1. 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?

They’re are 4 rows

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

They did not have a similar match in the other table

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     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 , by = 'customer_id')

How many rows are in the result?

They’re are 6 rows

Explain why this number differs from the inner join result.

This differs inner join because left join includes customer ids in the left table that arent in the right as well as ids in the left that match the right

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 , by = 'customer_id')

How many rows are in the result?

They’re are 6 rows

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

Customer ids 6 and 7 have NA for customer name and city because they appear in the right table, but their customer id is not shown in the left table.

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 , by = 'customer_id')

How many rows are in the result?

They’re are 8 rows

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

Customers 4,5,6 and 7 in rows 5,6,7 and 8 display information from only one table. This occurs because two of the customers do not have their IDs present in the left table and the other 2 do not have their IDs present in the right table.

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    150

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 , by = 'customer_id')

How many rows are in the result?

They’re are 3 rows

How does this result differ from the inner join result?

This result only displays 3 rows opposed to 4. It also only includes customer_id, name, and city.

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 , by = 'customer_id')

Which customers are in the result?

David and Eve

Explain what this result tells you about these customers.

The result displays David and Eve because they appear in the left table, but do not appear in the right table.

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.

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

I would use full_join to find all customers. This function would display all customer IDs present in both tables even if the ID only appears in one table.

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

I would use right_join to find customers who have placed orders. This function displays all customer IDs present in the orders table and the mathcing customer IDs from the right table.

Write the R code for both scenarios.

q7 <- list(full_join(customers , orders , by = 'customer_id') , right_join(customers , orders , by = 'customer_id'))

Display the result

print(q7)
## [[1]]
## # 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
## 
## [[2]]
## # 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

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

Display the result

  print(summary)
## # A tibble: 5 × 5
## # Groups:   customer_id, name [5]
##   customer_id name    city        total_orders total_amount
##         <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