Assignment 4

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?

They did not have a match in the other table

Display the result

## # 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.

Because it is including the employees with N/A as order ID product or amount as well

Display the result

## # 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.

customer_ids 6 and 7 have n/a for name and city because they may not have been bought by a specific named customer or bought for the company itself

Display the result

## # 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.

customer_id 4,5,6,7 all had information from only one table. The customers David and Eve dont have order information, and id 6,7 dont have information about the product they purchased.

Display the result

## # 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 all matching rows all matching rows from both of the tables, semi join returns unique customers who have a matching order

Display the result

## # 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?

David and Eve are the customers in the result

Explain what this result tells you about these customers.

They do not have a matching result in the orders table, they have no recorded orders

Display the result

## # 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?

You would use the left join because even if a customer does not have a matching order it will be printed as NA

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

You would use the inner join because it only includes customers who have placed orders not included the ones with orders

Write the R code for both scenarios.

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

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

Display the result

## # 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

## # 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.

q8 <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = sum(!is.na(order_id)), 
    total_spent = sum(amount, na.rm = TRUE)
  ) %>%
  mutate(total_orders = ifelse(is.na(total_spent), 0, total_orders))

Display the result

## # A tibble: 5 × 5
## # Groups:   customer_id, name [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                0           0
## 5           5 Eve     Phoenix                0           0