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

How many rows are in the result?

4

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

Because they did not have a match on the other table

Display the result
q1 <- inner_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

6

Explain why this number differs from the inner join result.

The number of rows is higher then the interjoin since the customers without orders are included

Display the result
q2 <- left_join(customers , orders, by = 'customer_id')

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.

How many rows are in the result?

7

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

6 and 7 because they are missing in the customers file

Display the result
q3 <- right_join(customers , orders , by = 'customer_id')

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.

How many rows are in the result?

8

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

Rows 5 and 6 originate only from the customer table because these customers have no recorded orders. Rows 7 and 8 originate only from the order table because these orders are not linked to any known customer, possibly due to missing or incorrect customer information.

Display the result
q4 <- full_join(customers, orders, by = "customer_id")

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.

How many rows are in the result?

3

How does this result differ from the inner join result?

It doesn’t show the amount of orders just the people who made a order

Display the result
q5 <- semi_join(customers,orders,by = "customer_id")

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.

Which customers are in the result?

“David and Eve”

Explain what this result tells you about these customers.

These customers have no orders

Display the result
q6 <- anti_join(customers, orders, by = "customer_id")

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?

Left Join because its used to see all customers and their orders even if thery have not placed any.

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

Inter Join because its used to see onloy customers who placed orders

Write the R code for both scenarios.
Display the result
q7 <- left_join(customers, orders, by = 'customer_id')

q7
## # 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
q7 <- inner_join(customers, orders, by = 'customer_id')

q7
## # 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

8 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.

q8 <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarise(
    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.
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