Setup

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.

a. How many rows are in the result?

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

c. Display the result

q1 <- inner_join(customers, orders)
## Joining with `by = join_by(customer_id)`

a. 4 rows

nrow(q1)
## [1] 4

b. Inner join returns where there is a match. Some customer IDs in both customers and orders do not match. This is why they are not included. 3 IDs match but there are 4 rows due to a duplicate customer ID.

c. Display

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.

a.How many rows are in the result?

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

c.Display the result

q2 <- left_join(customers, orders)
## Joining with `by = join_by(customer_id)`

a. 6 rows

nrow(q2)
## [1] 6

b. Left join uses all the rows from the customer IDs (even rows with no data) where as inner join only uses rows that have matching customer IDs.

c. Display

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.

a. How many rows are in the result?

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

c. Display the result

q3 <- right_join(customers, orders)
## Joining with `by = join_by(customer_id)`

a. 6 rows

nrow(q3)
## [1] 6

b. Customer IDs 6 and 7 have null for name and city because in the orders table there are no names and cites unlike customers table. So, when they right joined, they appear with no information in the table.

c. Display

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.

a. How many rows are in the result?

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

c. Display the result

q4 <- full_join(customers, orders)
## Joining with `by = join_by(customer_id)`
  1. 8 rows
nrow(q4)
## [1] 8

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

c. Display

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.

a. How many rows are in the result?

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

c. Display the result

q5 <- semi_join(customers, orders)
## Joining with `by = join_by(customer_id)`

a. 3 rows

nrow(q5)
## [1] 3

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

c. Display

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.

a. Which customers are in the result?

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

c. Display the result

q6 <- anti_join(customers, orders)
## Joining with `by = join_by(customer_id)`

a. 2 rows

nrow(q6)
## [1] 2

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

c. Display

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?

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

c. Write the R code for both scenarios.

d. Display the result

a. I would use left join. This because left join shows the entire customer base, no matter if they had placed an order or not.

b. I would use inner join. This is becasue it only shows rows in which there are matches between both tables, leaving all the people who placed orders.

c1. left join

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

c2. inner join

q7b <- customers %>%
  inner_join(orders, by = "customer_id")

d. Display

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
head(q7b)
## # 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