Assignment 4

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

a. How many rows are in the result?

4 rows

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

They are not included because they didn’t have a match on the other side of the table

c. 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.

a. How many rows are in the result?

6 rows

b. Explain why this number differs from the inner join result.

Left join includes all customers, even those without orders, so David and Eve are included in the customers table but not the orders table # Display the result

c. 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.

a. How many rows are in the result?

6 rows

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

Customer id’s 6 and 7 show up as NULL since there is no match in the data set

c. 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.

a. How many rows are in the result?

8 rows

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

Row 5, 6, 7, and 8 because there is no match between the data among all of the tables

c. 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.

a. How many rows are in the result?

3 riws

b. How does this result differ from the inner join result?

This result differs from the inner join result because it only includes customers who have placed orders, without including the order details

c. 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.

a. Which customers are in the result?

David and Eve

b. Explain what this result tells you about these customers.

This result tells us the customers have never placed an order

c. 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.

a. Which join would you use to find all customers, including those who haven’t placed any orders? Why?

Left join ensures all customers are included, even those who haven’t placed any orders

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

Inner Join since it filters out the customers who have no placed orders

c. Write the R code for both scenarios.

d. Display the result

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

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

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.

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