assignment 4

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

q1 <- inner_join(customers , orders)

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

a) How many rows are in the result?

nrow(q1)

## [1] 4

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

inner join returns when there is a match on both tables The customer data has only 3 customers in the order table where one customer has 2 orders

c) Display the result

head(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?

nrow(q2)

## [1] 6

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

In an inner join, only the records that have matching keys (customer_id in this case) in both tables are included. In the left join, all records from the left table (customers) are included, even if there are no matches in the right table (orders). This is why customers like David and Eve, who don’t have orders, are still present in the left join result.

c) Display the result

head(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?

nrow(q3)

## [1] 6

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

The customer_ids 6 and 7 have NULL for name and city because there are no matching records for these customer_ids in the customers table.

c) Display the result

head(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?

nrow(q4)

## [1] 8

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

Rows with NA in order_id correspond to customers who have no orders (David with customer_id 4 and Eve with customer_id 5). Rows with NA in name correspond to orders with no matching customer (orders with customer_id 6 and 7).

c) Display the result

head(q4)

## # 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

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?

nrow(q5)

## [1] 3

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

A semi join only returns rows from the customers table (the left table) and does not include any columns from the orders table. An inner join returns matching rows from both tables, so it includes columns from both customers and orders.

c) Display the result

head(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?

The customers in the result are those with customer_ids 4 and 5 (David and Eve). These are the customers in the customers table who have no matching customer_id in the orders table.

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

This result indicates that David and Eve have not placed any orders. They exist in the customers table but do not have a corresponding entry in the orders table, which means they haven’t made any purchases.

c) Display the result

head(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?

For this scenario, a left join is suitable. A left join will return all customers, including those who haven’t placed any orders, by including all records from the customers table and matching records from the orders table where possible. This way, customers without orders will appear with NA values in the order-related columns.

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

For this, an inner join is appropriate. An inner join will return only those customers who have corresponding entries in the orders table, thus showing only customers who have placed orders.

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

All Customers (including those without orders)

head(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

Customers who have placed orders:

head(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.

q8 <- q2 %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n_distinct(order_id, na.rm = TRUE),
    total_amount_spent = sum(amount, na.rm = TRUE)
  )

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

head(q8)

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

Michael Peters, Nicholas Cerimeli

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

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

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

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

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

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

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

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.