1. 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?

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

2. 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 are in the result?

The resulting data has 6 rows

Explain why this number differs from the inner join result.

The left join returns all rows from the customers table, even if they don’t have a matching order.

Display the 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

3. 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 are in the result?

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 NULL for customer name and city because the orders table has orders with these customer_ids, but there are no matching records with these IDs in the customers table.

Display the 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

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

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

How many rows are in the result?

The resulting data has 8 rows

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

Rows that consist of information from only one table include records of customers with no orders or orders without any matching customers. customer_ids 4 and 5 have no matching orders, so the full join results in only data from the customer table for these rows. Orders with customer_ids 6 and 7 have no matching customers, so the full join results in data solely from the orders table.

Display the 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

5. 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 are in the result?

The resulting data has 3 rows

How does this result differ from the inner join result?

This result differs from the inner join result because the semi join only includes rows from the left table (customers) that have a match in the right table, but it does not return any columns from the right table. It also doesn’t duplicate rows if a customer has multiple orders.

Display the 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

6. 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?

The resulting data has 2 rows

Explain what this result tells you about these customers.

This result tells us that David and Eve are the only customers without any orders.

Display the result

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

7. 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?

I would use a left join because a left join keeps all rows from the customers table, even if they do not have a matching order.

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

I would use a semi join or an inner join. A semi join is ideal in this scenario since it returns only the customers who have at least one matching order, without duplicating rows or including order details.

Write the R code for both scenarios.

all_customers <- left_join(customers, orders, by = "customer_id")
active_customers <- semi_join(customers, orders, by = "customer_id")

Display the result

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(active_customers)
## # 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

8. Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent. Include all customers, even those without orders. Hint: You’ll need to use a combination of joins and group_by/summarize operations.

customer_summary <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n_distinct(order_id[!is.na(order_id)]),
    total_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  )

customer_summary
## # A tibble: 5 × 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                0           0
## 5           5 Eve     Phoenix                0           0