Assignment 4

Inner Join (3 points) Perform an inner join between the customers and orders datasets.
## How many rows are in the result? # 4 rows ## Why are some customers or orders not included in the result? # Because, inner join returns data that is only a match on the other table by customer ID. ## Display the 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

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? # 6 ## Explain why this number differs from the inner join result. # Because left join returns matching data from the left data frame and the same from the right ##Display the 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

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? # 6 ## Which customer_ids in the result have NULL for customer name and city? Explain why. # customer_ids that have null are 6 & 7 because this represents the name & city of these two ID’s are unknown. ## Display the 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

Full Join (3 points) Perform a full join between customers and orders.
## How many rows are in the result? # 8 ## Identify any rows where there’s information from only one table. Explain these results. # rows with information from one table are 4, 5, 7 & 8. This means that some information is missing. ## Display the 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

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? # 3 ## How does this result differ from the inner join result? # inner join result had 4 rows ## Display the 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

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 in the result are David from Houston & Eve from Phoenix ## Explain what this result tells you about these customers. # it means these customers dont’t have any info for the product & the amount. ## Display the result

## # 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? # left join because it returns the customer names who made an order, including customers that haven’t ordered. ## Which join would you use to find only the customers who have placed orders? Why? # inner join because it returns customers who have an order. ## Write the R code for both scenarios. ## Display the result

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

    customers_with_orders <- customers %>%
      inner_join(orders, by = "customer_id")
    
    list("All Customers (Left Join)" = all_customers,
      "Customers with Orders (Inner Join)" = customers_with_orders)

## $`All Customers (Left Join)`
## # 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
## 
## $`Customers with Orders (Inner Join)`
## # 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

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.

## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_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                1           0
## 5           5 Eve     Phoenix                1           0

Assignment 4

Kai Ostaszewski, Michael

2025-02-18

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

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

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

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

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

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

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

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.