RSVP SQL final.sql

USE imdb;

/* Now that you have imported the data sets, let’s explore some of the tables. 
 To begin with, it is beneficial to know the shape of the tables and whether any column has null values.
 Further in this segment, you will take a look at 'movies' and 'genre' tables.*/



-- Segment 1:


-- Q1. Find the total number of rows in each table of the schema?
-- Type your code below:

SELECT 
    COUNT(*) AS movie_row_count
FROM
    movie;
-- Total number of rows in movie table : 7997
SELECT 
    COUNT(*) AS genre_row_count
FROM
    genre;
-- Total number of rows in genre table : 14662
SELECT 
    COUNT(*) AS director_mapping_row_count
FROM
    director_mapping;
-- Total number of rows in director_mapping table : 3867
SELECT 
    COUNT(*) AS names_row_count
FROM
    names;
-- Total number of rows in names table : 25735
SELECT 
    COUNT(*) AS ratings_row_count
FROM
    ratings;
-- Total number of rows in ratings table : 7997
SELECT 
    COUNT(*) AS role_mapping_row_count
FROM
    role_mapping;
-- Total number of rows in role_mapping table : 15615


-- Q2. Which columns in the movie table have null values?
-- Type your code below:

SELECT 
    (COUNT(*) - COUNT(id)) AS id_null_count,
    (COUNT(*) - COUNT(title)) AS title_null_count,
    (COUNT(*) - COUNT(year)) AS year_null_count,
    (COUNT(*) - COUNT(date_published)) AS date_published_null_count,
    (COUNT(*) - COUNT(duration)) AS duration_null_count,
    (COUNT(*) - COUNT(country)) AS country_null_count,
    (COUNT(*) - COUNT(worlwide_gross_income)) AS worlwide_gross_income_null_count,
    (COUNT(*) - COUNT(languages)) AS languages_null_count,
    (COUNT(*) - COUNT(production_company)) AS production_company_null_count
FROM
    movie;
		


-- Now as you can see four columns of the movie table has null values. Let's look at the at the movies released each year. 
-- Q3. Find the total number of movies released each year? How does the trend look month wise? (Output expected)

/* Output format for the first part:

+---------------+-------------------+
| Year			|	number_of_movies|
+-------------------+----------------
|	2017		|	2134			|
|	2018		|		.			|
|	2019		|		.			|
+---------------+-------------------+


Output format for the second part of the question:
+---------------+-------------------+
|	month_num	|	number_of_movies|
+---------------+----------------
|	1			|	 134			|
|	2			|	 231			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

-- First part solution
SELECT 
    year, COUNT(id) AS number_of_movies
FROM
    movie
GROUP BY year
ORDER BY year;
/* The number of movies has decreased (3052 to 2001) over years from 2017 to 2019 */

-- Second part solution
SELECT
    MONTH(date_published) AS month_num,
    COUNT(id) AS number_of_movies
FROM
    movie
GROUP BY MONTH(date_published)
ORDER BY MONTH(date_published);

-- The lowest number of movies are published in month of December 


/*The highest number of movies is produced in the month of March.
So, now that you have understood the month-wise trend of movies, let’s take a look at the other details in the movies table. 
We know USA and India produces huge number of movies each year. Lets find the number of movies produced by USA or India for the last year.*/
  
-- Q4. How many movies were produced in the USA or India in the year 2019??
-- Type your code below:

SELECT 
	COUNT(DISTINCT id) AS total_movies_produced
FROM
    movie
WHERE
    year = 2019
        AND (country REGEXP 'India' OR  country REGEXP 'USA');

-- Total 1059 Movies were produced by India Or USA in year 2019.

/* USA and India produced more than a thousand movies(you know the exact number!) in the year 2019.
Exploring table Genre would be fun!! 
Let’s find out the different genres in the dataset.*/

-- Q5. Find the unique list of the genres present in the data set?
-- Type your code below:

SELECT genre
FROM genre
GROUP BY genre;



/* So, RSVP Movies plans to make a movie of one of these genres.
Now, wouldn’t you want to know which genre had the highest number of movies produced in the last year?
Combining both the movie and genres table can give more interesting insights. */

-- Q6.Which genre had the highest number of movies produced overall?
-- Type your code below:

SELECT 
    g.genre, COUNT(m.id) AS movie_count
FROM genre AS g
	INNER JOIN movie AS m 
		ON g.movie_id = m.id
GROUP BY g.genre
ORDER BY COUNT(m.id) DESC
LIMIT 1;
-- Drama had the highest number of movies produced overall



/* So, based on the insight that you just drew, RSVP Movies should focus on the ‘Drama’ genre. 
But wait, it is too early to decide. A movie can belong to two or more genres. 
So, let’s find out the count of movies that belong to only one genre.*/

-- Q7. How many movies belong to only one genre?
-- Type your code below:

WITH one_genre AS
(
SELECT 
    m.id,COUNT(g.genre) AS genre_count
FROM genre AS g
	INNER JOIN movie AS m 
		ON g.movie_id = m.id
GROUP BY m.id
HAVING COUNT(g.genre) = 1
)
SELECT 
    COUNT(id) AS movie_count
FROM
    one_genre;

-- Number of movies belong to only one genre : 3289


/* There are more than three thousand movies which has only one genre associated with them.
So, this figure appears significant. 
Now, let's find out the possible duration of RSVP Movies’ next project.*/

-- Q8.What is the average duration of movies in each genre? 
-- (Note: The same movie can belong to multiple genres.)


/* Output format:

+---------------+-------------------+
| genre			|	avg_duration	|
+-------------------+----------------
|	thriller	|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

SELECT 
    g.genre, 
    ROUND(AVG(duration),2) AS avg_duration
FROM movie AS m
	INNER JOIN genre AS g 
		ON m.id = g.movie_id
GROUP BY g.genre;


/* Now you know, movies of genre 'Drama' (produced highest in number in 2019) has the average duration of 106.77 mins.
Lets find where the movies of genre 'thriller' on the basis of number of movies.*/

-- Q9.What is the rank of the ‘thriller’ genre of movies among all the genres in terms of number of movies produced? 
-- (Hint: Use the Rank function)


/* Output format:
+---------------+-------------------+---------------------+
| genre			|		movie_count	|		genre_rank    |	
+---------------+-------------------+---------------------+
|drama			|	2312			|			2		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:

WITH genre_rank_summary AS 
(
	SELECT 
		genre,
        COUNT(movie_id) AS movie_count,
		DENSE_RANK() 
				OVER(ORDER BY COUNT(movie_id) DESC) AS genre_rank
	FROM genre
	GROUP BY genre
)
SELECT * FROM genre_rank_summary
WHERE genre = "Thriller";


/*Thriller movies is in top 3 among all genres in terms of number of movies
 In the previous segment, you analysed the movies and genres tables. 
 In this segment, you will analyse the ratings table as well.
To start with lets get the min and max values of different columns in the table*/




-- Segment 2:




-- Q10.  Find the minimum and maximum values in  each column of the ratings table except the movie_id column?
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+
| min_avg_rating|	max_avg_rating	|	min_total_votes   |	max_total_votes 	 |min_median_rating|min_median_rating|
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+
|		0		|			5		|	       177		  |	   2000	    		 |		0	       |	8			 |
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+*/
-- Type your code below:

SELECT 
    MIN(avg_rating) 	AS min_avg_rating,
    MAX(avg_rating) 	AS max_avg_rating,
    MIN(total_votes) 	AS min_total_votes,
    MAX(total_votes) 	AS max_total_votes,
    MIN(median_rating)  AS min_median_rating,
    MAX(median_rating)  AS max_median_rating
FROM
    ratings;
    

    

/* So, the minimum and maximum values in each column of the ratings table are in the expected range. 
This implies there are no outliers in the table. 
Now, let’s find out the top 10 movies based on average rating.*/

-- Q11. Which are the top 10 movies based on average rating?
/* Output format:
+---------------+-------------------+---------------------+
| title			|		avg_rating	|		movie_rank    |
+---------------+-------------------+---------------------+
| Fan			|		9.6			|			5	  	  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:
-- It's ok if RANK() or DENSE_RANK() is used too


WITH rank_on_rating AS 
(
SELECT 
	title,
	avg_rating,
	DENSE_RANK() OVER(ORDER BY avg_rating DESC) AS movie_rank
FROM movie AS m
	INNER JOIN ratings r
		ON m.id = r.movie_id
)
SELECT *
FROM rank_on_rating
LIMIT 10; 



/* Do you find you favourite movie FAN in the top 10 movies with an average rating of 9.6? If not, please check your code again!!
So, now that you know the top 10 movies, do you think character actors and filler actors can be from these movies?
Summarising the ratings table based on the movie counts by median rating can give an excellent insight.*/

-- Q12. Summarise the ratings table based on the movie counts by median ratings.
/* Output format:

+---------------+-------------------+
| median_rating	|	movie_count		|
+-------------------+----------------
|	1			|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:
-- Order by is good to have

SELECT 
    median_rating, 
    COUNT(movie_id) AS movie_count
FROM
    ratings
GROUP BY median_rating
ORDER BY median_rating;



/* Movies with a median rating of 7 is highest in number. 
Now, let's find out the production house with which RSVP Movies can partner for its next project.*/

-- Q13. Which production house has produced the most number of hit movies (average rating > 8)??
/* Output format:
+------------------+-------------------+---------------------+
|production_company|movie_count	       |	prod_company_rank|
+------------------+-------------------+---------------------+
| The Archers	   |		1		   |			1	  	 |
+------------------+-------------------+---------------------+*/
-- Type your code below:
WITH production_company_summary 
AS
(
SELECT 
	production_company,
	COUNT(r.movie_id) AS movie_count,
    DENSE_RANK() 
		OVER(ORDER BY COUNT(r.movie_id) DESC) AS prod_company_rank
FROM movie AS m
	INNER JOIN ratings AS r
		ON m.id = r.movie_id
WHERE avg_rating >8 AND production_company IS NOT NULL
GROUP BY production_company
)
SELECT * 
FROM production_company_summary
WHERE prod_company_rank = 1;


-- It's ok if RANK() or DENSE_RANK() is used too
-- Answer can be Dream Warrior Pictures or National Theatre Live or both

-- Q14. How many movies released in each genre during March 2017 in the USA had more than 1,000 votes?
/* Output format:

+---------------+-------------------+
| genre			|	movie_count		|
+-------------------+----------------
|	thriller	|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

SELECT 
    g.genre, 
    COUNT(g.movie_id) AS movie_count
FROM genre AS g
	INNER JOIN movie AS m 
		ON g.movie_id = m.id
	INNER JOIN ratings AS r 
		ON g.movie_id = r.movie_id
WHERE MONTH(m.date_published) = 3
        AND m.year = 2017
        AND m.country = 'USA'
        AND r.total_votes > 1000
GROUP BY g.genre
ORDER BY COUNT(g.movie_id) DESC;

-- Drama genre has highest number of movies released during March 2017 in the USA had more than 1,000 votes

-- Lets try to analyse with a unique problem statement.
-- Q15. Find movies of each genre that start with the word ‘The’ and which have an average rating > 8?
/* Output format:
+---------------+-------------------+---------------------+
| title			|		avg_rating	|		genre	      |
+---------------+-------------------+---------------------+
| Theeran		|		8.3			|		Thriller	  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:

SELECT 
    m.title, 
    r.avg_rating, 
    g.genre
FROM movie AS m
	INNER JOIN ratings AS r 
		ON m.id = r.movie_id
	INNER JOIN genre AS g 
		ON m.id = g.movie_id
WHERE title REGEXP '^The' 
		AND avg_rating > 8
ORDER BY avg_rating DESC;



-- You should also try your hand at median rating and check whether the ‘median rating’ column gives any significant insights.
-- Q16. Of the movies released between 1 April 2018 and 1 April 2019, how many were given a median rating of 8?
-- Type your code below:

SELECT 
	median_rating,
    COUNT(id) AS movie_count
FROM movie AS m
	INNER JOIN ratings AS r 
		ON m.id = r.movie_id
WHERE date_published BETWEEN '2018-04-01' AND '2019-04-01'
		AND median_rating = 8 
GROUP BY median_rating;


-- Once again, try to solve the problem given below.
-- Q17. Do German movies get more votes than Italian movies? 
-- Hint: Here you have to find the total number of votes for both German and Italian movies.
-- Type your code below:

SELECT 
	country,
    SUM(total_votes) AS sum_total_votes
FROM movie AS m
	INNER JOIN ratings AS r
		ON m.id = r.movie_id
WHERE country IN ("Germany","Italy")
GROUP BY country;


-- Answer is Yes

/* Now that you have analysed the movies, genres and ratings tables, let us now analyse another table, the names table. 
Let’s begin by searching for null values in the tables.*/




-- Segment 3:



-- Q18. Which columns in the names table have null values??
/*Hint: You can find null values for individual columns or follow below output format
+---------------+-------------------+---------------------+----------------------+
| name_nulls	|	height_nulls	|date_of_birth_nulls  |known_for_movies_nulls|
+---------------+-------------------+---------------------+----------------------+
|		0		|			123		|	       1234		  |	   12345	    	 |
+---------------+-------------------+---------------------+----------------------+*/
-- Type your code below:

SELECT 
    SUM(CASE WHEN name IS NULL THEN 1 ELSE 0
		END) AS name_nulls,
    SUM(CASE WHEN height IS NULL THEN 1 ELSE 0
		END) AS height_nulls,
    SUM(CASE WHEN date_of_birth IS NULL THEN 1 ELSE 0
		END) AS date_of_birth_nulls,
    SUM(CASE WHEN known_for_movies IS NULL THEN 1 ELSE 0
		END) AS known_for_movies_nulls
FROM
    names;



/* There are no Null value in the column 'name'.
The director is the most important person in a movie crew. 
Let’s find out the top three directors in the top three genres who can be hired by RSVP Movies.*/

-- Q19. Who are the top three directors in the top three genres whose movies have an average rating > 8?
-- (Hint: The top three genres would have the most number of movies with an average rating > 8.)
/* Output format:

+---------------+-------------------+
| director_name	|	movie_count		|
+---------------+-------------------|
|James Mangold	|		4			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

WITH genre_rank_summary AS 
(
	SELECT 
		g.genre,
		r.avg_rating,
		COUNT(g.movie_id) AS movie_count,
		DENSE_RANK() 
			OVER(ORDER BY COUNT(g.movie_id) DESC) AS genre_rank
	FROM genre AS g
		INNER JOIN ratings AS r
			ON r.movie_id = g.movie_id
	WHERE r.avg_rating > 8
	GROUP BY g.genre 
),
-- top 3 genre from all genre ranks
top_3_genre 
AS 
(
	SELECT genre
    FROM genre_rank_summary
    WHERE genre_rank <=3
),
-- top directors names from top 3 genre
top_directors 
AS 
(
	SELECT 
		n.name AS director_name,
		COUNT(d.movie_id) AS movie_count,
        DENSE_RANK() 
				OVER(ORDER BY COUNT(d.movie_id) DESC) AS director_rank
        
	FROM names AS n
		INNER JOIN director_mapping AS d
			ON d.name_id = n.id 
		INNER JOIN genre AS g
			ON g.movie_id = d.movie_id
		INNER JOIN ratings AS r
			ON r.movie_id = g.movie_id, 
		top_3_genre AS t 
	WHERE g.genre IN (t.genre) AND r.avg_rating > 8
    GROUP BY n.name
)
-- top 3 directors in the top 3 genres based on their movie counts with an average rating > 8
SELECT 
	director_name, 
	movie_count 
FROM top_directors
WHERE director_rank <=3
LIMIT 3;



/* James Mangold can be hired as the director for RSVP's next project. Do you remeber his movies, 'Logan' and 'The Wolverine'. 
Now, let’s find out the top two actors.*/

-- Q20. Who are the top two actors whose movies have a median rating >= 8?
/* Output format:

+---------------+-------------------+
| actor_name	|	movie_count		|
+-------------------+----------------
|Christain Bale	|		10			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

WITH actor_ranking_summary AS
(
	SELECT 
		n.name AS actor_name,
		COUNT(rm.movie_id) AS movie_count,
        DENSE_RANK() OVER(ORDER BY COUNT(rm.movie_id) DESC) AS actor_rankings
	FROM names AS n
		INNER JOIN role_mapping AS rm
			ON n.id = rm.name_id
		INNER JOIN ratings AS r
			ON r.movie_id = rm.movie_id
	WHERE median_rating >=8
	GROUP BY n.name
)
SELECT 
	actor_name,
    movie_count
FROM actor_ranking_summary
WHERE actor_rankings <=2;

-- Top 2 actors are 'Mammootty' and 'Mohanlal'.

/* Have you find your favourite actor 'Mohanlal' in the list. If no, please check your code again. 
RSVP Movies plans to partner with other global production houses. 
Let’s find out the top three production houses in the world.*/

-- Q21. Which are the top three production houses based on the number of votes received by their movies?
/* Output format:
+------------------+--------------------+---------------------+
|production_company|vote_count			|		prod_comp_rank|
+------------------+--------------------+---------------------+
| The Archers		|		830			|		1	  		  |
|	.				|		.			|			.		  |
|	.				|		.			|			.		  |
+-------------------+-------------------+---------------------+*/
-- Type your code below:

SELECT 
	m.production_company, 
	SUM(r.total_votes) AS vote_count,
    DENSE_RANK() 
		OVER(ORDER BY SUM(r.total_votes) DESC) AS prod_comp_rank
FROM movie AS m
	INNER JOIN ratings AS r
		ON m.id = r.movie_id
GROUP BY m.production_company
LIMIT 3;

/*Yes Marvel Studios rules the movie world.
So, these are the top three production houses based on the number of votes received by the movies they have produced.

Since RSVP Movies is based out of Mumbai, India also wants to woo its local audience. 
RSVP Movies also wants to hire a few Indian actors for its upcoming project to give a regional feel. 
Let’s find who these actors could be.*/

-- Q22. Rank actors with movies released in India based on their average ratings. Which actor is at the top of the list?
-- Note: The actor should have acted in at least five Indian movies. 
-- (Hint: You should use the weighted average based on votes. If the ratings clash, then the total number of votes should act as the tie breaker.)

/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actor_name	|	total_votes		|	movie_count		  |	actor_avg_rating 	 |actor_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Yogi Babu	|			3455	|	       11		  |	   8.42	    		 |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

SELECT 
	n.name AS actor_name,
    SUM(r.total_votes) AS total_votes,
    COUNT(r.movie_id) AS movie_count,
    ROUND(SUM(r.avg_rating*r.total_votes)/SUM(total_votes),2) AS actor_avg_rating,
    DENSE_RANK() 
		OVER(ORDER BY ROUND(SUM(r.avg_rating*r.total_votes)/SUM(total_votes),2) DESC,r.total_votes DESC) 
			AS actor_rank
FROM names AS n
	INNER JOIN role_mapping AS rm
		ON n.id = rm.name_id
	INNER JOIN ratings AS r
		ON r.movie_id = rm.movie_id
	INNER JOIN movie AS m
		ON m.id = r.movie_id
WHERE country ="India" AND category ="actor"
GROUP BY n.name
HAVING COUNT(r.movie_id) >=5;


-- Top actor is Vijay Sethupathi

-- Q23.Find out the top five actresses in Hindi movies released in India based on their average ratings? 
-- Note: The actresses should have acted in at least three Indian movies. 
-- (Hint: You should use the weighted average based on votes. If the ratings clash, then the total number of votes should act as the tie breaker.)
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actress_name	|	total_votes		|	movie_count		  |	actress_avg_rating 	 |actress_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Tabu		|			3455	|	       11		  |	   8.42	    		 |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

WITH actress_rank_summary AS
(
SELECT 
	n.name AS actress_name,
    SUM(r.total_votes) AS total_votes,
    COUNT(r.movie_id) AS movie_count,
    ROUND(SUM(r.avg_rating*r.total_votes) / SUM(total_votes),2) AS actress_avg_rating,
    DENSE_RANK() 
		OVER(ORDER BY ROUND(SUM(r.avg_rating*r.total_votes)/SUM(total_votes),2) DESC,r.total_votes DESC) 
			AS actress_rank
FROM names AS n
	INNER JOIN role_mapping AS rm
		ON n.id = rm.name_id
	INNER JOIN ratings AS r
		ON r.movie_id = rm.movie_id
	INNER JOIN movie AS m
		ON m.id = r.movie_id
WHERE 
	country ="India" 
    AND category ="actress" 
    AND languages= 'hindi'
GROUP BY n.name
HAVING COUNT(r.movie_id) >=3
)
SELECT * FROM actress_rank_summary
WHERE actress_rank <=5 ;



/* Taapsee Pannu tops with average rating 7.74. 
Now let us divide all the thriller movies in the following categories and find out their numbers.*/


/* Q24. Select thriller movies as per avg rating and classify them in the following category: 

			Rating > 8: Superhit movies
			Rating between 7 and 8: Hit movies
			Rating between 5 and 7: One-time-watch movies
			Rating < 5: Flop movies
--------------------------------------------------------------------------------------------*/
-- Type your code below:

SELECT
	title AS movie_name_thriller,
    avg_rating,
(
	CASE
		 WHEN avg_rating > 8 THEN "Superhit movies"
		 WHEN avg_rating BETWEEN 7 AND 8 THEN "Hit movies"
		 WHEN avg_rating BETWEEN 5 AND 7 THEN "One-time-watch movies"
		 ELSE "Flop movies"
	END
) 		 AS avg_rating_category
			
FROM movie AS m
	INNER JOIN ratings AS r
		ON m.id = r.movie_id
	INNER JOIN genre AS g
		ON g.movie_id = r.movie_id
WHERE g.genre = "Thriller";



/* Until now, you have analysed various tables of the data set. 
Now, you will perform some tasks that will give you a broader understanding of the data in this segment.*/


-- Segment 4:


-- Q25. What is the genre-wise running total and moving average of the average movie duration? 
-- (Note: You need to show the output table in the question.) 
/* Output format:
+---------------+-------------------+---------------------+----------------------+
| genre			|	avg_duration	|running_total_duration|moving_avg_duration  |
+---------------+-------------------+---------------------+----------------------+
|	comdy		|			145		|	       106.2	  |	   128.42	    	 |
|		.		|			.		|	       .		  |	   .	    		 |
|		.		|			.		|	       .		  |	   .	    		 |
|		.		|			.		|	       .		  |	   .	    		 |
+---------------+-------------------+---------------------+----------------------+*/
-- Type your code below:
WITH movies_avg_duration
AS
(
	 SELECT     
		genre,
		ROUND(AVG(duration),2) AS avg_duration
	 FROM	genre g
			INNER JOIN movie m
					ON	g.movie_id = m.id
	 GROUP BY genre
)
SELECT *,
	SUM(avg_duration) 		OVER w    	AS running_total_duration,
	ROUND(AVG(avg_duration) OVER w,2)  	AS moving_avg_duration
FROM
	movies_avg_duration 
WINDOW w 	AS (ORDER BY avg_duration ROWS UNBOUNDED PRECEDING) ;


-- Round is good to have and not a must have; Same thing applies to sorting


-- Let us find top 5 movies of each year with top 3 genres.

-- Q26. Which are the five highest-grossing movies of each year that belong to the top three genres? 
-- (Note: The top 3 genres would have the most number of movies.)

/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| genre			|	year			|	movie_name		  |worldwide_gross_income|movie_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	comedy		|			2017	|	       indian	  |	   $103244842	     |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:


-- Top 3 Genres based on most number of movies
WITH top_3_genres
AS
(
	SELECT   
		genre,
		COUNT(movie_id) AS movie_count
	FROM     
		genre
	GROUP BY genre
	ORDER BY movie_count DESC
	LIMIT 3
),
-- Converting worldwide_gross_income datatype from 'varchar' to 'decimal'
-- Converting values in INR to dollars , Taking 1 USD = 80 INR
movie_gross_income
AS
(	
	SELECT 
		g.genre, 
        year, 
        title AS movie_name,
		CASE
			WHEN worlwide_gross_income LIKE 'INR%' 
				THEN Cast(Replace(worlwide_gross_income, 'INR', '') AS DECIMAL(12)) / 80
			WHEN worlwide_gross_income LIKE '$%' 
				THEN Cast(Replace(worlwide_gross_income, '$', '') AS DECIMAL(12))
			ELSE Cast(worlwide_gross_income AS DECIMAL(12))
		END AS worldwide_gross_income
	FROM   genre g
		   INNER JOIN movie m
				ON g.movie_id = m.id
		   
	WHERE  g.genre IN ( SELECT genre FROM top_3_genres )
	-- To avoid repetitions,group by for distinct movie titles.
	GROUP  BY movie_name
	ORDER  BY year
),
top_movies AS
(	
	SELECT *,
		   Dense_rank() 
				OVER(PARTITION BY year ORDER BY worldwide_gross_income DESC) AS movie_rank
	FROM   movie_gross_income
)
-- top 5 highest-grossing movies of each year from top 3 genres
SELECT 
		genre,
		year,
		movie_name,
		ROUND(worldwide_gross_income,0) AS worldwide_gross_income,
		movie_rank
FROM   	top_movies
WHERE  	movie_rank <= 5;


-- All values of worldwide_gross_income are in dollars
-- We converted values from INR into dollars by equation 1$ = 80 INR
-- world wide gross income data is changed to decimal from varchar for sorting / ranking purposes

/*
'The Fate of the Furious' is on top for year 2017,
'Bohemian Rhapsody' is ranked 1st in year 2018,
'Avengers: Endgame' is ranked 1st in year 2019
*/


-- Finally, let’s find out the names of the top two production houses that have produced the highest number of hits among multilingual movies.
-- Q27.  Which are the top two production houses that have produced the highest number of hits (median rating >= 8) among multilingual movies?
/* Output format:
+-------------------+-------------------+---------------------+
|production_company |movie_count		|		prod_comp_rank|
+-------------------+-------------------+---------------------+
| The Archers		|		830			|		1	  		  |
|	.				|		.			|			.		  |
|	.				|		.			|			.		  |
+-------------------+-------------------+---------------------+*/
-- Type your code below:
WITH prod_company_rank
AS 
(
	SELECT 
		production_company,
		COUNT(id) AS movie_count,
		Rank()
			OVER(ORDER BY COUNT(id) DESC) AS prod_comp_rank
			 FROM   movie AS m
					INNER JOIN ratings AS r
							ON m.id = r.movie_id
			 WHERE  median_rating >= 8
					AND production_company IS NOT NULL
					AND Position("," IN languages) > 0
			 GROUP  BY production_company
)
SELECT *
FROM   prod_company_rank
WHERE  prod_comp_rank BETWEEN 1 AND 2; 

-- Star Cinema and Twentieth Century Fox are top two production houses which produced the highest number of hits among multilingual movies

-- Multilingual is the important piece in the above question. It was created using POSITION(',' IN languages)>0 logic
-- If there is a comma, that means the movie is of more than one language


-- Q28. Who are the top 3 actresses based on number of Super Hit movies (average rating >8) in drama genre?
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actress_name	|	total_votes		|	movie_count		  |actress_avg_rating	 |actress_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Laura Dern	|			1016	|	       1		  |	   9.60			     |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

-- top 3 actresses based on movie count (avg_rating >8) in drama genre
WITH rank_actress
AS 
(
	SELECT 
		n.name					AS actress_name,
		SUM(total_votes)		AS total_votes,
		COUNT(rm.movie_id)		AS movie_count,
		avg_rating	AS actress_avg_rating,
		ROW_NUMBER()
				OVER(ORDER BY COUNT(r.movie_id) DESC) AS actress_rank
	FROM   names AS n
		INNER JOIN role_mapping AS rm
				ON n.id = rm.name_id
		INNER JOIN ratings AS r
				ON rm.movie_id = r.movie_id
		INNER JOIN genre AS g
				ON r.movie_id = g.movie_id
	 WHERE  category = "actress"
			AND genre = "drama"
            AND avg_rating > 8
	 GROUP  BY n.name
)
SELECT *
FROM   rank_actress
WHERE actress_rank <=3; 

-- Top 3 actresses are 'Parvathy Thiruvothu','Susan Brown','Amanda Lawrence'


/* Q29. Get the following details for top 9 directors (based on number of movies)
Director id
Name
Number of movies
Average inter movie duration in days
Average movie ratings
Total votes
Min rating
Max rating
total movie durations

Format:
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+
| director_id	|	director_name	|	number_of_movies  |	avg_inter_movie_days |	avg_rating	| total_votes  | min_rating	| max_rating | total_duration |
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+
|nm1777967		|	A.L. Vijay		|			5		  |	       177			 |	   5.65	    |	1754	   |	3.7		|	6.9		 |		613		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+

--------------------------------------------------------------------------------------------*/
-- Type you code below:
WITH director_details
AS
(
	 SELECT     
		dm.name_id,
		n.name,
		dm.movie_id,
		m.date_published,
		LEAD(date_published,1) 
				OVER(PARTITION BY name ORDER BY m.date_published,dm.movie_id) AS date_published_next,
		r.avg_rating,
		r.total_votes,
		m.duration
	 FROM	director_mapping dm
			INNER JOIN names n
				ON	dm.name_id = n.id
			INNER JOIN movie m
				ON	dm.movie_id = m.id
			INNER JOIN ratings r
				ON	m.id = r.movie_id
),
director_details_new
AS
(
	 SELECT *,
		DATEDIFF(date_published_next,date_published) AS inter_movie_days
	 FROM
		director_details
)
SELECT 
	name_id                        AS director_id,
	name                           AS director_name,
	COUNT(movie_id)                AS no_of_movies,
	ROUND(avg(inter_movie_days),2) AS avg_inter_movie_days,
	ROUND(avg(avg_rating),2)       AS avg_rating,
	SUM(total_votes)               AS total_votes,
	MIN(avg_rating)                AS min_rating,
	MAX(avg_rating)                AS max_rating,
	SUM(duration)                  AS total_duration
FROM     director_details_new
GROUP BY director_id
ORDER BY count(movie_id) DESC
LIMIT    9;