RSVP SQL CASE STUDY
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 TABLE_NAME, SUM(TABLE_ROWS) AS 'Total No. Of Rows'
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'imdb'
GROUP BY TABLE_NAME;
/* OR */
SELECT COUNT(*) AS No_of_row_director_mapping FROM director_mapping;
/*No_of_row_director_mapping 3867*/
SELECT COUNT(*) AS No_of_row_genre FROM genre;
/*No_of_row_genre 14662*/
SELECT COUNT(*) AS No_of_row_movie FROM movie;
/*No_of_row_movie 7997*/
SELECT COUNT(*) AS No_of_row_ratings FROM ratings;
/*No_of_row_ratings7997*/
SELECT COUNT(*) AS No_of_row_role_mapping FROM role_mapping;
/*No_of_row_role_mapping 15615*/
-- Q2. Which columns in the movie table have null values?
-- Type your code below:
SELECT column_name
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = 'imdb'
AND TABLE_NAME = 'movie';
SELECT
SUM(IF(id IS NULL, 1, 0 )) AS id_null_count,
SUM(IF(title IS NULL, 1, 0 )) AS title_null_count,
SUM(IF(year IS NULL, 1, 0 )) AS year_null_count,
SUM(IF(date_published IS NULL, 1, 0 )) AS date_published_null_count,
SUM(IF(duration IS NULL, 1, 0 )) AS duration_null_count,
SUM(IF(country IS NULL, 1, 0 )) AS country_null_count,
SUM(IF(worlwide_gross_income IS NULL, 1, 0 )) AS worlwide_gross_income_null_count,
SUM(IF(languages IS NULL, 1, 0 )) AS languages_null_count,
SUM(IF(production_company IS NULL, 1, 0 )) 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:
SELECT Year, COUNT(id) AS number_of_movies
FROM movie
GROUP BY Year
ORDER BY Year;
SELECT month(date_published) as month_num ,COUNT(id) AS number_of_movies
FROM movie
GROUP BY month_num
ORDER BY month_num ;
/*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(id) as Total_number_of_movies
FROM movie
WHERE (country LIKE '%USA%' OR country LIKE '%INDIA%' ) AND 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 DISTINCT(genre) AS Unique_genres_name FROM 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 genre, COUNT(movie_id)
AS Number_of_movies
FROM genre g
INNER JOIN movie m
ON g.movie_id = m.id
GROUP BY genre
ORDER BY Number_of_movies
DESC LIMIT 1;
/* 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 movies_with_one_genre
AS
(
SELECT movie_id
FROM genre
GROUP BY movie_id
HAVING Count(DISTINCT genre) = 1
)
SELECT Count(*) AS movies_with_one_genre FROM movies_with_one_genre;
/* 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(m.duration),2)
AS avg_duration
FROM genre AS g
INNER JOIN movie AS m
ON g.movie_id = m.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 thriller_genre_rank
AS
(
SELECT genre, COUNT(movie_id) as count_of_movie_id, DENSE_RANK()
OVER(ORDER BY COUNT(movie_id) DESC) genre_rank
FROM genre
GROUP BY genre
)
SELECT * FROM thriller_genre_rank 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_avg_rating,
MIN(median_rating) AS min_median_rating,
MAX(median_rating) AS max_avg_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:
-- Keep in mind that multiple movies can be at the same rank. You only have to find out the top 10 movies (if there are more than one movies at the 10th place, consider them all.)
SELECT title, avg_rating, DENSE_RANK()
OVER(ORDER BY avg_rating DESC) AS movie_rank
FROM movie AS m
INNER JOIN ratings AS r ON r.movie_id = m.id 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 movie_count DESC;
/* 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:
SELECT production_company, COUNT(id) AS movie_count,
DENSE_RANK() OVER(ORDER BY COUNT(id) DESC) prod_company_rank
FROM movie m INNER JOIN ratings r ON m.id = r.movie_id
WHERE r.avg_rating > 8 AND m.production_company
IS NOT NULL GROUP BY m.production_company;
-- 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(m.id) AS movie_count
FROM movie AS m
INNER JOIN genre AS g
ON m.id = g.movie_id
INNER JOIN ratings AS r
ON m.id = r.movie_id
WHERE m.year = 2017
AND MONTH(m.date_published) = 3
AND m.country
LIKE'USA'
AND r.total_votes>1000
GROUP BY g.genre
ORDER BY COUNT(m.id) DESC;
-- 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 m
INNER JOIN ratings r
ON m.id = r.movie_id
INNER JOIN genre g
ON m.id = g.movie_id
WHERE m.title
LIKE 'The%'
AND r.avg_rating > 8
ORDER BY r.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 COUNT(m.id)
AS no_of_movie_released, r.median_rating
FROM movie AS m
INNER JOIN ratings AS r
ON m.id = r.movie_id
WHERE m.date_published
BETWEEN '2018-04-01'AND '2019-04-01'
AND r.median_rating = 8
GROUP BY r.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 total_votes
FROM movie m
INNER JOIN ratings r
ON m.id = r.movie_id
WHERE country IN ('Germany' ,'Italy')
GROUP BY country
ORDER BY total_votes DESC;
-- 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 top_3_genres_movie_director
AS
(
SELECT genre, Count(m.id) AS movie_count ,
DENSE_Rank() OVER(ORDER BY Count(m.id) DESC) AS genre_rank
FROM movie m
INNER JOIN genre g ON g.movie_id = m.id
INNER JOIN ratings AS r ON r.movie_id = m.id
WHERE avg_rating > 8
GROUP BY genre limit 3
)
SELECT n.NAME AS director_name ,
Count(d.movie_id) AS movie_count
FROM director_mapping AS d
INNER JOIN genre g USING (movie_id)
INNER JOIN names AS n ON n.id = d.name_id
INNER JOIN top_3_genres_movie_director
USING (genre) INNER JOIN ratings
USING (movie_id) WHERE avg_rating > 8
GROUP BY NAME
ORDER BY movie_count DESC 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:
SELECT n.name AS actor_name,
COUNT(rm.movie_id) AS movie_count
FROM role_mapping rm
INNER JOIN names n
ON n.id = rm.name_id
INNER JOIN ratings r
ON r.movie_id = rm.movie_id
WHERE category="actor"
AND r.median_rating >= 8
GROUP BY n.name
ORDER BY movie_count DESC LIMIT 2;
/* 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(total_votes) AS vote_count,
DENSE_RANK() OVER(ORDER BY SUM(total_votes) DESC) AS prod_comp_rank
FROM movie m
INNER JOIN ratings r ON m.id = r.movie_id
GROUP BY m.production_company
ORDER BY vote_count DESC 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:
WITH rank_actors
AS
(
SELECT
NAME AS actor_name,
Sum(total_votes) AS total_votes,
Count(a.movie_id) AS movie_count,
Round(Sum(avg_rating * total_votes) / Sum(total_votes), 2) AS actor_avg_rating
FROM role_mapping a
INNER JOIN names b ON a.name_id = b.id
INNER JOIN ratings c ON a.movie_id = c.movie_id
INNER JOIN movie d ON a.movie_id = d.id
WHERE category = 'actor'
AND country LIKE '%India%'
GROUP BY name_id, NAME
HAVING Count(DISTINCT a.movie_id) >= 5
)
SELECT *, DENSE_Rank() OVER (ORDER BY actor_avg_rating DESC) AS actor_rank FROM rank_actors;
-- 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 rank_actress
AS
(
SELECT
NAME AS actress_name,
Sum(total_votes) AS total_votes,
Count(rm.movie_id) AS movie_count,
Round(Sum(avg_rating * total_votes) / Sum(total_votes), 2) AS actress_avg_rating ,
DENSE_RANK () OVER (ORDER BY Round(Sum(avg_rating * total_votes) / Sum(total_votes), 2) DESC) AS actress_rank
FROM role_mapping rm
INNER JOIN names n ON rm.name_id = n.id
INNER JOIN ratings r ON rm.movie_id = r.movie_id
INNER JOIN movie m ON rm.movie_id = m.id
WHERE category = 'actress'
AND country LIKE '%India%' AND languages LIKE '%HINDI%'
GROUP BY NAME
HAVING Count(DISTINCT rm.movie_id) >= 3
)
SELECT * FROM rank_actress;
/* 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. Consider thriller movies having at least 25,000 votes. Classify them according to their average ratings in
the following categories:
Rating > 8: Superhit
Rating between 7 and 8: Hit
Rating between 5 and 7: One-time-watch
Rating < 5: Flop
Note: Sort the output by average ratings (desc).
--------------------------------------------------------------------------------------------*/
/* Output format:
+---------------+-------------------+
| movie_name | movie_category |
+---------------+-------------------+
| Get Out | Hit |
| . | . |
| . | . |
+---------------+-------------------+*/
-- Type your code below:
SELECT
title As movie_name, 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 genre AS g ON m.id = g.movie_id
INNER JOIN ratings r ON r.movie_id = m.id
WHERE 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:
SELECT genre, ROUND(AVG(duration),2) AS avg_duration, SUM(AVG(duration))
OVER(ORDER BY genre) AS running_total_duration, AVG(AVG(duration))
OVER(ORDER BY genre) AS moving_avg_duration
FROM movie m INNER JOIN genre g ON m.id = g.movie_id
GROUP BY genre;
-- 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_genre AS
(
SELECT genre, COUNT(movie_id) as movie_count, RANK() OVER(ORDER BY COUNT(movie_id) DESC) genre_rank FROM genre GROUP BY genre limit 3 ),
find_rank AS( SELECT genre, year, title AS movie_name, worlwide_gross_income, RANK() OVER(ORDER BY worlwide_gross_income DESC) AS movie_rank
FROM movie AS m INNER JOIN genre AS g ON m.id=g.movie_id WHERE genre IN (SELECT genre FROM top_3_genre)
)
SELECT * FROM find_rank WHERE movie_rank<=5;
-- 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:
Select production_company, COUNT(id) as movie_count, ROW_NUMBER() OVER(ORDER BY COUNT(id) DESC) AS prod_comp_rank
FROM movie m
INNER JOIN ratings 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 LIMIT 2;
-- 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 the number of Super Hit movies (Superhit movie: average rating of movie > 8) in 'drama' genre?
-- Note: Consider only superhit movies to calculate the actress average ratings.
-- (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. If number of votes are same, sort alphabetically by actress name.)
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actress_name | total_votes | movie_count | actress_avg_rating | actress_rank |
+---------------+-------------------+---------------------+----------------------+-----------------+
|Laura Dern | 1016 | 1 | 9.6000 | 1 |
|. | . | . | . | . |
|. | . | . | . | . |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:
SELECT name as actress_name,
SUM(total_votes) AS total_votes,
COUNT(rm.movie_id) as movie_count,
Round(Sum(avg_rating * total_votes)/Sum(total_votes),2) AS actress_avg_rating,
DENSE_RANK() OVER(ORDER BY COUNT(rm.movie_id) DESC) AS actress_rank
FROM names n
INNER JOIN role_mapping rm
ON n.id = rm.name_id
INNER JOIN ratings r
ON r.movie_id = rm.movie_id
INNER JOIN genre g ON g.movie_id = r.movie_id
WHERE category="actress" AND avg_rating>8 AND g.genre="Drama"
GROUP BY name LIMIT 3;
/* 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 t_date_summary
AS
(
SELECT d.name_id,
NAME,
d.movie_id,
duration,
r.avg_rating,
total_votes,
m.date_published,
Lead(date_published,1)
OVER(PARTITION BY d.name_id ORDER BY date_published,movie_id )
AS next_date_published
FROM director_mapping AS d
INNER JOIN names AS n
ON n.id = d.name_id
INNER JOIN movie AS m
ON m.id = d.movie_id
INNER JOIN ratings AS r
ON r.movie_id = m.id
),
top_director_summary AS ( SELECT *, Datediff(next_date_published, date_published) AS date_difference FROM t_date_summary )
SELECT name_id AS director_id, NAME AS director_name, COUNT(movie_id) AS number_of_movies, ROUND(AVG(date_difference),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 top_director_summary
GROUP BY director_id
ORDER BY COUNT(movie_id) DESC limit 9;