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?

Since inner_join matches the customer IDs that are listed in both tables.

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?

6 rows result

Explain why this number differs from the inner join result.

Since left_join combines the orders dataset with the customers dataset even if there is missing information

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?

6 rows result

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

6 and 7 are null since there is no name or city data listed in the orders dataset.

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 nrow(q4) in the result

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

rows 4, 5, 6, 7. This happens since full_join connects any data from the two sets

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?

3 rows

How does this result differ from the inner join result?

semi_join only shows results that match in the left and right tables

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?

David and Eve

Explain what this result tells you about these customers.

David and Eve show up in only the customers dataset but not the orders dataset

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?

left_join since it gives me all customers in the data since you can find the amount of customers who did not place an order

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

inner_join since it would exclude customers who did not make orders.

Write the R code for both scenarios

q7.1 <- left_join(customers, orders, by = "customer_id")
q7.2 <- inner_join(customers, orders, by = "customer_id")

Display the result

q7.1

## # 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

q7.2

## # 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.

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

challenge

## # 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                0           0
## 5           5 Eve     Phoenix                0           0

Assignment 4

Chase Orciuch

2025-02-13

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.