1. 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?

nrow(q1)
## [1] 4

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

Inner join returns all rows from both tables where there is a match.Therefore the rows not returned 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?

nrow(q2)
## [1] 6

Explain why this number differs from the inner join result.

Left join returns all rows from the left table and matching rows from the right table.

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?

nrow(q3)
## [1] 6

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

Customer_ids 6 and 7 because they don’t exist in customers. Right join returns all rows from the right table and matching rows from the left 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?

nrow(q4)
## [1] 8

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

Customers 4 and 5 have no orders. Orders with customer_id 6 and 7 have no customer match. Full join returns all rows when there is a match in either left or right table.

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?

nrow(q5)
## [1] 3

How does this result differ from the inner join result?

Semi join returns all rows from the left table where there is a match in the right table.

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?

nrow(q6)
## [1] 2

Explain what this result tells you about these customers.

It shows employees without salary information.

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?

Left join because it will show all the customers in the dataset even without orders.

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

Inner join because it will only show customers that have value in the orders table.

Write the R code for both scenarios.

q7 <- left_join(customers , orders )
## Joining with `by = join_by(customer_id)`
q8 <- inner_join(customers , orders )
## Joining with `by = join_by(customer_id)`

Display the result

head(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
head(q8)
## # 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

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.

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

head(summary_result)
## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_amount
##         <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