Q1: 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?

Because they do not have a match in both tables, meaning they do not have a customer ID

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

Q2: 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

Explain why this number differs from the inner join result.

This includes customers without any purchases

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

Q3: 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

6 and 7.

Display the results

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

Q4: 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 1 table. Explain these results

Display the results

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

Q5: 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?

It does not include duplicate customers

Display the results

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

Q6: 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.

That David and Eve did not place any orders

Display the result

head(q6)
## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

Q7: Imagine you’re analyzing customer behavior

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

Left join because you want to see everything from both, customers and orders, tables

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

Inner join because it will result in rows where customers have placed orders, a match between the two tables

Write the R code for both scenarios.

q7a <- left_join(customers, orders, by = 'customer_id') 
q7b <-  inner_join(customers, orders, by = 'customer_id')

Display the results

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

Q8: 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.

q8 <- customers %>% 
    left_join(orders, by = 'customer_id') %>%  
    group_by(customer_id, name, city) %>% 
   summarize(
        total_orders = n(), 
        total_spent = sum(amount, na.rm = TRUE)) %>%  
  ungroup () 
## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.
head(q8)
## # A tibble: 5 × 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