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?
The data has 4 rows
Why are some customers or orders not included in the result?
Inner join returns where there is a match on both tables. The customer data has only 3 customers in the order table where 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.

Q2 <- left_join(customers, orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result? The data has 6 rows
Explain why this number differs from the inner join result. The left join includes all customers even those without order information.
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)

## Joining with `by = join_by(customer_id)`

How many rows are in the result? The data has 6 rows
Which customer_ids in the result have NULL for customer name and city? Explain why. The customer ids of 6 and 7 have NULL for customer name and city because those customer IDs are in the orders dataset but they are not included in the customers dataset.
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)

## Joining with `by = join_by(customer_id)`

How many rows are in the result? The data has 8 rows
Identify any rows where there’s information from only one table. Explain these results. In the customers table, customer IDs of 4 and 5 have information from only one table (customers) because they have not placed any orders. In the orders table, customer IDs 6 and 7 have information from only one table (orders) because they are not in the customers table.
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)

## Joining with `by = join_by(customer_id)`

How many rows are in the result? The data has 3 rows
How does this result differ from the inner join result? The results from the semi join returns rows from customers with matches in orders, showing only customer data. While the inner join returns rows with matches in both datasets, including all relevant data from both customers and orders.
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)

## Joining with `by = join_by(customer_id)`

Which customers are in the result?

Q6 %>%
  select(name)

## # A tibble: 2 × 1
##   name 
##   <chr>
## 1 David
## 2 Eve

Explain what this result tells you about these customers. These customers are part of the customers dataset but have not made any purchases reflected in the orders dataset.
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.

Which join would you use to find all customers, including those who haven’t placed any orders? Why? I would use the full join because it would combine the customer dataset and the orders dataset and return all rows from both datasets, so customers who have placed orders with their order detials and customers who haven’t placed orders with NA displayed in the order columns.
Which join would you use to find only the customers who have placed orders? Why? I would use the right join because then I will get a list of customers along with their order details for customers who actually placed orders.
Write the R code for both scenarios.

Q7A <- full_join(customers, orders, by = "customer_id")
Q7B <- right_join(customers, orders, by = "customer_id")

Display the result

head(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

head(Q7B)

## # 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

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.

EC <- full_join(customers, orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = sum(!is.na(order_id)),
    total_amount_spent = sum(amount, na.rm = TRUE)
  ) %>%
  arrange(name)

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

Assignment 4

Natalia A

2024-09-27