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

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

How many rows are in the result?

There are 4 rows

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

They did not have a match in the other table

Display the result

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?

There are 6 rows

Explain why this number differs from the inner join result.

The inner join had 4 rows because it only included customers who had matching orders. The left join includes all customers from the customers table, even if they do not have a matching order.

Display the result

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?

There are 6 rows

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

Customer_id 6 and customer_id 7 results have NULL for customer name and city. These IDs exist in the orders table but do not exist in the customers table. Since a right join keeps all rows from the orders table but there are no matching customer records, the name and city fields are filled with NA.

Display the result

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?

There are 8 rows.

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

There is information from only customers in rows 5 and 6. Their order_id, product, and amount are NA. There is information from only orders in rows 7 and 8. Their name and city are NA. A full join keeps all rows from both tables but if there is no match in one table, NA appears in the missing fields.

Display the result

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.

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

How many rows are in the result?

There are 3 rows

How does this result differ from the inner join result?

The inner join returns both matching rows from customers and corresponding columns from orders. The semi join returns only unique customers that have orders, without including order details. Duplicate customer entries are removed.

Display the result

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?

David and Eve are the only customers in the result

Explain what this result tells you about these customers.

These customers do not have any matching orders in the orders table meaning they have never placed an order but are registered in the system.

Display the result

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?

You would use left join because it ensures all customers are included even if they have not placed any orders.

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

You would use inner join because it ensures we only include customers who have at least one order and customers without orders will be excluded.

Write the R code for both scenarios.

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

order_customers <- inner_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(order_customers)
## # 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_orders <- left_join(customers, orders, by = "customer_id")

customer_summary <- customer_orders %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n(),  # Count number of rows per customer
    total_spent = sum(amount, na.rm = TRUE),  # Sum order amounts
    .groups = "drop"  # Remove grouping to avoid warning message
  ) %>%
  mutate(
    total_orders = ifelse(total_spent == 0, 0, total_orders),  # Ensure order count is 0 if no orders
    total_spent = replace_na(total_spent, 0)  # Replace NA with 0 for total spent
  )

head(customer_summary)
## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_spent
##         <dbl> <chr>   <chr>              <dbl>       <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