Question 1: Inner Join (3 points)

Perform an inner join between the customers and orders datasets.

Q1 <- inner_join(customers, orders, by = "customer_id")

a). How many rows are in the result?

There are 4 rows.

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

Only customers and orders with matching customer_id in both tables are included.

The customer table has only 3 customers in the orders table where one customer has 2 orders.

c). Display the result

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

Question 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")

a). How many rows are in the result?

There are 6 rows.

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

In the left join, all customers are included, even those without order information.

Customer_id 4 and 5 did not have an order_id but are still included as part of the customers table.

c). Display the result

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

Question 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")

a). How many rows are in the result?

There are 6 rows.

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

Customer_id 6 and 7 have null for name and city.

In right join, all order_id entries are included, even for customer_id 6 & 7 who are not in the customers table.

Customer_id 6 & 7 only have orders and not name and city customer information.

c). Display the result

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

Question 4: Full Join (3 points)

Perform a full join between customers and orders.

Q4 <- full_join(customers, orders, by = "customer_id")

a). How many rows are in the result?

There are 8 rows.

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

Customer_id 4 & 5 only have information from customers table and customer_id 6 & 7 only have information from orders table.

In full join, all rows from both tables are included and NA is marked where there’s no match between the tables for that customer_id.

c). Display the result

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

Question 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")

a). How many rows are in the result?

There are 3 rows.

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

In semi join, only customers with order information are returned, but order data isn’t included.

Inner join included the order information that matched with the customer_id but here order information does not matter.

That is also why customer_id 2 “Bob” is only shown once here, in this table we don’t care how many orders he has had, only if he has ordered at all.

c). Display the result

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

Question 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")

a). Which customers are in the result?

Customer_id 4 David & 5 Eve are included in the result.

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

These customers have customer information but they do not have any order information.

In anti join, only customers without order information are included.

c). Display the result

print (Q6)

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

Question 7: Practical Application (4 points)

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

To find all customers including those who haven’t placed any orders you would use the full join function.

The full join function would show all customers from both tables and would mark NA down for those who have not ordered.

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

To find only customers who have placed orders you would use the right join.

It will display all orders, even those that do not have customer information such as name and city (the customer_id will still be there).

c). Write the R code for both scenarios.

scenario_a <- full_join(customers, orders, by = "customer_id") %>% 
  select(customer_id, name, city) %>% 
  distinct()

scenario_b <- right_join(customers, orders, by = "customer_id") %>%
  select(customer_id, name, city) %>%
  distinct()

d). Display the result

print(scenario_a)

## # A tibble: 7 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago    
## 4           4 David   Houston    
## 5           5 Eve     Phoenix    
## 6           6 <NA>    <NA>       
## 7           7 <NA>    <NA>

print(scenario_b)

## # A tibble: 5 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago    
## 4           6 <NA>    <NA>       
## 5           7 <NA>    <NA>

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.

summary_data <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(name, city) %>%
  summarize(
    total_orders = n_distinct(order_id, na.rm = TRUE),
    total_amount_spent = sum(amount, na.rm = TRUE)
  ) %>% 
  ungroup()

print(summary_data)

## # A tibble: 5 × 4
##   name    city        total_orders total_amount_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

Assignment 4 - Table Joins

Anthony Fiatarone

2024-09-29

Question 1: Inner Join (3 points)

a). How many rows are in the result?

There are 4 rows.

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

Only customers and orders with matching customer_id in both tables are included.

The customer table has only 3 customers in the orders table where one customer has 2 orders.

c). Display the result

Question 2: Left Join (3 points)

a). How many rows are in the result?

There are 6 rows.

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

In the left join, all customers are included, even those without order information.

Customer_id 4 and 5 did not have an order_id but are still included as part of the customers table.

c). Display the result

Question 3: Right Join (3 points)

a). How many rows are in the result?

There are 6 rows.

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

Customer_id 6 and 7 have null for name and city.

In right join, all order_id entries are included, even for customer_id 6 & 7 who are not in the customers table.

Customer_id 6 & 7 only have orders and not name and city customer information.

c). Display the result

Question 4: Full Join (3 points)

a). How many rows are in the result?

There are 8 rows.

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

Customer_id 4 & 5 only have information from customers table and customer_id 6 & 7 only have information from orders table.

In full join, all rows from both tables are included and NA is marked where there’s no match between the tables for that customer_id.

c). Display the result

Question 5: Semi Join (3 points)

a). How many rows are in the result?

There are 3 rows.

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

In semi join, only customers with order information are returned, but order data isn’t included.

Inner join included the order information that matched with the customer_id but here order information does not matter.

That is also why customer_id 2 “Bob” is only shown once here, in this table we don’t care how many orders he has had, only if he has ordered at all.

c). Display the result

Question 6: Anti Join (3 points)

a). Which customers are in the result?

Customer_id 4 David & 5 Eve are included in the result.

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

These customers have customer information but they do not have any order information.

In anti join, only customers without order information are included.

c). Display the result

Question 7: Practical Application (4 points)

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

To find all customers including those who haven’t placed any orders you would use the full join function.

The full join function would show all customers from both tables and would mark NA down for those who have not ordered.

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

To find only customers who have placed orders you would use the right join.

It will display all orders, even those that do not have customer information such as name and city (the customer_id will still be there).

c). Write the R code for both scenarios.

d). Display the result

Challenge Question (3 points)