SQL Data Analysis Exercises – Stack Exchange

-- Basic Analytics 
-- 1. How many post were made in each year
SELECT COUNT(*) number_of_posts, YEAR(CreationDate) creation_year
FROM posts 
GROUP BY YEAR(CreationDate) 

-- 2. How many votes were made in each day of the week (Sunday, Monday, Tuesday, etc.) ? 
SELECT COUNT(*) AS 'Number of Posts',  DATENAME(DW, CreationDate) AS 'Day of the Week'
FROM votes 
GROUP BY DATENAME(DW, CreationDate) 

-- 3. List all comments created on 2012-12-19
SELECT * 
FROM comments 
WHERE CAST(creationDate AS DATE) = '2012-12-19'

-- Or
SELECT * 
FROM comments 
WHERE DATEDIFF(DAY, CreationDate, '2012-12-19') = 0


-- 4. List all users under the age of 33, living in London 
SELECT * 
FROM users 
WHERE age < 33
AND   location LIKE '%London%'

-- Advanced Analytics
-- 1. Display the number of votes for each post title 
SELECT pst.Title, COUNT(*) number_of_votes 
FROM posts pst JOIN votes vt 
ON   pst.Id = vt.postID
GROUP BY pst.Title
ORDER BY COUNT(*) DESC 

-- 2. Display comments created by users living in the same location as the post creator 
SELECT pst.Id AS 'post_id',
       pst.Title AS 'post_title',
	   pst.OwnerUserID AS 'created_by_user',
	   usr_p.Id AS 'user_id', 
	   usr_p.DisplayName AS 'creator_user_name',
	   usr_p.location AS 'creator_location',
	   cmt.UserId AS 'commentor_id', 
	   usr_c.DisplayName AS 'commentor_user_name',
	   usr_c.location AS 'commentor_location'
FROM posts pst JOIN users usr_p 
ON   pst.OwnerUserID = usr_p.Id
			   JOIN comments cmt
ON   cmt.postId = pst.Id
			   JOIN users usr_c
ON   cmt.UserID = usr_c.Id
WHERE usr_c.location = usr_p.location

-- 3. How many users have never voted ? 
WITH "VOTE-CTE" AS  
	(
	SELECT id FROM users
	EXCEPT 
	SELECT userID FROM votes  
	)
SELECT COUNT(*)
FROM "VOTE-CTE"

-- 4. Display all posts having the highest amount of comments 
WITH "TOP-COMMENT-POSTS"
AS
	(
	SELECT	pst.Title, 
			COUNT(*) AS 'number_of_comments' , 
			DENSE_RANK() OVER (ORDER BY COUNT(*) DESC) AS 'comment_count_ranking'
	FROM posts pst JOIN comments cmt
	ON   pst.id = cmt.postid
	GROUP BY pst.Title
	)
SELECT Title 
FROM "TOP-COMMENT-POSTS"
WHERE comment_count_ranking = 1  

-- 5. For each post, how many votes are coming from users living in Canada ?
--    What's their percentage of the total number of votes 
SELECT	pst.Title, 
		COUNT(*) number_of_votes, 
		SUM(CASE WHEN usr.location LIKE '%canada%' THEN 1 ELSE 0 END) AS 'votes_from_canada',
		CAST(SUM(CASE WHEN usr.location LIKE '%canada%' THEN 1 ELSE 0 END) AS FLOAT) /
		CAST(COUNT(*) AS FLOAT) AS 'votes_percentage'
FROM posts pst JOIN votes vt 
ON   pst.Id = vt.postID
			   JOIN users usr
ON   vt.UserID = usr.id
GROUP BY pst.Title
ORDER BY COUNT(*) DESC 

-- 6. How many hours in average, it takes to the first comment to be posted after a creation of a new post
;WITH "COMMENTS-TIMING-CTE"
AS 
	(
	SELECT  pst.id AS 'post_id',
			pst.Title AS 'post_title',
			pst.creationDate AS 'post_creation_date',
			cmt.creationDate AS 'comment_creation_date',
			DENSE_RANK() OVER (PARTITION BY pst.id ORDER BY cmt.creationDate) AS 'comment_rank'
	FROM posts pst JOIN comments cmt 
	ON   pst.Id = cmt.postID
	)
SELECT AVG(DATEDIFF(HOUR, post_creation_date, comment_creation_date)) AS 'avg_num_of_hours'
FROM "COMMENTS-TIMING-CTE"
WHERE comment_rank = 1

-- 7. Whats the most common post tag ? 
--     Note, each post may have 1 or more tags. 
--     The goal of this question is to find the most common *single* tag
;WITH "CTE-TAGS-SEP" (Tags) AS
(
	SELECT CAST(Tags AS VARCHAR(MAX)) 
	FROM Posts
	UNION ALL
	SELECT STUFF(Tags, 1, CHARINDEX('><' , Tags), '') 
	FROM "CTE-TAGS-SEP"
	WHERE Tags  LIKE '%><%'
), "CTE-TAGS-COUNTER" AS 
(   
	SELECT CASE WHEN Tags LIKE '%><%' THEN LEFT(Tags, CHARINDEX('><' , Tags)) 
	            ELSE Tags 
			END AS 'Tags'
	FROM "CTE-TAGS-SEP"
)
SELECT TOP 1 COUNT(*), Tags
FROM "CTE-TAGS-COUNTER"
GROUP BY Tags 
ORDER BY COUNT(*) DESC 

-- Solution by YB
SELECT REPLACE(REPLACE(REPLACE(Tags, '><', ','), '<',''),'>','') 'TAGS' INTO #TABI1
FROM posts
SELECT VALUE, COUNT(*)
FROM #TABI1
CROSS APPLY string_split(TAGS, ',')
GROUP BY VALUE
ORDER BY COUNT(*) DESC

-- 8. Create a pivot table displaying how many posts were created for each year (Y axis) and each month (X axis)
SELECT *   
FROM (  
	SELECT YEAR(CreationDate) AS 'Year', DATENAME(MONTH,CreationDate) AS 'Month', id 
	FROM posts
  ) AS S  
PIVOT   
     (   
    COUNT(id) 
    FOR  [Month] IN ([January], [February], [March], [April], [May], [June], [July], [August], [September], [October], [November], [December]) 
   ) AS PVT
ORDER BY [Year]