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?
Inner join returns only rows that match with the Customer Id.The Id 2 is used twice so it is seen in two of the rows.
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? There are 6 rows.
Explain why this number differs from the inner join result. Inner join returns the rows that have matches with the customer Id, whereas left join returns all of the rows from the left table and matches them with rows from the right 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)

## Joining with `by = join_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 7 have NULL for customer name and city because they are recorded in the orders table but are not in the customers table. When these joined, only customer Ids that were present in the customers table are present in the new 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)

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

How many rows are in the result? There are 6 rows. b)Identify any rows where there’s information from only one table. Explain these results. Rows 5,6,7, and 8 include information from only one table. Full join returns all rows where there is a match in the left or the right 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? There are 3 rows.
How does this result differ from the inner join result? Semi join returns all the rows from the left table where there is a match in the right table. There is one less row in this result because Bobs orders are combined since the semi join is just returning the customer Id, name, and city which are all the same for Bob’s two 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? David and Eve are in the result.
Explain what this result tells you about these customers. Since anti join returns all variables from the left table where there is no match from the right, David and Eve do not have any orders recorded in the order table.
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 left join because the customers are all listed in the left table so it will have to return all of the customers regardless of if they have placed an order.
Which join would you use to find only the customers who have placed orders? Why? I would use right joining because that takes the orders table and combines it only when there is a match form the customers table. The customers that have not placed orders will have N/A in the category for order number indicating they have no orders. c)Write the R code for both scenarios.

a7 <- left_join(customers , orders)

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

b7 <- right_join(customers , orders)

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

Display the result

head(a7)

## # 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(b7)

## # 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.

Assignment 4

Olivia Burghardt

2024-10-04