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?

There are 4.

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)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

There are 6 rows in the result.

Explain why this number differs from the inner join result.

A left join keeps all the rows from the customers table, even if there are no matching row in the orders table. Customers 4 and 5 appear with NA values because they have not placed any orders. This differs from the inner join because it only keeps rows where there is a match in both tables.

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

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)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

There are 6 rows in the results

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

customer_id 6 & 7 have NA for name and city because those customer_ids appear in the orders table but do not exist 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

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

q4 <- full_join(customers, orders)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

There are 8 rows in the result.

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

Customers 4 & 5 have NA for order columns because they have no orders. Orders with customer_id 6 & 7 have NA for customer name/city because those customers are not in the customer dataset.

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

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)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

There are 3 rows in the result.

How does this result differ from the inner join result?

A semi join returns only the matching rows from the left table (customers) and does not include any columns from orders. It also does not duplicate customer rows when a customer has multiple orders (so customer 2 will only appear once here).

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

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)
## Joining with `by = join_by(customer_id)`

Which customers are in the result?

# The customers in the result are David and Eve

Explain what this result tells you about these customers.

# These customers are in the customers table but have no matching orders, meaning they haven’t placed any orders in the orders data set.

Display the 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?

I would use a left join because it keeps every customer and fills in NA for customers with no orders.

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

I would use a semi join because it returns only customers who have at least one order and avoids duplicates and extra order columns.

Write the R code for both scenarios.

q7b <- semi_join(customers, orders)
## Joining with `by = join_by(customer_id)`

Display the result

head(q7b)
## # 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 (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.

q8 <- customers %>% 
  left_join(orders, by = "customer_id") %>% 
  group_by(customer_id, name, city) %>% 
  summarise(
    total_orders = sum(!is.na(order_id)),
    total_amount_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  )

head(q8)
## # 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