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

How many rows are in the result?

4 rows.

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

Because, they did not have a match on the other table.

Display the result
q1 <- inner_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

6 rows.

Explain why this number differs from the inner join result.

Because, left join includes all customers, even those without orders: David and Eve are included in the customers table, but not in the orders table.

Display the result
q2 <- left_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

6 rows.

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

6 and 7 have NULL because they are included in the orders table but do not exist in the customers table.

Display the result
q3 <- full_join(customers , orders , by = 'customer_id')

q3
## # A tibble: 8 × 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
## 7           6 <NA>    <NA>             105 Camera     600
## 8           7 <NA>    <NA>             106 Printer    150

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

How many rows are in the result?

8 rows.

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

The last four rows include information from only one table: David and Eve (customer_id = 4 and 5) have NA for order_id, product, and amount because they are not included in the orders table, only the customers table. Orders 105 and 106 (customer_id = 6 and 7) have NA for name and city because they are not included in the customers table, only the orders table.

Display the result
q4 <- full_join(customers , orders , by = 'customer_id')

q4
## # A tibble: 8 × 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
## 7           6 <NA>    <NA>             105 Camera     600
## 8           7 <NA>    <NA>             106 Printer    150

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

How many rows are in the result?

3 rows.

How does this result differ from the inner join result?

Because, it only include customers who have place orders, without including order details: David and Eve are not included because they have no orders. Camera and printer are not included because they have no customers. Bob is only included once, even though he has multiple orders.

Display the result
q5 <- semi_join(customers , orders , by = 'customer_id')

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.

Which customers are in the result?

David and Eve.

Explain what this result tells you about these customers.

This shows that David and Eve did not place any orders.

Display the result
q6 <- anti_join(customers , orders , by = 'customer_id')

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?

Use left join because it includes all customers from the customers table, even if they didn’t place orders. The customers with no orders will still be included, they will just have NA for the order details.

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

Use inner join because it only includes customers who have a matching order in the orders table, and customers without any orders will not be included.

Write the R code for both scenarios.
Display the result
q7_all_customers <- left_join(customers, orders, by = 'customer_id')

q7_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
q7_customers_placed_order <- inner_join(customers, orders, by = 'customer_id')

q7_customers_placed_order
## # 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

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(), 
    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                1           0
## 5           5 Eve     Phoenix                1           0