Assignment #4

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 do 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.

Left join keeps all customers even if they haven’t placed any orders.

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.

The customer id with NULL for customer name and city exist in the orders table but not in the customers table.

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 was a customer with no order or an order with no matching customer.

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?

Semi join returns customers who have orders but doesn’t show the order details, while inner join returns customers and order details.

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?

People who have no orders in the order table.

Explain what this result tells you about these customers.

Customers are registered but have not placed an order.

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?

Use a left join to find all customers. It keeps every customer even those who have no matching order.

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

Use semi join to find only customers who have placed orders. It shows just the customers with at least one order.

Write the R code for both scenarios.

q7 <- left_join(customers, orders, by = "customer_id"); semi_join(customers, orders, by = "customer_id")

## # 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

Display the result

q7

## # 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

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.

q8 <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, city) %>%
  summarize(total_orders = n(), total_spent = sum(amount, na.rm = TRUE)) %>%
  arrange(desc(total_spent))

## `summarise()` has grouped output by 'customer_id'. You can override using the
## `.groups` argument.

Display the result

q8

## # A tibble: 5 × 4
## # Groups:   customer_id [5]
##   customer_id city        total_orders total_spent
##         <dbl> <chr>              <int>       <dbl>
## 1           2 Los Angeles            2        2300
## 2           1 New York               1        1200
## 3           3 Chicago                1         300
## 4           4 Houston                1           0
## 5           5 Phoenix                1           0