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

q1 <- inner_join(customers , orders)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 4 rows

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

Because the customers and orders that are no included did not have a 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

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?

The resulting data has 6 rows

Explain why this number differs from the inner join result.

The data is joined primarily from the data on the left, therefore, although some customer_id’s are not linked to an order_id they will still be included as they are collected from the left table first.

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

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?

The resulting data has 6 rows

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

It is a right join adn thesefore all order_id’s are inlcuded. some order_id’s do not link with an existing customer_id bust are still included in the table due to the type of join

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

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

 q4 <- full_join(customers, orders)
## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 8 rows

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

 customers_only <- q4 %>%
   filter(is.na(order_id))

customers_only
## # A tibble: 2 × 6
##   customer_id name  city    order_id product amount
##         <dbl> <chr> <chr>      <dbl> <chr>    <dbl>
## 1           4 David Houston       NA <NA>        NA
## 2           5 Eve   Phoenix       NA <NA>        NA
orders_only <- q4 %>%
  filter(is.na(name))

orders_only
## # A tibble: 2 × 6
##   customer_id name  city  order_id product amount
##         <dbl> <chr> <chr>    <dbl> <chr>    <dbl>
## 1           6 <NA>  <NA>       105 Camera     600
## 2           7 <NA>  <NA>       106 Printer    150

full joins bring together all data despite if it has data relevent to both tables therefore all data in both customers and orders tables will be inlcuded

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?

The resulting data has 3 rows

How does this result differ from the inner join result?

semi join returns only rows from customers, it also only does not duplicate rows nor customers who have not ordered

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

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?

q6 %>%
  select(customer_id, name)
## # A tibble: 2 × 2
##   customer_id name 
##         <dbl> <chr>
## 1           4 David
## 2           5 Eve

Explain what this result tells you about these customers.

Anti join returns customers who do not have matching rown in 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

Practical Application (4 points) Imagine you’re analyzing customer behavior.

q7 <- left_join(customers, orders, by = "customer_id")

the left join keeps every customer in the dataset

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

q7b <- inner_join(customers, orders, by = "customer_id")

this shows only the real customers who have placed orders

Display the result

head(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
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

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.

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

head(challenge_table)
## # 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                0           0
## 5           5 Eve     Phoenix                0           0