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

####How many rows are in the result?

Q1_rows <- nrow(Q1)
print(paste("Number of rows in the result:", Q1_rows))
## [1] "Number of rows in the result: 4"

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

#In an inner join, both datasets must have matching values in the key column for rows to appear in the final result. If there is no match in either dataset, that row is omitted.

####Display the result

print(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

Q2 <- customers %>%
  left_join(orders, by = "customer_id")

How many rows are in the result?

Q2_rows <- nrow(Q2)
print(paste("Number of rows in the result:", Q2_rows))
## [1] "Number of rows in the result: 6"

Explain why this number differs from the inner join

Left join preserves all rows from the left table, which is why it has more rows than the inner join, where only rows with matching customer id’s are retained

Display the result

print(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

Q3 <- customers %>%
  right_join(orders, by = "customer_id")

How many rows are in the result?

Q3_rows <- nrow(Q3)
print(paste("Number of rows in the result:", Q3_rows))
## [1] "Number of rows in the result: 6"

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

Customer_id 6 and 7 are not found in the customers dataset, so the name and city columns for these customer_ids will be NULL.

Display the result

print(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.

Q4 <- customers %>%
  full_join(orders, by = "customer_id")

How many rows are in the result?

Q4_rows <- nrow(Q4)
print(paste("Number of rows in the result:", Q4_rows))
## [1] "Number of rows in the result: 8"

Identify any rows where theres information from omly one table. Explain these results.

Rows where customer_id is present in customers but not in orders will have NA for order information Rows where customer_id is present in orders but not in customers will have NA for name and city

Display the result

print(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

Q5 <- customers %>%
  semi_join(orders, by = "customer_id")

How many rows are in the result?

Q5_rows <- nrow(Q5)
print(paste("Number of rows in the result:", Q5_rows))
## [1] "Number of rows in the result: 3"

How does this differ from the inner join result?

In a semi join, only the rows from the left table (customers) are returned, No columns from the right table (orders) are included. It shows the customers who have orders, not order details

display the result

print(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

Q6 <- customers %>% 
  anti_join(orders, by = "customer_id")

which customers are in the results?

Customers in the customer table but not in the order table

explain what this result tells you about these customers

Customers in the result have not placed any orders

display the result

print(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 since it keeps all rows from the customers table, regardless of status on the order table

customers_with_orders <- customers %>%
  left_join(orders, by = "customer_id")

print (customers_with_orders)
## # 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

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

Inner join because it only displays rows that have a match in both tables, meaning it will only include customers who were also included in the orders table

customers_with_orders <- customers %>%
  inner_join(orders, by = "customer_id")

print(customers_with_orders)
## # 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

customer_orders <- customers %>%
     left_join(orders, by = "customer_id")
summary_result <- customer_orders %>%
    group_by(name, city) %>%
     summarize(
        total_orders = n_distinct(order_id),
        total_amount_spent = sum(amount, na.rm = TRUE)
      )
## `summarise()` has grouped output by 'name'. You can override using the
## `.groups` argument.