Assignment 4

Dataset 1: Customers

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

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.

t

q1 <- inner_join(customers, orders)

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

a. How many rows are in the result?

nrow(q1)

## [1] 4

## a. 4 rows

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

## b. Interjoin returns only when there is a match

c. Display the resul

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

2. 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)`

a.How many rows are in the result?

# 6 rows
nrow(q2)

## [1] 6

b. Explain why this number differs from the inner join result.

##left join uses all the rows, interjoin only uses rows that match 

c. 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

3. 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)`

a. How many rows are in the result?

## 6 rows
nrow(q3)

## [1] 6

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

##ID 6 and 7 have NULL for customer name and city because in the orders table, there are no names and cities unlike the customers table. Because of that, when they right joined, they have no info in the table 

c. 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

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

q4 <- full_join(customers, orders)

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

a. How many rows are in the result?

## 8 Rows
nrow(q4)

## [1] 8

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

## In Customers 3 and 4 order ID, product and amount appears N/A. For customers 6 and 7, name and city appear N/A. This is becasue in the full join, all of both tables join together meaning all ther parts from customers and orders which don't commonly appear on both tables will still show in the full join.

c. 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

5. 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)`

a. How many rows are in the result?

## 3 rows 
nrow(q5)

## [1] 3

b. How does this result differ from the inner join result?

## The table differs from the inner join because the semi join uses 1 row for the customer ID 2 and doesn't have an amount, product or order ID row.

c. 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

6. 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)`

a. Which customers are in the result?

## 2 rows 
nrow(q6)

## [1] 2

b. Explain what this result tells you about these customers.

## The table tells us about the two customers who have not placed orders yet

c. Display the result

head(q6)

## # 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

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

##I would use a outer join because it displays the entire customer base, regardless of whether they have placed an order or not

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

##I would use an right join because it only displays rows where there are matches between both tables, showing only the customers who have placed orders

c. Write the R code for both scenarios.

q7a <- customers %>%
  left_join(orders, by = "customer_id")

d. 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