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.
q1 <- inner_join(customers, orders)## Joining with `by = join_by(customer_id)`a. How many rows are in the result?
The data has 4 rows
b. Why are some customers or orders not included in the
result?
Inner join returns where there is a match on both tables The customer
data only has 3 customers in the order table where one customer has 2
orders.
c. 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 3002. 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?
The data has 6 rows
b. Explain why this number differs from the inner join
result.
Inner join returns where there is a match on both tables The customer
data only has 3 customers in the order table where one customer has 2
orders.
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> NA3. 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?
The data has 6 rows
b. Which customer_ids in the result have NULL for customer
name and city? Explain why.
Customer_id 6 and 7 because they don’t have any name or city data. This
is because we take the data from the order table and match it to what we
have from customers.
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 1504. 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?
The data has 8 rows
b. Identify any rows where there’s information from only one
table. Explain these results.
Rows 5, 6, 7, and 8 because rows 5 and 6 don’t have matching data from
the orders table and rows 7 and 8 don’t have matching data from the
customers table.
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> NA5. 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?
The data has 3 rows
b. How does this result differ from the inner join
result?
It doesn’t join the orders table for semi join because there are no true
data matches.
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 Chicago6. 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, by = "customer_id")a. Which customers are in the result?
The data has 2 rows
b. Explain what this result tells you about these
customers.
Anti join tells us about the data that doesn’t match in both the
customers and orders tables which is customer_id 4 and 5, David and Eve.
The customers table is first so it will display any data where
customer_id doesn’t appear in orders table.
c. Display the result
head(q6)## # A tibble: 2 × 3
## customer_id name city
## <dbl> <chr> <chr>
## 1 4 David Houston
## 2 5 Eve Phoenix7. 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?
Left join because it returns the rows from the customers table that
matches rows from the orders table.
b. Which join would you use to find only the customers who
have placed orders? Why?
Inner join because it returns only the rows that have a match in both
customers and orders tables.
c. Write the R code for both scenarios.
PracticalApplication1 <- result_left_join <- customers %>%
left_join(orders, by = "customer_id")
PracticalApplication2 <- result_inner_join <- customers %>%
inner_join(orders, by = "customer_id")d. Display the result
head(PracticalApplication1)## # 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> NA8. 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.
summary_data <- customers %>%
left_join(orders, by = "customer_id") %>%
group_by(customer_id, name, city) %>%
summarize(total_orders = n_distinct(order_id), total_spent = sum(amount, na.rm = TRUE))## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.print(summary_data)## # A tibble: 5 × 5
## # Groups: customer_id, name [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