Assignment 4

R Markdown

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?

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

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)

## 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.

It includes the customers with no match in the other 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

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)

## 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_id 6 and customer_id 7. This is because there are entries for order_id 105 and order_id 106, but no customer_ids to match, represented by “N/A”.

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)

## 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.

Rows 5-8 have information from only one table. Rows 5 and 6 have information only for customer_id, meaning that they have made no orders. Rows 7 and 8 have informatipon only for order_id, meaning that no customers ordered these items.

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)

## 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?

the inner join result produces 4 rows. Bob is displayed twice because he has made 2 orders.

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)

## Joining with `by = join_by(customer_id)`

Which customers are in the result?

David and Eve.

Explain what this result tells you about these customers.

#####David and Eve have placed no orders.

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?

I would use a left join because it would result in the customers with no orders being presented with N/A, keeping them included in the table.

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

I would use an inner join because it would result in only customers who have a matching order entry.

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

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

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

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.

q9 <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%  # Group by customer details
  summarize(
    total_orders = n_distinct(order_id, na.rm = TRUE),  # Count unique orders, avoiding NA
    total_spent = sum(amount, na.rm = TRUE)  # Sum total amount spent, treating NA as 0
  )

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

q9

## # A tibble: 5 × 5
## # Groups:   customer_id, name [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

Jack Casamassima

2025-02-18

R Markdown

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?

They did not have a match in the other table.

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.

It includes the customers with no match in the other table

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.

customer_id 6 and customer_id 7. This is because there are entries for order_id 105 and order_id 106, but no customer_ids to match, represented by “N/A”.

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.

Rows 5-8 have information from only one table. Rows 5 and 6 have information only for customer_id, meaning that they have made no orders. Rows 7 and 8 have informatipon only for order_id, meaning that no customers ordered these items.

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?

the inner join result produces 4 rows. Bob is displayed twice because he has made 2 orders.

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?

David and Eve.

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?

I would use a left join because it would result in the customers with no orders being presented with N/A, keeping them included in the table.

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

I would use an inner join because it would result in only customers who have a matching order entry.

Write the R code for both scenarios.

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.