Inner Join (3 points) Perform an inner join between the customers and orders datasets.

q1 <- inner_join(customers , orders)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 4 rows

Why are some customers or orders not included in the result?

Because the customers and orders that are not included did not have a match in the other table

Display the result

head(q1)
## # A tibble: 4 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300

Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

q2 <- left_join(customers, orders, by = "customer_id")
How many rows?

The resulting data has 6 rows

Why does this differ from the inner join?

The left join keeps all customers, even those without orders, while the inner join only keeps matches.

Display result

head(q2)
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

q3 <- right_join(customers, orders, by = "customer_id")

How many rows?

The resulting data has 6 rows.

Which customer_ids in the result have NULL for customer name and city? Explain why.

Customer_ids 6 and 7 have NA values because they exist in orders but not in customers.

Display result

head(q3)
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           6 <NA>    <NA>             105 Camera     600
## 6           7 <NA>    <NA>             106 Printer    150

Full Join (3 points) Perform a full join between customers and orders

q4 <- full_join(customers, orders, by = "customer_id")

How many rows?

The resulting data has 6 rows

Identify any rows where there’s information from only one table. Explain these results.

Customers 4 and 5 only appear in customers, and customer_ids 6 and 7 only appear in orders.

Display result

head(q4)
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

q5 <- semi_join(customers, orders, by = "customer_id")

How many rows?

The resulting data has 3 rows

How does this differ from the inner join?

A semi join only returns matching rows from the customers table, while an inner join returns matching rows from both tables and includes order details.

Display result

head(q5)
## # A tibble: 3 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago

Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table

q6 <- anti_join(customers, orders, by = "customer_id")

Which customers are in the result?

David and Eve

Explanation:

The anti join returns customers who do not have matching customer_id values in the orders table. These customers have not placed any orders.

Display result

head(q6)
## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

Practical Application (4 points) Imagine you’re analyzing customer behavior.

Which join would you use to find all customers, including those who haven’t placed any orders? Why?

Use a left join because it keeps all customers, even if they do not have matching orders.

Which join would you use to find only the customers who have placed orders? Why?

Use a semi join because it returns only customers who have matching orders.

Display results

all_customers <- left_join(customers, orders, by = "customer_id")
customers_with_orders <- semi_join(customers, orders, by = "customer_id")
head(all_customers)
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA
head(customers_with_orders)
## # A tibble: 3 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago

Challenge Question: Create summary of total orders and total amount spent for each customer

summary_table <- left_join(customers, orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarise(
    total_orders = n(),
    total_spent = sum(amount, na.rm = TRUE)
  )
## `summarise()` has regrouped the output.
## ℹ Summaries were computed grouped by customer_id, name, and city.
## ℹ Output is grouped by customer_id and name.
## ℹ Use `summarise(.groups = "drop_last")` to silence this message.
## ℹ Use `summarise(.by = c(customer_id, name, city))` for per-operation grouping
##   (`?dplyr::dplyr_by`) instead.
head(summary_table)
## # A tibble: 5 × 5
## # Groups:   customer_id, name [5]
##   customer_id name    city        total_orders total_spent
##         <dbl> <chr>   <chr>              <int>       <dbl>
## 1           1 Alice   New York               1        1200
## 2           2 Bob     Los Angeles            2        2300
## 3           3 Charlie Chicago                1         300
## 4           4 David   Houston                1           0
## 5           5 Eve     Phoenix                1           0