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)`

How many rows are in the result?

nrow(q1)
## [1] 4

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

Because the customers and orders that are not included did not have a match in the other table

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)
## 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.

The inner join function only displays data that has a match while the left join function displays all of the data from the left table, whether there is a match or not, and the matches from the right.

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)
## 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_ids 6 & 7 result NULL because they do not exist in the customers dataset, so there are no matching results for that data. Only order ids 1, 2, 3, 4, and 5 have matching customer ids that show up in the customers table, which is why those orders have names attached to them.

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)
## 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&6 show null data in columns order_id, product, and amount. This is because the customers in those rows do not have matching data from the orders table. Rows 7&8 show null data in columns name and city. This is because the customer ids in those rows do not have matching data from the customers table.

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)
## 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?

This differs from the inner join because a semi join displays only the customers that have matching orders and keeps each customer listed once. The inner join functions displays one row for every matching customer and order pair. This means that customers will appear multiple times if they have placed multiple orders.

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

Which customers are in the result?

select(q6, name)
## # A tibble: 2 × 1
##   name 
##   <chr>
## 1 David
## 2 Eve

Explain what this result tells you about these customers.

This result tells me that David and Eve do not have any matching data in the customers and orders tables.

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.

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 this would display all data from the left table, which is customers. A left join will display data even if it doesn’t have a match in the right table.

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

I would use an inner join to find only the customers who have places orders. This is because the inner join will only show the customers who placed orders, not all orders like a right join.

Write the R code for both scenarios.

q7 <- left_join(customers , orders)
## Joining with `by = join_by(customer_id)`
q7 <- inner_join(customers , orders)
## Joining with `by = join_by(customer_id)`

Display the result

head(q7)
## # 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.

bonus <- full_join(customers , orders) %>% 
  group_by(customer_id , name , city) %>% 
  summarise(total_orders = n_distinct(order_id),
            total_amount_spent = sum(amount , na.rm = TRUE))
## Joining with `by = join_by(customer_id)`
## `summarise()` has regrouped the output.

Display the results

head(bonus)
## # A tibble: 6 × 5
## # Groups:   customer_id, name [6]
##   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                1                  0
## 5           5 Eve     Phoenix                1                  0
## 6           6 <NA>    <NA>                   1                600