Assignment #4

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

Q1 <- inner_join(customers, orders, by = "customer_id")

a. How many rows are in the result?

There are 4 rows.

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

Some customers and orders are not included in the result because the inner join only includes customers who have placed an order. Due to David (4) and Eve (5) not having corresponding orders in the orders data set, they are excluded. The same goes for customer_ids 6 and 7, because they do not have a match in the customers data set.

c. 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")

a. How many rows are in the result?

There are 6 rows.

b. Explain why this number differs from the inner join result.

This number differs from the previous result because the left join includes all customers, regardless of whether they placed an order or not.

c. 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")

a. How many rows are in the result?

There are 6 rows.

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

Customer_ids 6 and 7 have null values for customer name and city. This is because these ids are only displayed in the orders data set, and do not have a corresponding match in the customers data set.

c. 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")

a. How many rows are in the result?

There are 8 rows.

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

Customers 4 and 5 only have data from one table because they have not placed any orders, therefore they do not show up in the orders data set. This also includes customer_ids 6 and 7 because these customer_ids do no exist in the customers data set.

c. 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")

a. How many rows are in the result?

There are 3 rows.

b. How does this result differ from the inner join result?

This result differs from the inner join result because a semi join only includes columns from the customers data set, and filters out the customers who have not placed any orders. The inner join inlcudes all columns from both tables.

c. 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")

a. Which customers are in the result?

David and Eve are the two customers in the result.

b. Explain what this result tells you about these customers.

This result tells us that David and Eve did not make any orders.

c. 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.

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

In order to find all customers, including those who have not placed any orders, the best join to use is the left join.

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

In order to find only the customers who have placed orders, the best join to use is the inner join.

c. Write the R code for both scenarios.

Q7a <- left_join(customers, orders, by = "customer_id")

Q7b <- inner_join(customers, orders , by = "customer_id")

d. Display the result

Q7a

## # 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

Q7b

## # 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

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.

# Step 1: Perform a left join to include all customers
Q8 <- left_join(customers, orders, by = "customer_id")

# Step 2: Summarize data to calculate total orders and total amount spent per customer
summary_data <- Q8 %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n_distinct(order_id, na.rm = TRUE),
    total_amount_spent = sum(amount, na.rm = TRUE)  
  ) %>%
  ungroup()

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

print(summary_data)

## # 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

Assignment #4

Derek Williams, James Cornell

2025-02-20

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

a. How many rows are in the result?

There are 4 rows.

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

c. Display the result

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

a. How many rows are in the result?

There are 6 rows.

b. Explain why this number differs from the inner join result.

This number differs from the previous result because the left join includes all customers, regardless of whether they placed an order or not.

c. Display the result

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

a. How many rows are in the result?

There are 6 rows.

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

Customer_ids 6 and 7 have null values for customer name and city. This is because these ids are only displayed in the orders data set, and do not have a corresponding match in the customers data set.

c. Display the result

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

a. How many rows are in the result?

There are 8 rows.

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

Customers 4 and 5 only have data from one table because they have not placed any orders, therefore they do not show up in the orders data set. This also includes customer_ids 6 and 7 because these customer_ids do no exist in the customers data set.

c. Display the result

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

a. How many rows are in the result?

There are 3 rows.

b. How does this result differ from the inner join result?

This result differs from the inner join result because a semi join only includes columns from the customers data set, and filters out the customers who have not placed any orders. The inner join inlcudes all columns from both tables.

c. Display the result

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

a. Which customers are in the result?

David and Eve are the two customers in the result.

b. Explain what this result tells you about these customers.

This result tells us that David and Eve did not make any orders.

c. Display the result

7. Practical Application (4 points) Imagine you’re analyzing customer behavior.

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

In order to find all customers, including those who have not placed any orders, the best join to use is the left join.

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

In order to find only the customers who have placed orders, the best join to use is the inner join.

c. Write the R code for both scenarios.

d. Display the result

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.