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

How many rows are in the result?

4 rows

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

Because they did not have a match in the other table

Display the result
q1 <- inner_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

6 rows

Explain why this number differs from the inner join result.

Because inner join returns rows from both tables with a match while left join only returns rows with a match from the customer table

Display the result
q2 <- left_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

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 they are not in the customer table

Display the result
q3 <- right_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

8 rows

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

Rows 5 and 6 only have information from the customer table and rows 7 and 8 only have information from the orders table. Customer IDs 4 and 5 have no order history and Customer IDs 6 and 7 have no customer information

Display the result
q4 <- full_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

3 rows

How does this result differ from the inner join result?

This result does not show any order information and therefore only has one row for Bob

Display the result
q5 <- semi_join(customers , orders , by = 'customer_id')

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.

Which customers are in the result?

David and Eve

Explain what this result tells you about these customers.

These customers have no order information

Display the result
q6 <- anti_join(customers , orders , by = 'customer_id' )

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?

Full Join since the customer table has customers with no orders and the order table has orders with a customer ID with no name

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

Right Join since it would only use rows from the customer table who have order information. ##### Write the R code for both scenarios.

q7p1 <- full_join(customers , orders , by = 'customer_id')
q7p2 <- right_join(customers, orders , by = 'customer_id')
Display the result
q7p1
## # 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
q7p2
## # 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.

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