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?

## [1] 4

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

Becuase they do 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

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?

## [1] 6

Explain why this number differs from the inner join result.

This number of rows is different from the inner join result because it shows the values tha produce NA results

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

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?
## [1] 6

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

Customer ids 6 and 7 both have NULL for customer name and city because there is not customer data for them in the database.

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

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?

## [1] 8

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

Rows 5,6,7, and 8 have information from only one table. Rows 5 and 6 are missing order_id information for customers with the customer_id of 4 and 5. Rows 7 and 8 are missing customer_id information withorders withe the order_id of 105 and 106.

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

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?

## [1] 3

How does this result differ from the inner join result?

This result does not duplicate any same results while the inner join has duplicates.

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

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?

David and Eve

Explain what this result tells you about these customers.

They have never purchased anything.

Display the result

q6
## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

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?

The only customers that count are the ones that appear in the customer table. To find these customers I would use left join because this includes all customers, and those without orders will still appear with their order details as NA.

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

To find only customers who have placed orders I would use inner join because this will return only customers who have matching orders, and exclude those who haven’t placed any.

Write the R code for both scenarios.

left_join(customers, orders, by = "customer_id")
## # 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
inner_join(customers, orders , by = 'customer_id') 
## # 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. Hint: You’ll need to use a combination of joins and group_by/summarize operations.

challenge <- left_join(customers, orders, by = "customer_id") %>%
  group_by(name, city) %>%  
  summarize(
    total_orders = sum(!is.na(order_id)), 
    total_spent = sum(amount, na.rm = TRUE)
  ) %>%
  ungroup() 
challenge 
## # A tibble: 5 × 4
##   name    city        total_orders total_spent
##   <chr>   <chr>              <int>       <dbl>
## 1 Alice   New York               1        1200
## 2 Bob     Los Angeles            2        2300
## 3 Charlie Chicago                1         300
## 4 David   Houston                0           0
## 5 Eve     Phoenix                0           0