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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 Rows

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 rows

Explain why this number differs from the inner join result.

The number of rows is higher than the inner join since 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? 6

####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.

’Row 5 and 6 have customers with no order, and rows 7 and 8 have orderes with no matching customers.

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?

The result differed because the inner join showed all of his orders placed, which shows Bob’s two orders, but in this result it only shows Bob once because it only goes by customers not orders.

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 they have not placed any

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

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

Inner Join, because its used to see only customers who placed orders

q8 <- inner_join(customers, orders, by = "customer_id")
q8
## # 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

####Write the R code for both scenarios.Display the result Results Above

9. 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.

q9 <- 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.
q9
## # 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