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

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

a. How many rows are in the result?

There are 4 rows

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

Because they do not have a match in the other table.

c. 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')

a. How many rows are in the result?

There are 6 rows.

b. Explain why this number differs from the inner join result.

There are matches for customers in the left table that do not have matches in the right table but since it is a left join they are still visible with the results in the rights table as NA.

c. 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')

a. How many rows are in the result?

There are 6 rows.

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

6 and 7 have NULL because there is not data for their orders but since we are right joining the left table to the right table they still appear in the result.

c. 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')

a. How many rows are in the result?

There are 8 rows.

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

In rows 5 and 6 there is only information from the left table. In rows 7 and 8 there is only information from the right table. When you do a full join it will show the data in both tables, whether there is matching data in the other table or not.

c. 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')

a. How many rows are in the result?

There are 3 rows.

b. How does this result differ from the inner join result?

This is different because it does not repeat the data from the second customer 2 and it only shows the right table.

c. 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')

a. Which customers are in the result?

David and Eve.

b. Explain what this result tells you about these customers.

This tells us that David and Eve do not have any information in the right table.

c. 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.

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

You would use the full join because it returns all rows from both tables.

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

You would use right join because it returns only the rows that have information in the right table, which is the orders information.

c. Write the R code for both scenarios.

q7a <- full_join(customers, orders, by = 'customer_id')
q7b <- right_join(customers , orders , by = 'customer_id')

d. Display the result

q7a
## # 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
q7b
## # 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

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_summary <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = sum(!is.na(order_id)),  # Counts only valid orders
    total_amount_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  )

print(customer_orders_summary)
## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_amount_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