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

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)`

Joining with `by = join_by(customer_id)`

a. How many rows are in the result? The data has 4 rows.

b. Why are some customers or orders not included in the result? They did not have a match in the other table.

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

2. 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)`

Joining with `by = join_by(customer_id)`

a. How many rows are in the result? 6 rows

b. Explain why this number differs from the inner join result. Inner join returns where there is a match on both tables and the customer data only has 3 customers in the order table where one customer has 2 orders.

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

3. 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)`

Joining with `by = join_by(customer_id)`

a. How many rows are in the result? There are 6 rows of data.

b. Which customer_ids in the result have NULL for customer name and city? Explain why. Customer_id 6 and 7 because they don’t have any name or city data. This is because we take the data from the order table and match it to what we have from customers.

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

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

q4 <- full_join(customers, orders)

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

Joining with `by = join_by(customer_id)`

a. How many rows are in the result? 8 rows.

b. Identify any rows where there’s information from only one table. Explain these results. Rows 5, 6, 7, and 8 because rows 5 and 6 don’t have matching data from the orders table and rows 7 and 8 don’t have matching data from customers table.

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

5. 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)`

Joining with `by = join_by(customer_id)`

a. How many rows are in the result? The data has 3 rows

b. How does this result differ from the inner join result? It doesn’t join the orders table for semi join because there are no true data matches.

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

6. 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")

a. Which customers are in the result? 2 rows.

b. Explain what this result tells you about these customers. Anti join tells us about the data that doesn’t match in both the customers and orders tables which is customer_id 4 and 5, the customers names are David and Eve. The customers table is first so it will display any data where customer_id doesn’t appear in orders table.

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

7. 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? Left join as left join returns the rows from the customers table that match the corresponding rows in the orders table.

b. Which join would you use to find only the customers who have placed orders? Why? Inner join because it returns only the rows that have a match in both customers and orders tables.

c. Write the R code for both scenarios.

PracticalApplication1 <- result_left_join <- customers %>%
  left_join(orders, by = "customer_id")
PracticalApplication2 <- result_inner_join <- customers %>%
  inner_join(orders, by = "customer_id")

d.Display this result.

head(PracticalApplication1)

## # 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. 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.

summary_data <- customers %>%
  left_join(orders, by = "customer_id") %>%
group_by(customer_id, name, city) %>%
  summarize(total_orders = n_distinct(order_id), total_spent = sum(amount, na.rm = TRUE))

## `summarise()` has grouped output by 'customer_id', 'name'. You can override
## using the `.groups` argument.

`summarise()` has grouped output by ‘customer_id’, ‘name’. You can override

using the `.groups` argument.

print(summary_data)

## # 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                1           0
## 5           5 Eve     Phoenix                1           0

Assignment4Hurley

Sean Hurley

2025-02-18

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

Joining with `by = join_by(customer_id)`

2. Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.**

Joining with `by = join_by(customer_id)`

3. Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

Joining with `by = join_by(customer_id)`

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

Joining with `by = join_by(customer_id)`

5. Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

Joining with `by = join_by(customer_id)`

6. Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

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

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.

`summarise()` has grouped output by ‘customer_id’, ‘name’. You can override

using the `.groups` argument.

Assignment4Hurley

Sean Hurley

2025-02-18

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

Joining with by = join_by(customer_id)

2. Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.**

Joining with by = join_by(customer_id)

3. Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

Joining with by = join_by(customer_id)

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

Joining with by = join_by(customer_id)

5. Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

Joining with by = join_by(customer_id)

6. Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

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

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.

summarise() has grouped output by ‘customer_id’, ‘name’. You can override

using the .groups argument.

Joining with `by = join_by(customer_id)`

Joining with `by = join_by(customer_id)`

Joining with `by = join_by(customer_id)`

Joining with `by = join_by(customer_id)`

Joining with `by = join_by(customer_id)`

`summarise()` has grouped output by ‘customer_id’, ‘name’. You can override

using the `.groups` argument.