk = walk * population/100) %>% 
# Count weighted by the new column
count(state, wt = population_walk, sort = TRUE)

2.5 The group by, summarize and ungroup verbs (video)
2.6 Summarizing
The summarize() verb is very useful for collapsing a large dataset into a single observation.

counties_selected <- counties %>%
select(county, population, income, unemployment)

Instruction:
Summarize the counties dataset to find the following columns: min_population (with the smallest population), max_unemployment (with the maximum unemployment), and average_income (with the mean of the income variable).

# Summarize to find minimum population, maximum unemployment, and average income
counties_selected %>%
summarise(min_population = min(population), 
          max_unemployment = max(unemployment),
          average_income = mean(income))

2.7 Summarizing by state
Another interesting column is land_area, which shows the land area in square miles. Here, you’ll summarize both population and land area by state, with the purpose of finding the density (in people per square miles).

counties_selected <- counties %>%
select(state, county, population, land_area)

Instruction 1:
Group the data by state, and summarize to create the columns total_area (with total area in square miles) and total_population (with total population).

# Group by state and find the total area and population
counties_selected %>%
group_by(state) %>%
summarise(total_area = sum(land_area), total_population = sum(population))

Instruction 2:
Add a density column with the people per square mile, then arrange in descending order.

# Add a density column, then sort in descending order
counties_selected %>%
group_by(state) %>%
summarize(total_area = sum(land_area),
          total_population = sum(population)) %>%
mutate(density = total_population/total_area) %>%
arrange(desc(density))

2.8 Summarizing by state and region
You can group by multiple columns instead of grouping by one. Here, you’ll practice aggregating by state and region, and notice how useful it is for performing multiple aggregations in a row.

counties_selected <- counties %>%
select(region, state, county, population)

Instruction 1:
Summarize to find the total population, as a column called total_pop, in each combination of region and state.

# Summarize to find the total population
counties_selected %>%
group_by(region, state) %>%
summarize(total_pop = sum(population))

Instruction 2:
Notice the tibble is still grouped by region; use another summarize step to calculate two new columns: the average state population in each region (average_pop) and the median state population in each region (median_pop).

# Calculate the average_pop and median_pop columns 
counties_selected %>%
group_by(region, state) %>%
summarize(total_pop = sum(population)) %>%
summarize(average_pop = mean(total_pop),
          median_pop = median(total_pop))

2.9 The top_n verb (video)
2.10 Selecting a country from each region
Previously, you used the walk column, which offers a percentage of people in each county that walk to work, to add a new column and count to find the total number of people who walk to work in each county.
Now, you’re interested in finding the county within each region with the highest percentage of citizens who walk to work.

counties_selected <- counties %>%
select(region, state, county, metro, population, walk)

Instruction:
Find the county in each region with the highest percentage of citizens who walk to work.

# Group by region and find the greatest number of citizens who walk to work
counties_selected %>%
group_by(region)%>%
top_n(1,walk)

2.11 Finding the highest-income state in each region
You’ve been learning to combine multiple dplyr verbs together. Here, you’ll combine group_by(), summarize(), and top_n() to find the state in each region with the highest income.
When you group by multiple columns and then summarize, it’s important to remember that the summarize “peels off” one of the groups, but leaves the rest on. For example, if you group_by(X, Y) then summarize, the result will still be grouped by X.

counties_selected <- counties %>%
select(region, state, county, population, income)

Instruction:
Calculate the average income (as average_income) of counties within each region and state (notice the group_by() has already been done for you). Find the highest income state in each region.

counties_selected %>%
group_by(region, state) %>%
# Calculate average income
summarize(average_income = mean(income))%>%
# Find the highest income state in each region
top_n(1,average_income)

2.12 Using summarize, top_n, and count together
In this chapter, you’ve learned to use five dplyr verbs related to aggregation: count(), group_by(), summarize(), ungroup(), and top_n(). In this exercise, you’ll use all of them to answer a question: In how many states do more people live in metro areas than non-metro areas?
Recall that the metro column has one of the two values “Metro” (for high-density city areas) or “Nonmetro” (for suburban and country areas).

counties_selected <- counties %>%
select(state, metro, population)

Instruction 1:
For each combination of state and metro, find the total population as total_pop.

# Find the total population for each combination of state and metro
counties_selected %>%
group_by(state, metro) %>%
summarize(total_pop = sum(population))

Instruction 2:
Extract the most populated row from each state, which will be either Metro or Nonmetro.

# Extract the most populated row for each state
counties_selected %>%
group_by(state, metro) %>%
summarize(total_pop = sum(population)) %>%
top_n(1, total_pop)

Instruction 3:
Ungroup, then count how often Metro or Nonmetro appears to see how many states have more people living in those areas.

# Count the states with more people in Metro or Nonmetro areas
counties_selected %>%
group_by(state, metro) %>%
summarize(total_pop = sum(population)) %>%
top_n(1, total_pop) %>%
ungroup() %>%
count(metro)

3. Selecting and Transforming Data
3.1 Selecting (video)
3.2 Selecting columns
Using the select verb, we can answer interesting questions about our dataset by focusing in on related groups of verbs. The colon (:) is useful for getting many columns at a time.
Instruction:
Use glimpse() to examine all the variables in the counties table. Select the columns for state, county, population, and (using a colon) all five of those industry-related variables; there are five consecutive variables in the table related to the industry of people’s work: professional, service, office, construction, and production. Arrange the table in descending order of service to find which counties have the highest rates of working in the service industry.

# Glimpse the counties table
glimpse(counties)

counties %>%
# Select state, county, population, and industry-related columns
select(state, county, population,professional, service, office, construction, production)%>%
# Arrange service in descending order 
arrange(desc(service))

3.3 Select helpers
In the video you learned about the select helper starts_with(). Another select helper is ends_with(), which finds the columns that end with a particular string.
Instruction:
Select the columns state, county, population, and all those that end with work. Filter just for the counties where at least 50% of the population is engaged in public work.

counties %>%
# Select the state, county, population, and those ending with "work"
select(state, county, population, ends_with('work'))%>%
# Filter for counties that have at least 50% of people engaged in public work
filter(public_work >= 50)

3.4 The renames verb (video)
3.5 Renaming a column after count
The rename() verb is often useful for changing the name of a column that comes out of another verb, such as count(). In this exercise, you’ll rename the n column from count() (which you learned about in Chapter 2) to something more descriptive.
Instruction 1:
Use count() to determine how many counties are in each state.

# Count the number of counties in each state
counties %>%
count(state)

Instruction 2:
Notice the n column in the output; use rename() to rename that to num_counties.

# Rename the n column to num_counties
counties %>%
count(state)  %>%
rename(num_counties = n)

3.6 Rename a column as part of a select
rename() isn’t the only way you can choose a new name for a column: you can also choose a name as part of a select().
Instruction:
Select the columns state, county, and poverty from the counties dataset; in the same step, rename the poverty column to poverty_rate.

# Select state, county, and poverty as poverty_rate
counties %>%
select(state, county, poverty_rate = poverty)

3.7 The transmute verb (video)
3.8 Choosing among verbs
3.9 Using transmute
As you learned in the video, the transmute verb allows you to control which variables you keep, which variables you calculate, and which variables you drop.
Instruction:
Keep only the state, county, and population columns, and add a new column, density, that contains the population per land_area. Filter for only counties with a population greater than one million. Sort the table in ascending order of density.

counties %>%
# Keep the state, county, and populations columns, and add a density column
transmute(state, county, population, density = population/land_area)%>%
# Filter for counties with a population greater than one million 
filter(population > 1000000)%>%
# Sort density in ascending order 
arrange(density)

3.10 Matching verbs to their definitions
3.11 Choosing among the four verbs
In this chapter you’ve learned about the four verbs: select, mutate, transmute, and rename. Here, you’ll choose the appropriate verb for each situation. You won’t need to change anything inside the parentheses.
Instruction:
Choose the right verb for changing the name of the unemployment column to unemployment_rate. Choose the right verb for keeping only the columns state, county, and the ones containing poverty. Calculate a new column called fraction_women with the fraction of the population made up of women, without dropping any columns. Keep only three columns: the state, county, and employed/population, which you’ll call employment_rate.

# Change the name of the unemployment column
counties %>%
rename(unemployment_rate = unemployment)

# Keep the state and county columns, and the columns containing poverty
counties %>%
select(state, county, contains("poverty"))

# Calculate the fraction_women column without dropping the other columns
counties %>%
mutate(fraction_women = women/population)

# Keep only the state, county, and employment_rate columns
counties %>%
transmute(state, county, employment_rate = employed/population)

4. Case Study: The babynames Dataset
4.1 The babynames data (video)
4.2 Filtering and arranging for one year
The dplyr verbs you’ve learned are useful for exploring data. For instance, you could find out the most common names in a particular year.
Instruction:
Filter for only the year 1990. Sort the table in descending order of the number of babies born.

babynames %>%
# Filter for the year 1990
filter(year == 1990)%>%
# Sort the number column in descending order 
arrange(desc(number))

4.3 Using top_n with babynames
You saw that you could use filter() and arrange() to find the most common names in one year. However, you could also use group_by and top_n to find the most common name in every year.
Instruction:
Use group_by and top_n to find the most common name for US babies in each year.

# Find the most common name in each year
babynames %>%
group_by(year)%>%
top_n(1, number)

4.4 Visualizing names with ggplots
The dplyr package is very useful for exploring data, but it’s especially useful when combined with other tidyverse packages like ggplot2.
Instruction 1:
Filter for only the names Steven, Thomas, and Matthew, and assign it to an object called selected_names.

# Filter for the names Steven, Thomas, and Matthew 
selected_names <- babynames %>%
filter(name %in% c("Steven","Thomas","Matthew"))

Instruction 2:
Visualize those three names as a line plot over time, with each name represented by a different color.

# Plot the names using a different color for each name
ggplot(selected_names, aes(x = year, y = number, color = name)) +
geom_line()

4.5 Grouped mutates (video)
4.6 Finding the year each name is most common
In an earlier video, you learned how to filter for a particular name to determine the frequency of that name over time. Now, you’re going to explore which year each name was the most common.
To do this, you’ll be combining the grouped mutate approach with a top_n.
Instruction:
Complete the code so that it finds the year each name is most common.

# Find the year each name is most common 
babynames %>%
group_by(year) %>%
mutate(year_total = sum(number)) %>%
ungroup() %>%
mutate(fraction = number/year_total) %>%
group_by(name) %>%
top_n(1, fraction)

4.7 Adding the total and maximum for each name
In the video, you learned how you could group by the year and use mutate() to add a total for that year.
In these exercises, you’ll learn to normalize by a different, but also interesting metric: you’ll divide each name by the maximum for that name. This means that every name will peak at 1.
Once you add new columns, the result will still be grouped by name. This splits it into 48,000 groups, which actually makes later steps like mutates slower.
Instruction 1:
Use a grouped mutate to add two columns:
name_total, with the total number of babies born with that name in the entire dataset. name_max, with the highest number of babies born in any year.

# Add columns name_total and name_max for each name
babynames %>%
group_by(name) %>%
mutate(name_total = sum(number),
       name_max = max(number))

Instruction 2:
Add another step to ungroup the table. Add a column called fraction_max, with the number in the year divided by the maximum for that name.

babynames %>%
group_by(name) %>%
mutate(name_total = sum(number),
       name_max = max(number)) %>%
# Ungroup the table 
ungroup() %>%
# Add the fraction_max column containing the number by the name maximum 
mutate(fraction_max = number/name_max)

4.8 Visualizing the normalized change in popularity
You picked a few names and calculated each of them as a fraction of their peak. This is a type of “normalizing” a name, where you’re focused on the relative change within each name rather than the overall popularity of the name.
In this exercise, you’ll visualize the normalized popularity of each name. Your work from the previous exercise, names_normalized, has been provided for you.

names_normalized <- babynames %>%
                     group_by(name) %>%
                     mutate(name_total = sum(number),
                            name_max = max(number)) %>%
                     ungroup() %>%
                     mutate(fraction_max = number/name_max)

Instruction:
Filter the names_normalized table to limit it to the three names Steven, Thomas, and Matthew. Visualize fraction_max for those names over time.

# Filter for the names Steven, Thomas, and Matthew
names_filtered <- names_normalized %>%
filter(name %in% c('Steven', 'Thomas', 'Matthew'))

# Visualize these names over time
ggplot(names_filtered, aes(x = year, y = fraction_max, color = name)) + 
geom_line()

4.9 Window function (video)
4.10 Using ratios to describe the frequency of a name
In the video, you learned how to find the difference in the frequency of a baby name between consecutive years. What if instead of finding the difference, you wanted to find the ratio?
You’ll start with the babynames_fraction data already, so that you can consider the popularity of each name within each year.
Instruction:
Arrange the data in ascending order of name and then year. Group by name so that your mutate works within each name. Add a column ratio containing the ratio between each year.

babynames_fraction %>%
# Arrange the data in order of name, then year 
arrange(name, year) %>%
# Group the data by name
group_by(name) %>%
# Add a ratio column that contains the ratio between each year 
mutate(ratio = fraction/lag(fraction))

4.11 Biggest jumps in a name
Previously, you added a ratio column to describe the ratio of the frequency of a baby name between consecutive years to describe the changes in the popularity of a name. Now, you’ll look at a subset of that data, called babynames_ratios_filtered, to look further into the names that experienced the biggest jumps in popularity in consecutive years.

babynames_ratios_filtered <- babynames_fraction %>%
                     arrange(name, year) %>%
                     group_by(name) %>%
                     mutate(ratio = fraction/lag(fraction)) %>%
                     filter(fraction >= 0.00001)

Instruction:
From each name in the data, keep the observation (the year) with the largest ratio; note the data is already grouped by name. Sort the ratio column in descending order. Filter the babynames_ratios_filtered data further by filtering the fraction column to only display results greater than or equal to 0.001.

babynames_ratios_filtered %>%
# Extract the largest ratio from each name 
top_n(1,ratio) %>%
# Sort the ratio column in descending order 
arrange(desc(ratio)) %>%
# Filter for fractions greater than or equal to 0.001
filter(fraction >= 0.001)

4.12 Congratulations!