Assignment 4

# Dataset 1: Customers
customers <- tibble(
  customer_id = c(1, 2, 3, 4, 5),
  name = c("Alice", "Bob", "Charlie", "David", "Eve"),
  city = c("New York", "Los Angeles", "Chicago", "Houston", "Phoenix")
)

# Dataset 2: Orders
orders <- tibble(
  order_id = c(101, 102, 103, 104, 105, 106),
  customer_id = c(1, 2, 3, 2, 6, 7),
  product = c("Laptop", "Phone", "Tablet", "Desktop", "Camera", "Printer"),
  amount = c(1200, 800, 300, 1500, 600, 150)
)

print(customers)

## # A tibble: 5 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago    
## 4           4 David   Houston    
## 5           5 Eve     Phoenix

print(orders)

## # A tibble: 6 × 4
##   order_id customer_id product amount
##      <dbl>       <dbl> <chr>    <dbl>
## 1      101           1 Laptop    1200
## 2      102           2 Phone      800
## 3      103           3 Tablet     300
## 4      104           2 Desktop   1500
## 5      105           6 Camera     600
## 6      106           7 Printer    150

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

innerjoin <- 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?

Some customers are not included in the result because they didn’t make any orders; there are no matches in the other table

Display the result

innerjoin

## # 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.

leftjoin <- 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.

The two extra rows are from the customers table. They are included in the result even though they don’t appear in the orders table because they are on the left table (customers).

Display the result

leftjoin

## # 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.

rightjoin <- 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_ids 6 and 7 have null for customer name and city because there is no match on the left table (customers). They only exist in the orders table.

Display the result

rightjoin

## # 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.

fulljoin <- 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.

The order_id, product, and amount columns for customer_id 4 and 5 are null because they only exist in the customers table; there is no match in the orders table. The name and city for customer_id 6 and 7 are null because they only exist in the orders table but not the customers table; there is no match in the customers table.

Display the result

fulljoin

## # 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.

semijoin <- 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?

There are only three rows, customer_id 2 only shows up once, and the information on the orders table are not shown. It only returned rows from the left table where there is a match in the right table. This only shows the names and cities of the people who made orders and are a match in the orders table.

Display the result

semijoin

## # 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.

antijoin <- 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.

These customers’ customer_ids don’t show up in the orders table - they didn’t make any orders.

Display the result

antijoin

## # 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?

I would use full_join because it shows how many people made orders based on the customer_ids made. In total, there are 7 unique customer_ids so there must be 7 unique customers. It also shows David and Eve’s names even if they didn’t make any orders.

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

I would use right_join because it shows the number and the names of customers who placed orders. Although customer_ids 6 and 7 don’t have a name or city, it shows how many orders were made. We can also use their customer_id to identify them.

Write the R code for both scenarios.

fulljoin <- full_join(customers,
                      orders,
                      by = "customer_id")

rightjoin <- right_join(customers, 
                        orders,
                        by = "customer_id")

Display the result

fulljoin

## # 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

rightjoin

## # 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

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.

# Create a full list of customer IDs from both datasets
all_customers <- customers %>%
  select(customer_id, name, city) %>%
  full_join(orders %>% select(customer_id), by = "customer_id") %>%
  distinct()

# Join with orders to get order details
customer_summary <- all_customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n(),
    total_amount_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  ) %>%
  # Replace NA names and cities for unknown customers
  mutate(
    name = ifelse(is.na(name), paste0("Unknown_", customer_id), name),
    city = ifelse(is.na(city), "Unknown City", city)
  )

# Print the summary
print(customer_summary)

## # A tibble: 7 × 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                 1                  0
## 5           5 Eve       Phoenix                 1                  0
## 6           6 Unknown_6 Unknown City            1                600
## 7           7 Unknown_7 Unknown City            1                150

Assignment 4 - Joins

Rhynner Franz Rodriguez

2025-02-18

1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

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?

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

Write the R code for both scenarios.

Display the result

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.