Assignment 4

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.

q1 <- inner_join(customers , orders)

## Joining with `by = join_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?
An inner join returns where there is a match in both tables. The customer data has only 3 customers in the orderwhere one customer has 2 orders.
Display the result

head(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.

# Perform a left join
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.
The number of rows in a left join is higher than in an inner join because the left join keeps all rows from the left table (in this case, the customers table), even if there’s no matching data in the orders table.
Display the result

head(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.

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_id 6 and customer_id 7 have NULL for customer name and city. These are the customer_ids that appear in orders but do not exist in the customers table, which is why their name and city are NULL. They represent orders placed by customers who are not listed in the customers table.
Display the result

head(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.

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.
Rows 5, 6, 7, and 8 have information from only one table. These rows occur when a customer_id exists in one table but not the other.
Display the result

head(q4)

## # 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

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?
The semi join includes only the columns from the customers table for matching customer_ids. In contrast, the inner join includes columns from both tables for matching rows. The row count will be the same as the inner join, but the semi join doesn’t include information about the orders (like order_id, product, or amount).
Display the result

head(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.

q6 <- anti_join(customers, orders, by = "customer_id")

Which customers are in the result?
David and Eve
Explain what this result tells you about these customers.
The result tells us that these customers are registered in the customers table but have not made any purchases recorded in the orders table. Essentially, these customers haven’t placed any orders.
Display the result

head(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.

q7 <- left_join(customers, orders, by = "customer_id")

Which join would you use to find all customers, including those who haven’t placed any orders? Why?
You would use a left join for this scenario. A left join will return all customers from the customers table, along with any matching orders from the orders table. If a customer hasn’t placed any orders, their order details will be NA, but they will still be included in the result.
Which join would you use to find only the customers who have placed orders? Why?
You would use a semi join for this scenario. A semi join returns only the customers who have placed orders, providing a clean list of those customers without duplicating order details. This is efficient when you only need customer information without any order data.
Write the R code for both scenarios.

all_customers_with_orders <- left_join(customers, orders, by = "customer_id")
customers_with_orders <- semi_join(customers, orders, by = "customer_id")

Display the result

head(q7)

## # 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

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

summary_result <- customer_orders_summary %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = sum(!is.na(order_id)),  # Count only non-NA order_id
    total_amount_spent = sum(amount, na.rm = TRUE)  # Sum total amount spent
  ) %>%
  ungroup()

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

head(summary_result)

## # A tibble: 5 × 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

2024-09-27