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

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.

Because all customers are included in the table

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

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 6 and 7 have NULL they are present in orders table but not 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

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

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

customer_id = 6 and customer_id = 7 are in the orders table but not in the customers table, so their name and city are NA

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

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?

Inner join duplicates customers if they have multiple orders, while semi join only keeps unique customer records.

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

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

Explain what this result tells you about these customers.

It tells us the customer id number and what city they are from

Display the result

q6

## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

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

Left Join because it ensures that all customers from the customers table are included in the result, even if they have never placed an order.

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

Inner join because it only returns records where there is a match between customers and orders on customer_id # Write the R code for both scenarios.

Find all customers, including those who haven’t placed any orders (Left Join)

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

Find only the customers who have placed orders (Inner Join)

customers_with_orders <- inner_join(customers, orders, by = "customer_id")
print(customers_with_orders)

## # 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.

customer_summary <- left_join(customers, orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n(),
    total_spent = sum(amount, na.rm = TRUE),  
    .groups = "drop"
  ) %>%
  mutate(
    total_orders = ifelse(is.na(total_spent), 0, total_orders),
    total_spent = replace_na(total_spent, 0)
  )

print(customer_summary)

## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_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                1           0
## 5           5 Eve     Phoenix                1           0

Assignment4

Kathryn Young

2025-02-18

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

How many rows are in the result?

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

Display the result

Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

How many rows are in the result?

Explain why this number differs from the inner join result.

Display the result

Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

How many rows are in the result?

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

Display the result

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

How many rows are in the result?

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

Display the result

Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

How many rows are in the result?

How does this result differ from the inner join result?

Display the result

Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

Which customers are in the result?

Explain what this result tells you about these customers.

Display the result

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

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

Find all customers, including those who haven’t placed any orders (Left Join)

Find only the customers who have placed orders (Inner Join)

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.