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?

Becuase they don’t have matching records

print(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?

There are 6 rows

Explain why this number differs from the inner join result.

Left Join keeps all customers, even if they don’t have an order.

print(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?

There are 6 rows

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

Becuase these customer id’s do not have any data for name and city

print(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 8 rows

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

A Full Join includes all records from both tables, even if there is no matching record in the other table.

print(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?

There are 3 rows

How does this result differ from the inner join result?

Inner join includes duplicate customers rows, semi join does not.

print(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?

Customers 4 and 5 becuase they have no matching orders

Explain what this result tells you about these customers.

These customers have no placed any orders. They might have signed up to receive info about the product but have not made a purchase

print(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?

To find all customers we would use left_join becuase it ensures that all customers are included, even if they have no matching order records.

q7a <- left_join(customers, orders, by = 'customer_id')

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

To only find customers that have placed orders we use inner join becuase it keeps only customers that have matching orders.

q7b <- inner_join(customers, orders, by = 'customer_id')
print(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
print(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

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.

challenge_question <- left_join(customers, orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarise(
    total_orders = sum(!is.na(order_id)),                 
    total_amount = sum(amount, na.rm = TRUE),  
    .groups = 'drop'
)
print(challenge_question)
## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_amount
##         <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