library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
# Dataset 1: Customers
customers <- tibble(
  customer_id = c(1, 2, 3, 4, 5),
  name = c("Alice", "Bob", "Charlie", "David", "Eve"),
  city = c("New York", "Los Angeles", "Chicago", "Houston", "Phoenix")
)

# Dataset 2: Orders
orders <- tibble(
  order_id = c(101, 102, 103, 104, 105, 106),
  customer_id = c(1, 2, 3, 2, 6, 7),
  product = c("Laptop", "Phone", "Tablet", "Desktop", "Camera", "Printer"),
  amount = c(1200, 800, 300, 1500, 600, 150)
)
  1. 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)`

a) How many rows are in the result?
There are 4 rows.
b) Why are some customers or orders not included in the result?
Inner join returns where there is a match on both tables. The customer data has only 3 customers in the order table where one customer has 2 orders.
c) 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
  1. 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)
## Joining with `by = join_by(customer_id)`

a) How many rows are in the result?
There are 6 rows.
b) Explain why this number differs from the inner join result.
This table differs from the inner join table because this table includes David and Eve.
c) 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
  1. 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)
## Joining with `by = join_by(customer_id)`

a) How many rows are in the result?
There are 6 rows.
b) Which customer_ids in the result have NULL for customer name and city? Explain why. Customer id 6 and 7 have null for customer name and city because there are only 5 customers in our customers table.
c) 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
  1. Full Join (3 points) Perform a full join between customers and orders.
q4 <- full_join(customers , orders)
## Joining with `by = join_by(customer_id)`

a) How many rows are in the result?
There are 8 rows.
b) Identify any rows where there’s information from only one table. Explain these results.
Rows 5,6,7,8 have NA values meaning there is a lack of information from both the customer and order tables.
c) Display the result

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
  1. 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)
## Joining with `by = join_by(customer_id)`

a) How many rows are in the result?
There are 3 rows.
b) How does this result differ from the inner join result?
The inner join code included 4 rows (one more than the semi join), two for Bob because he has two devices. There are also three additional columns order_id, product, and amount.
c) 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
  1. 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)
## Joining with `by = join_by(customer_id)`

a) Which customers are in the result?
There are 2 rows.
b) Explain what this result tells you about these customers.
This result tells us the two customers whose Customer ID appears in the customers data but not in orders data.
c) Display the result

head(q6)
## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix
  1. Practical Application (4 points) Imagine you’re analyzing customer behavior.
    a) Which join would you use to find all customers, including those who haven’t placed any orders? Why?
    I would use a left join to find all customers as it would take customer ID from the left data table labeled customers.
    b) Which join would you use to find only the customers who have placed orders? Why?
    I would use inner join because it will match corresponding customer ID’s between customers and orders.
    c) Write the R code for both scenarios.
q7 <- left_join(customers , orders)
## Joining with `by = join_by(customer_id)`
q8 <- inner_join(customers, orders)
## Joining with `by = join_by(customer_id)`

d) 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(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
  1. 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.

Step 1: Perform a left join to include all customers, even those without orders

customer_orders <- left_join(customers, orders)
## Joining with `by = join_by(customer_id)`

Step 2: Group by customer to calculate total number of orders and total amount spent

customer_summary <- customer_orders %>%
  group_by(customer_id, name, city) %>%
  summarize(
    total_orders = sum(!is.na(order_id)),  # Count only non-NA order_ids
    total_spent = sum(amount, na.rm = TRUE)  # Sum the amount, ignoring NA values
  )
## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.
customer_summary
## # A tibble: 5 × 5
## # Groups:   customer_id, name [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