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?

  nrow(q1)

## [1] 4

Why are some customers or orders not included in the result?

Because the customers and orders that are not included

did not have a match in the other table.

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?

nrow(q2)

## [1] 6

Explain why this number differs from the inner join result.

All customers are included, the left table is matched with rows from the right table. Customers without orders have NA values.

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?

  nrow(q3)

## [1] 6

Which customer_ids in the result have NULL for customer name and city? Explain why.

Customer_ids 6 & 7 exist in orders but don’t have corresponding customers.

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?

  nrow(q4)

## [1] 8

Identify any rows where there’s information from only one table. Explain these results.

Customer_ids 4 & 5 are only from one table since there are no matches for them in the other table,

meaning that they did not place an order.

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?

  nrow(q5)

## [1] 3

How does this result differ from the inner join result?

Returns only customers with mathcing orders, does not duplicate rows or include other columns.

Includes Alice, Bob, and Charlie’s 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 shown

Explain what this result tells you about these customers.

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.

Which join would you use to find all customers, including those who haven’t placed any orders? Why?

Left join because we want to keep the data from all the customers, even if they haven’t placed an order.

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

Inner join because we only want customers who have matching records in the orders table.

Write the R code for both scenarios. Display result.

all_customers <- left_join(customers, orders)

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

head(all_customers)

## # 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_orderes <- inner_join(customers, orders)

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

head(customers_orderes)

## # 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