Assignment 4

Inner Join (3 points) Perform an inner join between the customers and orders datasets.

q1 <- inner_join(customers, orders, by = "customer_id")

How many rows are in the result?

There are 4 rows.

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

Some customers or orders are not included in the result because they do not have a match in the other table.

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, by = "customer_id")

How many rows are in the result?

There are 6 rows.

Why is this number different from inner join result?

This number of rows is different from the inner join result because it shows the values that produce NA results.

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, by = "customer_id")

How many rows are in the result?

There are 6 rows.

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

customer_ids 6 and 7 both have NULL results for customer name and city because there is no customer_id data for 6 and 7 in the Customers database

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, by = "customer_id")

How many rows are in the result?

There are 8 rows.

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

Rows 5, 6, 7, and 8 all are missing information. Rows 5 and 6 are missing order_id information for customers with the customer_id of 4 and 5. Rows 7 and 8 are missing customer_id information with orders with the order_id of 105 and 106.

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    150

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, by = "customer_id")

How many rows are in the result?

There are 3 rows.

How does this result differ from the inner join result?

This result does not duplicate any same results while the inner join has duplicates.

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, by = "customer_id")

Which customers are in the result?

Customers David (customer_id of 4) and Eve (customer_id of 5)

Explain what this result tells you about these customers.

This result means that David and Eve have missing information in the orders database

q6

## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

Practical Application:

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

The result that includes all customers, even those who haven’t placed an order, would be left join because it keeps all the rows from the customers table and fills missing order information with NA.

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

The result that inlcudes only customers who have placed orders, would be inner join because this keeps only matching records from both tables.

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

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

Challenge Question: 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.

challenge <- left_join(customers, orders, by = "customer_id") %>%
  group_by(name, city) %>%  
  summarize(
    total_orders = sum(!is.na(order_id)), 
    total_spent = sum(amount, na.rm = TRUE)
  ) %>%
  ungroup() 
challenge

## # A tibble: 5 × 4
##   name    city        total_orders total_spent
##   <chr>   <chr>              <int>       <dbl>
## 1 Alice   New York               1        1200
## 2 Bob     Los Angeles            2        2300
## 3 Charlie Chicago                1         300
## 4 David   Houston                0           0
## 5 Eve     Phoenix                0           0

Assignment 4

Madison Ryder

Inner Join (3 points) Perform an inner join between the customers and orders datasets.

How many rows are in the result?

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

Some customers or orders are not included in the result because they do not have a match in the other table.

Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

How many rows are in the result?

There are 6 rows.

Why is this number different from inner join result?

This number of rows is different from the inner join result because it shows the values that produce NA results.

Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

How many rows are in the result?

There are 6 rows.

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

customer_ids 6 and 7 both have NULL results for customer name and city because there is no customer_id data for 6 and 7 in the Customers database

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

How many rows are in the result?

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

Rows 5, 6, 7, and 8 all are missing information. Rows 5 and 6 are missing order_id information for customers with the customer_id of 4 and 5. Rows 7 and 8 are missing customer_id information with orders with the order_id of 105 and 106.

Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

How many rows are in the result?

How does this result differ from the inner join result?

This result does not duplicate any same results while the inner join has duplicates.

Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

Which customers are in the result?

Customers David (customer_id of 4) and Eve (customer_id of 5)

Explain what this result tells you about these customers.

This result means that David and Eve have missing information in the orders database

Practical Application:

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

The result that includes all customers, even those who haven’t placed an order, would be left join because it keeps all the rows from the customers table and fills missing order information with NA.

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

The result that inlcudes only customers who have placed orders, would be inner join because this keeps only matching records from both tables.

Challenge Question: 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.