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)`a) How many rows are in the result? The data has 4 rows
nrow(q1)## [1] 44 rows come up in the resulting table.
b) Why are some customers or orders not included in the result?
Innerjoin returns results only when there is a match on both tables involved. The customer table has five individual customers and customer IDs; the orders table only includes three of those customers and IDs. So, the other customers’ orders (customers 6 and 7) are not going to show up.
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 300Left 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? Six rows come up in the result. b) Explain why this number differs from the inner join result. The left join function keeps all rows from the left table (which was customers), so all five customers were shown in the new table result. In the inner join result, only matching rows are retained. Additionally, since customer 2, Bob, had two orders, the table produced a new row for each order. 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> NARight 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. Customers 6 and 7 have NULL for customer name and city because their ids are not associated with an address in the original customer dataset. 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 150Full 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. The rows with customers 4,5,6 and 7 are incomplete; customers 4 and 5 have an address registered in the customers dataset, but have not placed any orders according to the orders set. The opposite is true for customers 6 and 7, who each ordered something but are not in the customers dataset with a name or address. Full join just includes all rows from both tables, with the NAs filling in missing data. c) Display the result.
print(q4)## # A tibble: 8 × 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
## 7 6 <NA> <NA> 105 Camera 600
## 8 7 <NA> <NA> 106 Printer 150Semi 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? The semi join table only includes rows from the customers that have a match in the orders table. No columns from the right table (orders) are included. 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 ChicagoAnti 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? The customers that didn’t place orders were in the result. b) Explain what this result tells you about these customers. This result tells me that these customers did not place orders, which is pretty cool to be able to do with anti join. c) Display the result.
head(q6)## # A tibble: 2 × 3
## customer_id name city
## <dbl> <chr> <chr>
## 1 4 David Houston
## 2 5 Eve PhoenixPractical 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’d use left join because it takes all the customers. b) Which join would you use to find only the customers who have placed orders? Why? I’d use inner join, since it returns customers that also show up on the orders table (which means they’ve placed an order). c) Write the R code for both scenarios.
left <- left_join(customers, orders)## Joining with `by = join_by(customer_id)`inner <- inner_join(customers, orders)## Joining with `by = join_by(customer_id)`d) Display the result.
print(left)## # 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> NAprint(inner)## # 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 300Challenge 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
summary <- full_join(customers, orders) %>%
group_by(customer_id,name, city) %>%
summarise(
total_orders_placed = n_distinct(order_id, na.rm = TRUE),
total_amount_spent = sum(amount, na.rm = TRUE)
)## Joining with `by = join_by(customer_id)`
## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.print(summary)## # A tibble: 7 × 5
## # Groups: customer_id, name [7]
## customer_id name city total_orders_placed total_amount_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 0 0
## 5 5 Eve Phoenix 0 0
## 6 6 <NA> <NA> 1 600
## 7 7 <NA> <NA> 1 150