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

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.

Becasue left join includes all customers, even the ones without orders

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

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.

6 and 7 are null becase the have orders but do not exist 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

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 Customers table and rows 7 and 8 only have information from the Order table.

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

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 differs because it only shows name and city while inner join showed order_id, name, product and amount

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

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.

This shows us that David and Eve did not place any orders

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

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?

You would use left join because that would ensure that all customers appear, even if they do not have any orders.

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

You would use inner join because because it would only use rows from the customer table which have order information

Write the R code for both scenarios.

q7p1 <- left_join(customers, orders, by = “customer_id”) q7p1

q7p2 <- inner_join(customers, orders, by = “customer_id”) q7p2

Display the result
q7p1 <- left_join(customers, orders, by = "customer_id")
  q7p1
## # 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
q7p2 <- inner_join(customers, orders, by = "customer_id")
  q7p2
## # 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.

customer_summary <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = n(), 
    total_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  )
customer_summary
## # A tibble: 5 × 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                1           0
## 5           5 Eve     Phoenix                1           0