library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

Dataset 1: Customers

customers <- tibble(
  customer_id = c(1, 2, 3, 4, 5),
  name = c("Alice", "Bob", "Charlie", "David", "Eve"),
  city = c("New York", "Los Angeles", "Chicago", "Houston", "Phoenix")
)

Dataset 2: Orders

orders <- tibble(
  order_id = c(101, 102, 103, 104, 105, 106),
  customer_id = c(1, 2, 3, 2, 6, 7),
  product = c("Laptop", "Phone", "Tablet", "Desktop", "Camera", "Printer"),
  amount = c(1200, 800, 300, 1500, 600, 150)
)

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

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

How many rows are in the result?

 nrow(joined_data)

## [1] 4

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

Customers 4 and 5 are missing from the result because they do not have any matching orders in the orders dataset.

Orders 105 and 106 are missing because they belong to customer IDs 6 and 7, which do not exist in the customers dataset.

Display the result

print(joined_data)

## # 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.

left_joined_data <- left_join(customers, orders, by = "customer_id")

How many rows are in the result?

nrow(left_joined_data)

## [1] 6

Explain why this number differs from the inner join result.

The left join keeps all customers, even if they don’t have a matching order.

Display the result

print(left_joined_data)

## # 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.

right_joined_data <- right_join(customers, orders, by = "customer_id")

How many rows are in the result?

nrow(right_joined_data)

## [1] 6

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

Customer IDs 6 and 7 will have NA for name and city. These IDs exist in the orders dataset but do not exist in the customers dataset.

Display the result

print(right_joined_data)

## # 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.

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

How many rows are in the result?

nrow(full_joined_data)

## [1] 8

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

Customer IDs 4 (David) and 5 (Eve) exist in customers but have no orders.Customer IDs 6 and 7 exist in orders but have no corresponding entry in customers.

These results happen because the full join retains all records, even if there’s no match in the other table.

Display the result

print(full_joined_data)

## # 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.

semi_joined_data <- semi_join(customers, orders, by = "customer_id")

How many rows are in the result?

nrow(semi_joined_data)

## [1] 3

How does this result differ from the inner join result?

The semi join only returns the unique customers who have at least one order, without including order details, resulting in 3 rows.

Display the result

print(semi_joined_data)

## # 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.

anti_joined_data <- anti_join(customers, orders, by = "customer_id")

Which customers are in the result?

Customers David (ID 4) and Eve (ID 5) are in the result because they do not appear in the orders dataset.

Explain what this result tells you about these customers.

These customers exist in the customers dataset but have never placed an order in the orders dataset.

Display the result

print (anti_joined_data)

## # 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?

I would use left join as it finds all customers who have and have not placed orders.

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

I would use semi join as it filters out customers who have not placed orders

Write the R code for both scenarios.

all_customers <- left_join(customers, orders, by = "customer_id")

active_customers <- semi_join(customers, orders, by = "customer_id")

Display the result

print(all_customers)

## # 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

print(active_customers)

## # 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

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

## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.

print(customer_summary)

## # A tibble: 5 × 5
## # Groups:   customer_id, name [5]
##   customer_id name    city        total_orders total_amount_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

Assignment 4

Sam Elliott

2025-02-20

Dataset 1: Customers

Dataset 2: Orders

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

How many rows are in the result?

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

Customers 4 and 5 are missing from the result because they do not have any matching orders in the orders dataset.

Orders 105 and 106 are missing because they belong to customer IDs 6 and 7, which do not exist in the customers dataset.

Display the result

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?

Explain why this number differs from the inner join result.

The left join keeps all customers, even if they don’t have a matching order.

Display the result

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?

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

Customer IDs 6 and 7 will have NA for name and city. These IDs exist in the orders dataset but do not exist in the customers dataset.

Display the result

Full Join (3 points) Perform a full join between customers and orders.

How many rows are in the result?

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

Customer IDs 4 (David) and 5 (Eve) exist in customers but have no orders.Customer IDs 6 and 7 exist in orders but have no corresponding entry in customers.

These results happen because the full join retains all records, even if there’s no match in the other table.

Display the result

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?

How does this result differ from the inner join result?

The semi join only returns the unique customers who have at least one order, without including order details, resulting in 3 rows.

Display the result

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?

Customers David (ID 4) and Eve (ID 5) are in the result because they do not appear in the orders dataset.

Explain what this result tells you about these customers.

These customers exist in the customers dataset but have never placed an order in the orders dataset.

Display the result

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?

I would use left join as it finds all customers who have and have not placed orders.

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

I would use semi join as it filters out customers who have not placed orders

Write the R code for both scenarios.

Display the result

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.