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)
)

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

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

How many rows are in the result?

nrow(inner_join_result)

## [1] 4

Why are some customers or orders not included in the result?
Inner join only displays rows where the customer_id matches and is present in both databases.

Display the result

   print(inner_join_result)

## # 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

2 Left Join Perform a left join with customers as the left table and orders as the right table.

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

How many rows are in the result?

nrow(left_join_result)

## [1] 6

Explain why this number differs from the inner join result.
This number is different because it includes all the customers from the customer database rather than just the matching customers from both databases.

Display the result

 print(left_join_result)

## # 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

3 Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

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

How many rows are in the result?

    nrow(right_join_result)

## [1] 6

Which customer_ids in the result have NULL for customer name and city? Explain why.
The customer_ids 6 and 7 have null as customer name and city, this is because it includes all the information from the order database rather than just the matching order data from both databases.

Display the result

print(right_join_result)

## # 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

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

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

How many rows are in the result?

   nrow(full_join_result)

## [1] 8

Identify any rows where there’s information from only one table. Explain these results.
Rows 5 and 6 only have information from only one table, the customer table, you can tell this by the NA displayed under the order_id, product, and amount but the display of the name and city from the customer table. Rows 7 and 8 only have information from one table, the order table, you can tell this by the information showing NA for the name and city information but having the order data.

Display the result

print(full_join_result)

## # 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

5 Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

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

How many rows are in the result?

  nrow(semi_join_result)

## [1] 3

How does this result differ from the inner join result?
Semi join only returns all matching rows in the left table where there is a match with the right, it yields one value per match. Inner join returns all matched rows in both tables and can have multiple values per customer_id.

Display the result

print(semi_join_result)

## # 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

6 Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

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

Which customers are in the result?
Customer_id 4 and 5, their names are David and Eve and they are the only two customers in the result.

Explain what this result tells you about these customers.
This tells us that these customers in the left tables do not have matches in the right table so they appear in the anti join.

Display the result

print( anti_join_result)

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

7 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?

You would use a left join to find all customers including those who haven’t placed order because this function joins all the customer information from the left customer table and then corresponding information from the order table, it will display NA if they have not placed an order.

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

You would use an inner join to find only customers who have placed orders because this function joins both tables only where the customer_id matches in both tables, this ensure each customer has at least one match.

Write the R code for both scenarios

 left_join_result <- left_join(customers, orders, by = "customer_id")
  inner_join_result <- inner_join(customers, orders, by = "customer_id")

Display the result

 print(left_join_result)

## # 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(inner_join_result)

## # 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

8 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.

   summary_result <- customers %>%
      left_join(orders, by = "customer_id") %>%
      group_by(name, city) %>%
      summarize(
        total_orders = n(),                     
        total_amount_spent = sum(amount, na.rm = TRUE),
        .groups = "drop"               
      )

Display the result

  print(summary_result)

## # 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                1                  0
## 5 Eve     Phoenix                1                  0

Assignment 4

Stephanie Reap

2024-10-08

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

2 Left Join Perform a left join with customers as the left table and orders as the right table.

3 Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

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

5 Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

6 Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

7 Practical Application (4 points) Imagine you’re analyzing customer behavior.

8 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.