Assignment 4

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

q1 <- inner_join(customers,orders,by = "customer_id")

a)How many rows are in the result:

Ans: 4 rows

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

Ans: Because only 4 of the customer id’s match

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

a) How many rows are in the result:

Ans: 6 rows

b) Explain why this number differs from the inner join result:

Ans: because it includes all of the rows from the left data dataset (5) and one extra matching from the right data set.

c)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<- right_join(customers, orders, by ="customer_id")

a) How many rows are in the result:

Ans: 6 rows

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

Ans: Id’s 6 & 7 because these numbers are within the right data set but have no match from the left which is the set that has name and city information

c) 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 <- full_join(customers, orders, by ="customer_id")

a) How many rows are in the result:

Ans: 8 rows

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

Ans: Id #’s 4,5,6, and 7 are missing information - 4&5 are missing order_id and product. This means there is no match and the row comes from the left dataset only. - 6&7 are missing name and city info. This means there is no match and the row comes from the right dataset only.

c) 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 <- semi_join(customers, orders, by ="customer_id")

a) How many rows are in the result:

Ans: 3 rows

b) How does this result differ from the inner join result:

Ans: It contains one less row and only name and city information. This is because it only outputs left rows wit a right row match ID, and only the left row information.

c) 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 <- anti_join(customers, orders, by ="customer_id")

a) Which customers are in the result:

Ans: Id# 4 and 5

b) Explain what this result tells you about these customers.

Ans: These customer’s id#’s are in the customers (left) dataset but not the right dataset, and have no match

c) 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.

a) Which join would you use to find all customers, including those who haven’t placed any orders? Why?

Ans: left_Join because it gives all customer name and ID’s regardless of if they have a corresponding order ID.

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

Ans: semi_join because it oututs a customer from the left data set only when they have a matching id with an order from the right dataset

c) Write the R code for both scenarios.

left_join(customers, orders, by = “customer_id”)
semi_join(customers, orders, by =“customer_id”)

d) 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

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

Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent.

Include:

customer name, city, total # of orders, total amt spent

CQnums <- full_join(customers, orders, by = "customer_id")%>% #gives ttl # orders & ttl amt spent
  
group_by(customer_id) %>% #groups customers with same id

  summarise(
    
    order_count = sum(!is.na(order_id)), #counts how many rows are in each group
    total_amount_spent = sum(amount) # creates new column based on sum of amt of the rows within each group
  )



CQ_ans <- full_join(customers, CQnums, by= "customer_id")
print (CQ_ans)

## # A tibble: 7 × 5
##   customer_id name    city        order_count total_amount_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                 NA
## 5           5 Eve     Phoenix               0                 NA
## 6           6 <NA>    <NA>                  1                600
## 7           7 <NA>    <NA>                  1                150

Assignment 4

Conrad Tobiassen

2025-02-13

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

a)How many rows are in the result:

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

c) Display the result:

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

a) How many rows are in the result:

b) Explain why this number differs from the inner join result:

c)Display the result

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

a) How many rows are in the result:

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

c) Display the result

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

a) How many rows are in the result:

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

c) Display the result

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

a) How many rows are in the result:

b) How does this result differ from the inner join result:

c) Display the result

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

a) Which customers are in the result:

b) Explain what this result tells you about these customers.

c) Display the result

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?

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

c) Write the R code for both scenarios.

d) Display the result

Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent.

Include:

customer name, city, total # of orders, total amt spent