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Join Implementation Part 1

The focus for this first part was on generating a base table for the category insights.

The table below would be the final output table for the category insights, that is why is important to get the base table first.

customer_id category_ranking category_name rental_count average_comparison percentile category_percentage
1 1 Classics 6 4 1 19
1 2 Comedy 5 4 2 16
2 1 Sports 5 3 7 19
2 2 Classics 4 2 11 15

From the table above I noticed that the columns of average_comparison, percentile and category_percentage needed the rental_count column for its calculations.

Then the focus would be on generating table that includes the customer_id, category_name and rental_count.

customer_id category_name rental_count
1 Classics 6
1 Comedy 5
2 Sports 5
2 Classics 4

To be able to generate the outout from above, table joins were requiered since the customer_id and the category_name data was not in the same table.

Table Journey

When looking at the ERD, the table joins would be from table 1 to table 5.

erd

The table joining journey for the catagory insights base table:

Join Journey Part Start End Foreign Key
Part 1 rental inventory inventory_id
Part 2 inventory film film_id
Part 3 film film_category film_id
Part 4 film_category category category_id

Each start and end point was analyzed in depth to proceed each joining journey with confidence.

Before joining the tables, i had to see which table join i should use by answering the following questions:

1. What is the purpose of joining these two tables?

a. What contextual hypotheses do we have about the data?

b. How can we validate these assumptions?

2. What is the distribution of foreign keys within each table?

3. How many overlapping and missing unique foreign key values are there between the two tables?

The questions were answered first for the rental and inventory tables which is the join journey part 1.

Join Journey Part 1

1. What is the purpose of joining these two tables?

The rental table does not track the records at the film_id but tracks the records with the inventory_id.

rental_id rental_date inventory_id customer_id return_date staff_id last_update
1 2005-05-24T22:53:30.000Z 367 130 2005-05-26T22:04:30.000Z 1 2006-02-15T21:30:53.000Z
2 2005-05-24T22:54:33.000Z 1525 459 2005-05-28T19:40:33.000Z 1 2006-02-15T21:30:53.000Z

The join with the inventory table helps to get the film_id of each rental because it matches up each rental record with its equivalent film_id value from the inventory table.

a. & b. What contextual hypotheses do we have about the data and how can we validate it?

The rental table contains every single rental for each customer so each valid rental record in the rental table should have a relevant inventory_id record.

Every item in the inventory table should have a unique inventory_id but there may also be multiple copies of a specific film.

Hypothesis 1:

The number of unique inventory_id records will be equal in both rental and inventory tables

Rental results:

SELECT 
  COUNT(DISTINCT inventory_id)
FROM dvd_rentals.rental;
count
4580

Inventory results:

SELECT 
  COUNT(DISTINCT inventory_id)
FROM dvd_rentals.inventory;
count
4581

Findings

There was one additional inventory_id value in the inventory table.

Hypothesis 2:

There will be a multiple records per unique inventory_id in the rental table

-- generate group by counts on the target_column_values column
WITH counts_base AS(
  SELECT 
    inventory_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.rental
  GROUP BY inventory_id
)
-- summarize the group by counts above by grouping again on the row_counts from counts_base CTE part
SELECT 
  row_counts,
  COUNT(inventory_id) AS count_inventory_ids
FROM counts_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_inventory_ids
1 4
2 1126
3 1151
4 1160
5 1139

Findings

There were multiple records per inventory_id value in the rental table.

Hypothesis 3:

There will be multiple inventory_id records per unique film_id value in the inventory table

-- generate group by counts on the target_column_values column
WITH counts_base AS(
  SELECT 
    film_id,
    COUNT(DISTINCT inventory_id) AS unique_record_counts
  FROM dvd_rentals.inventory
  GROUP BY film_id
)
-- summarize the group by counts above by grouping again on the row_counts from counts_base CTE part
SELECT 
  unique_record_counts,
  COUNT(film_id) AS count_film_ids
FROM counts_base
GROUP BY unique_record_counts
ORDER BY unique_record_counts;
unique_record_counts count_film_ids
2 133
3 131
4 183
5 136
6 187
7 116
8 72

Findings

There were multiple unique inventory_id per film_id value in the inventory table.

2. What is the distribution of foreign keys within each table?

rental distribution analysis on inventory_id foreign key:

-- first generate group by counts on the inventory_id column
WITH counts_base AS(
  SELECT 
    inventory_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.rental
  GROUP BY inventory_id
)
-- summarize the group by counts above by grouping again on the row_counts from counts_base CTE part
SELECT 
  row_counts,
  COUNT(inventory_id) AS count_foreign_key_values
FROM counts_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_foreign_key_values
1 4
2 1126
3 1151
4 1160
5 1139

inventory distribution analysis on inventory_id foreign key

-- first generate group by counts on the inventory_id column
WITH counts_base AS(
  SELECT 
    inventory_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.inventory
  GROUP BY inventory_id
)
-- summarize the group by counts above by grouping again on the row_counts from counts_base CTE part
SELECT 
  row_counts,
  COUNT(inventory_id) AS count_foreign_key_values
FROM counts_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_foreign_key_values
1 4581

Findings

In the rental table there were multiple row counts for some values of inventory_id which meant there is a 1 to many relationship for the inventory_id in the rental table.

In the inventory table there was only 1 table row record per inventory_id which means there is a 1-to-1 relationship.

3. How many overlapping and missing unique foreign key values are there between the two tables?

Rental Table

By using an anti join, the foreign keys which only exist in the rental table and not in the inventory table can be seen.

-- how many foreign keys only exist in the left table and not in the right?
SELECT
  COUNT(DISTINCT inventory_id)
FROM dvd_rentals.rental
WHERE NOT EXISTS (
  SELECT inventory_id
  FROM dvd_rentals.inventory
  WHERE rental.inventory_id = inventory.inventory_id
);
count
0

Findings

The inventory_id values from the rental table were also in the inventory table as well.

Inventory Table

The unique foreign keys that only exists in the inventory table and not in the rental table were also checked.

SELECT
  COUNT(DISTINCT inventory_id)
FROM dvd_rentals.inventory
WHERE NOT EXISTS (
  SELECT inventory_id
  FROM dvd_rentals.rental
  WHERE inventory.inventory_id = rental.inventory_id
);
count
1

There was one inventory_id value which is not in the rental table.

This one value was checked:

SELECT *
FROM dvd_rentals.inventory
WHERE NOT EXISTS (
  SELECT inventory_id
  FROM dvd_rentals.rental
  WHERE inventory.inventory_id = rental.inventory_id
);
inventory_id film_id store_id last_update
5 1 2 2006-02-15T05:09:17.000Z

The inventory_id was linked to a specific film. It could be that the film might have never been rented by a customer.

I also got the count of foreign key values from the intersection of both tables using a left semi join or a WHERE EXISTS

SELECT 
  COUNT(DISTINCT inventory_id)
FROM dvd_rentals.rental
WHERE EXISTS (
  SELECT inventory_id
  FROM dvd_rentals.inventory
  WHERE rental.inventory_id = inventory.inventory_id 
);
count
4580

Findings

This was the same number of the output from hypothesis 1 which was the distinct count from the rental table.

After this analysis the conclusion was that it does not matter if a left join or a inner join is used to join the rental and inventory tables.

Left Join or Inner Join?

Both left and inner joins were implemented to validate that the raw row counts output would be same for both joins.

DROP TABLE IF EXISTS left_rental_join;
CREATE TEMP TABLE left_rental_join AS
SELECT 
  rental.customer_id,
  rental.inventory_id,
  inventory.film_id
FROM dvd_rentals.rental
LEFT JOIN dvd_rentals.inventory
  ON rental.inventory_id = inventory.inventory_id;
  
DROP TABLE IF EXISTS inner_rental_join;
CREATE TEMP TABLE inner_rental_join AS 
SELECT 
  rental.customer_id,
  rental.inventory_id,
  inventory.film_id
FROM dvd_rentals.rental
INNER JOIN dvd_rentals.inventory
  ON rental.inventory_id = inventory.inventory_id;
  
-- check the counts for each output
(
  SELECT
    'left join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT inventory_id) AS unique_key_values
  FROM left_rental_join
)

UNION 

(
  SELECT
    'inner join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT inventory_id) AS unique_key_values
  FROM inner_rental_join
)
join_type record_count unique_key_values
inner join 16044 4580
left join 16044 4580

As seen in the table above it did not matter which join was used, the output was the same for the inner and left join.

Next Table Journies

All of these questions and analysis were done for every join journey part. To not make the main part too long, the rest of the analysis is in the appendix.

Final Ouput

The table joins to create the base table for the category insights is shown below

DROP TABLE IF EXISTS complete_join_dataset;
CREATE TEMP TABLE complete_join_dataset AS
SELECT 
  rental.customer_id,
  inventory.film_id,
  film.title,
  film_category.category_id,
  category.name AS category_name
FROM dvd_rentals.rental
INNER JOIN dvd_rentals.inventory
  ON rental.inventory_id = inventory.inventory_id
INNER JOIN dvd_rentals.film
  ON inventory.film_id = film.film_id
INNER JOIN dvd_rentals.film_category
  ON film.film_id = film_category.film_id
INNER JOIN dvd_rentals.category
  ON film_category.category_id = category.category_id;
  
SELECT *
FROM complete_join_dataset
LIMIT 5;
customer_id film_id title category_id category_name
130 80 BLANKET BEVERLY 8 Family
459 333 FREAKY POCUS 12 Music
408 373 GRADUATE LORD 3 Children
333 535 LOVE SUICIDES 11 Horror
222 450 IDOLS SNATCHERS 3 Children

Join Implementation Part 2

The base table for the actor insights was generated for the next join implementation part.

Please click on the link below to go to the next part.

forthebadge

Appendix

Join Journey Part 2

1. What is the purpose of joining these two tables?

Match the films on film_id to obtain the title of each film.

a. & b. What contextual hypotheses do we have about the data and how can we validate it?

Hypothesis 1:

There will be multiple records per unique film_id in the inventory table since one specific film might have multiple copies to be rented at the store.

WITH counts_base AS (
  SELECT
    film_id,
    COUNT(film_id) AS row_counts
  FROM
    dvd_rentals.inventory
  GROUP BY film_id
)
SELECT
  row_counts,
  COUNT(DISTINCT film_id) AS unique_film_id_values
FROM counts_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts unique_film_id_values
2 133
3 131
4 183
5 136
6 187
7 116
8 72

Findings

There were multiple records per film_id value in the inventory table.

Hypothesis 2:

In the film table there should be a 1-to-1 relationship for the film_id since it would not make sense to have duplicate film_id values.

WITH counts_base AS (
  SELECT
    film_id,
    COUNT(film_id) AS row_counts
  FROM
    dvd_rentals.film
  GROUP BY film_id
)
SELECT
  row_counts,
  COUNT(DISTINCT film_id) AS unique_film_id_values
FROM counts_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts unique_film_id_values
1 1000

Findings

There is a 1-to-1 relationship in the film table.

2. What is the distribution of foreign keys within each table?

This was done with the validation of the hypotheses above.

3. How many overlapping and missing unique foreign key values are there between the two tables?

By using an anti join, the foreign keys which only exist in the inventory table and not in the film table can be seen.

Inventory Table

SELECT
  COUNT(DISTINCT film_id)
FROM dvd_rentals.inventory
WHERE NOT EXISTS (
  SELECT film_id
  FROM dvd_rentals.film
  WHERE inventory.film_id = film.film_id
);
count
0

Findings

The film_id values from the inventory table exist in the film table.

Film Table

The unique foreign keys that only exists in the film table and not in the inventory table were also checked.

SELECT
  COUNT(DISTINCT film_id)
FROM dvd_rentals.film
WHERE NOT EXISTS (
  SELECT film_id
  FROM dvd_rentals.inventory
  WHERE film.film_id = inventory.film_id 
);
count
42

Findings

There were 42 foreign key values which are not in the inventory table.

The total count of distinct foreign key values was checked using a left semi join between the inventory and the film tables.

SELECT 
  COUNT(DISTINCT film_id)
FROM dvd_rentals.inventory
WHERE EXISTS (
  SELECT film_id
  FROM dvd_rentals.film
  WHERE inventory.film_id = film.film_id
);
count
958

Findings

Once the join was performed 958 distinct film_id values are expected.

Left Join or Inner Join?

The left and inner joins were implemented to validate that the raw counts are the same for both joins

DROP TABLE IF EXISTS left_inventory_join;
CREATE TEMP TABLE left_inventory_join AS 
SELECT 
  inventory.inventory_id,
  inventory.film_id,
  film.title
FROM dvd_rentals.inventory
LEFT JOIN dvd_rentals.film
  ON inventory.film_id = film.film_id;
  
DROP TABLE IF EXISTS inner_inventory_join;
CREATE TEMP TABLE inner_inventory_join AS 
SELECT 
  inventory.inventory_id,
  inventory.film_id,
  film.title
FROM dvd_rentals.inventory
LEFT JOIN dvd_rentals.film
  ON inventory.film_id = film.film_id;
  
(
  SELECT
    'left join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT film_id) AS unique_key_values 
  FROM left_inventory_join
)  

UNION ALL 

(
  SELECT 
    'inner join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT film_id) AS unique_key_values
  FROM inner_inventory_join  
)
join_type record_count unique_key_values
inner join 4581 958
left join 4581 958

As seen in the table above, no matter which join is used the output will be the same.

Join Journey Part 3

1. What is the purpose of joining these two tables?

Match the films on film_id to obtain the category_id of each film.

a. & b. What contextual hypotheses do we have about the data and how can we validate it?

Hypothesis 1:

The number of unique film_id records will be equal in both film and film_category tables

Film Table

SELECT 
  COUNT(DISTINCT film_id)
FROM dvd_rentals.film;
count
1000

Film_category Table

SELECT 
  COUNT(DISTINCT film_id)
FROM dvd_rentals.film_category;
count
1000

Findings

Both of the tables had the same distinct film_id values.

Hypothesis 2:

In the film_category table there should be a 1-to-1 relationship for the film_id.

SELECT 
  film_id,
  COUNT(*) AS value_count
FROM dvd_rentals.film_category
GROUP BY film_id
ORDER BY values_count DESC
LIMIT 5;
film_id value_count
273 1
51 1
951 1
839 1
652 1

Findings

There is a 1-to-1 relationship in the film_category table because the value count of the film_id values appeared only once.

2. What is the distribution of foreign keys within each table?

Film Table

WITH count_base AS (
  SELECT 
    film_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.film
  GROUP BY film_id
)

SELECT 
  row_counts,
  COUNT(DISTINCT film_id) AS count_foreign_key_values
FROM count_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts unique_film_id_values
1 1000

Film_category Table

WITH count_base AS (
  SELECT 
    film_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.film_category
  GROUP BY film_id
)

SELECT 
  row_counts,
  COUNT(DISTINCT film_id) AS count_foreign_key_values
FROM count_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts unique_film_id_values
1 1000

Findings

There is a 1-to-1 relationship for the film_id in both film and film_category tables.

3. How many overlapping and missing unique foreign key values are there between the two tables?

Film Table

By using an anti join, the foreign keys which only exist in the film table and not in the film_category table can be seen.

SELECT 
  COUNT(DISTINCT film_id)
FROM dvd_rentals.film
WHERE NOT EXISTS(
  SELECT 
    film_id
  FROM dvd_rentals.film_category
  WHERE film.film_id = film_category.film_id
);
count
0

Film_category Table

The unique foreign keys that only exists in the film_category table and not in the film table were also checked.

SELECT 
  COUNT(DISTINCT film_id)
FROM dvd_rentals.film_category
WHERE NOT EXISTS(
  SELECT 
    film_id
  FROM dvd_rentals.film
  WHERE film_category.film_id = film.film_id
);
count
0

Findings

The same film_id values that were in the film table were also in the film_category and vice versa.

Left Join or Inner Join?

The left and inner joins were implemented to validate that the raw row counts output was the same.

DROP TABLE IF EXISTS left_film_join;
CREATE TEMP TABLE left_film_join AS 
SELECT 
  film.film_id,
  film.title,
  film_category.category_id
FROM dvd_rentals.film
LEFT JOIN dvd_rentals.film_category
  ON film.film_id = film_category.film_id;
  
DROP TABLE IF EXISTS inner_film_join;
CREATE TEMP TABLE inner_film_join AS 
SELECT 
  film.film_id,
  film.title,
  film_category.category_id
FROM dvd_rentals.film
INNER JOIN dvd_rentals.film_category
  ON film.film_id = film_category.film_id;
  
(
  SELECT
    'left join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT film_id) AS unique_key_values 
  FROM left_film_join
)  

UNION ALL 

(
  SELECT 
    'inner join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT film_id) AS unique_key_values
  FROM inner_film_join  
)
join_type record_count unique_key_values
left join 1000 1000
inner join 1000 1000

As seen in the table above, the output was the same for both joins.

Join Journey Part 4

1. What is the purpose of joining these two tables?

Match the category_id to obtain the name of each category.

a. & b. What contextual hypotheses do we have about the data and how can we validate it?

Hypothesis 1:

There will be a multiple records per unique category_id in the film_category table

WITH count_base AS (
  SELECT 
    category_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.film_category
  GROUP BY category_id
)

SELECT 
  row_counts,
  COUNT(DISTINCT category_id) AS count_foreign_key_values
FROM count_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_foreign_key_values
51 1
56 1
57 2
58 1
60 1
61 2
62 1
63 1
64 1
66 1
68 1
69 1
73 1
74 1

Findings

There were multiple records per unique category_id in the film_category table.

Hypothesis 2:

In the category table there should be a 1-to-1 relationship for the category_id.

SELECT 
  category_id,
  COUNT(*) AS value_count
FROM dvd_rentals.category
GROUP BY category_id
ORDER BY values_count DESC
LIMIT 5;
category_id value_count
10 1
6 1
13 1
2 1
4 1

Findings

There is a 1-to-1 relationship in the category table because the value count of the category_id values appeared only once.

2. What is the distribution of foreign keys within each table?

Film_category Table

WITH count_base AS (
  SELECT 
    category_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.film_category
  GROUP BY category_id
)

SELECT 
  row_counts,
  COUNT(DISTINCT category_id) AS count_foreign_key_values
FROM count_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_foreign_key_values
51 1
56 1
57 2
58 1
60 1
61 2
62 1
63 1
64 1
66 1
68 1
69 1
73 1
74 1

Category Table

WITH count_base AS (
  SELECT 
    category_id,
    COUNT(*) AS row_counts
  FROM dvd_rentals.category
  GROUP BY category_id
)

SELECT 
  row_counts,
  COUNT(DISTINCT category_id) AS count_foreign_key_values
FROM count_base
GROUP BY row_counts
ORDER BY row_counts;
row_counts count_foreign_key_values
1 16

Findings

There is a 1-to-many relationship for the category_id in the film_category table and there is a 1-to-1 relationship for the category_id in the category table.

3. How many overlapping and missing unique foreign key values are there between the two tables?

Film_category Table

By using an anti join, the foreign keys which only exist in the film_category table and not in the category table can be seen.

SELECT 
  COUNT(DISTINCT category_id)
FROM dvd_rentals.film_category
WHERE NOT EXISTS(
  SELECT 
    category_id
  FROM dvd_rentals.category
  WHERE film_category.category_id = category.category_id
);
count
0

Category Table

The unique foreign keys that only exists in the category table and not in the film_category table were also checked.

SELECT 
  COUNT(DISTINCT category_id)
FROM dvd_rentals.category
WHERE NOT EXISTS(
  SELECT 
    category_id
  FROM dvd_rentals.film_category
  WHERE category.category_id = film_category.category_id
);
count
0

Findings

The same category_id values that were in the film_category table were also in the category table and vice versa.

Left Join or Inner Join?

The left and inner joins were implemented to validate that the raw row counts output was the same.

DROP TABLE IF EXISTS left_film_category_join;
CREATE TEMP TABLE left_film_category_join AS 
SELECT 
  film_category.film_id,
  film_category.category_id,
  category.name
FROM dvd_rentals.film_category
LEFT JOIN dvd_rentals.category
  ON film_category.category_id = category.category_id;
  
DROP TABLE IF EXISTS inner_film_category_join;
CREATE TEMP TABLE inner_film_category_join AS 
SELECT 
  film_category.film_id,
  film_category.category_id,
  category.name
FROM dvd_rentals.film_category
INNER JOIN dvd_rentals.category
  ON film_category.category_id = category.category_id;
  
(
  SELECT
    'left join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT category_id) AS unique_key_values 
  FROM left_film_category_join
)  

UNION ALL 

(
  SELECT 
    'inner join' AS join_type,
    COUNT(*) AS record_count,
    COUNT(DISTINCT category_id) AS unique_key_values
  FROM inner_film_category_join  
)
join_type record_count unique_key_values
left join 1000 16
inner join 1000 16

As seen in the table above, the output was the same for both joins.