Profiling and Analyzing the Yelp Dataset.txt

Data Scientist Role Play: Profiling and Analyzing the Yelp Dataset

Part 1: Yelp Dataset Profiling and Understanding

1. Profile the data by finding the total number of records for each of the tables below:
	
i. Attribute table = 10000

SELECT COUNT(*)
FROM attribute;

ii. Business table = 10000

SELECT COUNT(*)
FROM business;

iii. Category table = 10000

SELECT COUNT(*)
FROM category;

iv. Checkin table = 10000

SELECT COUNT(*)
FROM checkin;

v. elite_years table = 10000

SELECT COUNT(*)
FROM elite_years;

vi. friend table = 10000

SELECT COUNT(*)
FROM friend;

vii. hours table = 10000

SELECT COUNT(*)
FROM hours;

viii. photo table = 10000

SELECT COUNT(*)
FROM photo;

ix. review table = 10000

SELECT COUNT(*)
FROM review;

x. tip table = 10000

SELECT COUNT(*)
FROM tip;

xi. user table = 10000

SELECT COUNT(*)
FROM user;


2. Find the total distinct records by either the foreign key or primary key for each table. If two foreign keys are listed in the table, please specify which foreign key.

i. Business = 10000

SELECT COUNT(DISTINCT id)
FROM business;

ii. Hours = 1562

SELECT COUNT(DISTINCT business_id)
FROM hours;

iii. Category = 2643

SELECT COUNT(DISTINCT business_id)
FROM category;

iv. Attribute = 1115

SELECT COUNT(DISTINCT business_id)
FROM attribute;

v. Review = 10000, 8090, 9581

SELECT COUNT(DISTINCT id)
	,COUNT(DISTINCT business_id)
	,COUNT(DISTINCT user_id)
FROM review;

vi. Checkin = 493

SELECT COUNT(DISTINCT business_id)
FROM checkin;

vii. Photo = 10000, 6493

SELECT COUNT(DISTINCT id)
	,COUNT(DISTINCT business_id)
FROM photo;

viii. Tip = 537, 3979

SELECT COUNT(DISTINCT user_id)
	,COUNT(DISTINCT business_id)
FROM tip;

ix. User = 10000

SELECT COUNT(DISTINCT id)
FROM user;

x. Friend = 11

SELECT COUNT(DISTINCT user_id)
FROM friend;

xi. Elite_years = 2780

SELECT COUNT(DISTINCT user_id)
FROM elite_years;


Note: Primary Keys are denoted in the ER-Diagram with a yellow key icon.	



3. Are there any columns with null values in the Users table? Indicate "yes," or "no."

	Answer: No
	
	
	SQL code used to arrive at answer:
	
		
	SELECT *
	FROM user
	WHERE id IS NULL
		OR name IS NULL
		OR review_count IS NULL
		OR yelping_since IS NULL
		OR useful IS NULL
		OR funny IS NULL
		OR cool IS NULL
		OR fans IS NULL
		OR average_stars IS NULL
		OR compliment_hot IS NULL
		OR compliment_more IS NULL
		OR compliment_profile IS NULL
		OR compliment_cute IS NULL
		OR compliment_list IS NULL
		OR compliment_note IS NULL
		OR compliment_plain IS NULL
		OR compliment_cool IS NULL
		OR compliment_funny IS NULL
		OR compliment_writer IS NULL
		OR compliment_photos IS NULL
	
4. For each table and column listed below, display the smallest (minimum), largest (maximum), and average (mean) value for the following fields:

	i. Table: Review, Column: Stars
	
		min: 1  	max: 5  	avg: 3.7082
		
		SELECT min(stars)
			,max(stars)
			,avg(stars)
		FROM review;

	ii. Table: Business, Column: Stars
	
		min: 1.0	max: 5.0	avg: 3.6549
		
		SELECT min(stars)
			,max(stars)
			,avg(stars)
		FROM business;
		
	
	iii. Table: Tip, Column: Likes
	
		min: 0  	max: 2  	avg: 0.0144
		
		SELECT min(likes)
			,max(likes)
			,avg(likes)
		FROM tip;
		
	
	iv. Table: Checkin, Column: Count
	
		min: 1		max: 53		avg: 1.9414
		
		SELECT min(count) AS min
			,max(count) AS max
			,avg(count) AS avg
		FROM checkin;
		
	
	v. Table: User, Column: Review_count
	
		min: 0		max: 2000		avg: 24.2995
		
		SELECT min(review_count)
			,max(review_count)
			,avg(review_count)
		FROM user;


5. List the cities with the most reviews in descending order:

	SQL code used to arrive at answer:
	
	SELECT city
			,sum(review_count)
	FROM business
	GROUP BY city
	ORDER BY sum(review_count) DESC
	
	Copy and Paste the Result Below:
		
		+-----------------+-------------------+
	| city            | sum(review_count) |
	+-----------------+-------------------+
	| Las Vegas       |             82854 |
	| Phoenix         |             34503 |
	| Toronto         |             24113 |
	| Scottsdale      |             20614 |
	| Charlotte       |             12523 |
	| Henderson       |             10871 |
	| Tempe           |             10504 |
	| Pittsburgh      |              9798 |
	| Montréal        |              9448 |
	| Chandler        |              8112 |
	| Mesa            |              6875 |
	| Gilbert         |              6380 |
	| Cleveland       |              5593 |
	| Madison         |              5265 |
	| Glendale        |              4406 |
	| Mississauga     |              3814 |
	| Edinburgh       |              2792 |
	| Peoria          |              2624 |
	| North Las Vegas |              2438 |
	| Markham         |              2352 |
	| Champaign       |              2029 |
	| Stuttgart       |              1849 |
	| Surprise        |              1520 |
	| Lakewood        |              1465 |
	| Goodyear        |              1155 |
	+-----------------+-------------------+
	(Output limit exceeded, 25 of 362 total rows shown)
	
6. Find the distribution of star ratings to the business in the following cities:

i. Avon

SQL code used to arrive at answer:

SELECT stars
	,COUNT(stars) AS count
FROM business
WHERE city = 'Avon'
GROUP BY stars


Copy and Paste the Resulting Table Below (2 columns – star rating and count):

+-------+-------+
| stars | count |
+-------+-------+
|   1.5 |     1 |
|   2.5 |     2 |
|   3.5 |     3 |
|   4.0 |     2 |
|   4.5 |     1 |
|   5.0 |     1 |
+-------+-------+


ii. Beachwood

SQL code used to arrive at answer:

SELECT stars
	,COUNT(stars) AS count
FROM business
WHERE city = 'Beachwood'
GROUP BY stars

Copy and Paste the Resulting Table Below (2 columns – star rating and count):
		
+-------+-------+
| stars | count |
+-------+-------+
|   2.0 |     1 |
|   2.5 |     1 |
|   3.0 |     2 |
|   3.5 |     2 |
|   4.0 |     1 |
|   4.5 |     2 |
|   5.0 |     5 |
+-------+-------+


7. Find the top 3 users based on their total number of reviews:
		
	SQL code used to arrive at answer:
	
	SELECT name
			,sum(review_count)
	FROM user
	GROUP BY name
	ORDER BY sum(review_count) DESC limit 3
		
	Copy and Paste the Result Below:
		
	+--------+-------------------+
	| name   | sum(review_count) |
	+--------+-------------------+
	| Nicole |              2397 |
	| Sara   |              2253 |
	| Gerald |              2034 |
	+--------+-------------------+
	

8. Does posing more reviews correlate with more fans? 

	No, posing more reviews does not correlate with more fans.

	Please explain your findings and interpretation of the results:
	
	SELECT name
		,sum(review_count)
		,sum(fans)
	FROM user
	GROUP BY name
	ORDER BY sum(review_count) DESC
	
	+-----------+-------------------+-----------+
	| name      | sum(review_count) | sum(fans) |
	+-----------+-------------------+-----------+
	| Nicole    |              2397 |       200 |
	| Sara      |              2253 |        72 |
	| Gerald    |              2034 |       256 |
	| Lisa      |              2021 |       207 |
	| Mark      |              1945 |       156 |
	| Jennifer  |              1862 |        86 |
	| Jen       |              1744 |       148 |
	| Michelle  |              1737 |       133 |
	| Melissa   |              1678 |       104 |
	| William   |              1639 |       142 |
	| Karen     |              1625 |       123 |
	| Justin    |              1572 |        78 |
	| Rachel    |              1519 |        79 |
	| Daniel    |              1441 |        75 |
	| Sarah     |              1436 |       100 |
	| Chris     |              1419 |        52 |
	| Jeff      |              1410 |        45 |
	| Mike      |              1400 |       119 |
	| Yuri      |              1341 |        76 |
	| Dan       |              1333 |        44 |
	| Christine |              1309 |       187 |
	| John      |              1264 |        46 |
	| Jessica   |              1248 |       116 |
	| .Hon      |              1246 |       101 |
	| Matt      |              1239 |        62 |
	+-----------+-------------------+-----------+
	
	for example, sara has 2253 reviews and only 72 fans, 
	while sarah has significantly fewer reviews (1436) and more fans 100.
	
9. Are there more reviews with the word "love" or with the word "hate" in them?

	Answer:
	SQL code used to arrive at answer:
	
	SELECT count(*) AS hate_count
	FROM review
	WHERE TEXT LIKE '%hate%'

	+------------+				
	| hate_count |
	+------------+
	|        232 |
	+------------+
	
	SELECT count(*) AS love_count
	FROM review
	WHERE TEXT LIKE '%love%'
	
	+------------+
	| love_count |
	+------------+
	|       1780 |
	+------------+

	
10. Find the top 10 users with the most fans:

	SQL code used to arrive at answer:
	
	SELECT fans
		,name
	FROM user
	GROUP BY fans
	ORDER BY fans DESC limit 10
	
	Copy and Paste the Result Below:

	+------+-----------+
	| fans | name      |
	+------+-----------+
	|  503 | Amy       |
	|  497 | Mimi      |
	|  311 | Harald    |
	|  253 | Gerald    |
	|  173 | Christine |
	|  159 | Lisa      |
	|  133 | Cat       |
	|  126 | William   |
	|  124 | Fran      |
	|  120 | Lissa     |
	+------+-----------+
		

Part 2: Inferences and Analysis

1. Pick one city and category of your choice and group the businesses in that city or category by their overall star rating. Compare the businesses with 2-3 stars to the businesses with 4-5 stars and answer the following questions. Include your code.
	
i. Do the two groups you chose to analyze have a different distribution of hours?
	Yes, they have different distribution hours. For example, in the Food category in the city of Las Vegas, restaurants with 2-3 stars have longer hours than those with 4-5 stars.

ii. Do the two groups you chose to analyze have a different number of reviews?
     Yes, they have a different number of reviews.    

iii. Are you able to infer anything from the location data provided between these two groups? Explain.
	No, I couldn't infer anything because two groups have different zip codes.

SQL code used for analysis:

		SELECT b.name
				,c.category
				,b.city
				,b.postal_code AS zipcode
				,h.hours
				,CASE 
					WHEN stars BETWEEN 2 AND 3 THEN '2-3 stars'
					WHEN stars BETWEEN 4 AND 5 THEN '4-5 stars'
					END AS rating
				,b.review_count AS reviews
		FROM business b
		INNER JOIN hours h ON b.id = h.business_id
		INNER JOIN category c ON c.business_id = b.id
		WHERE city = 'Las Vegas' AND category = 'Food' AND rating IN ('2-3 stars','4-5 stars')
		GROUP BY name
		ORDER BY stars DESC
		
2. Group business based on the ones that are open and the ones that are closed. What differences can you find between the ones that are still open and the ones that are closed? 
List at least two differences and the SQL code you used to arrive at your answer.
		
i. Difference 1: The average number of reviews was higher for open businesses than for closed businesses.
         
ii. Difference 2: The number of distinct business_id for the open businesses is 5 times higher than for the closed businesses and therefore the average number of reviews for the open businesses is higher.
         
SQL code used for analysis:

	SELECT count(DISTINCT id)
			,count(DISTINCT city)
			,avg(stars)
			,avg(review_count)
			,is_open
	FROM business
	GROUP BY is_open
	
3. For this last part of your analysis, you are going to choose the type of analysis you want to conduct on the Yelp dataset and are going to prepare the data for analysis.

Ideas for analysis include: Parsing out keywords and business attributes for sentiment analysis, clustering businesses to find commonalities or anomalies between them, predicting the overall star rating for a business,
predicting the number of fans a user will have, and so on. These are just a few examples to get you started, so feel free to be creative and come up with your own problem you want to solve. 
Provide answers, in-line, to all of the following:
	
i. Indicate the type of analysis you chose to do:
    I chose to analyse whether the businesses like Coffee & Tea having the arrtibutes like 'WheelchairAccessible','OutdoorSeating' ,'DogsAllowed' and 'BusinessAcceptsCreditCards'  
	anyway relate to the number of stars or the review counts that they get.  
         
ii. Write 1-2 brief paragraphs on the type of data you will need for your analysis and why you chose that data:
    I used 3 tables for my analysis: Company, Category and Attribute. 
	I chose the name of the business, its category, the state in which it is run, the characteristics it has, its ratings and the number of its ratings. 
	I have variables like  
		1)name, state, stars , review_count from  the table business, 
		2)category from the table category and
		3)name ,value from the attribute table.
	To connect all the 3 tables. I used the primary keys like id from business, business_id from category and 
	business_id from attribute table.
                 
	I wanted to know having a free wifi or OutdoorSeating or having BusinessAcceptsCreditCards or having WheelchairAccessible or having any combination of 3 or all the attributes 
	contributes to good rating or having more reviews in particular. 
	but what I found out was, the two businesses have same attributes, but are in two different states. The business in OH clearly has more reviews and thus more stars 
                  
iii. Output of your finished dataset:
         
	+-------------+----------------------------+-------+-------+-------+--------------+
	| business    | attribute                  | value | state | stars | review_count |
	+-------------+----------------------------+-------+-------+-------+--------------+
	| Koko Bakery | WiFi                       | free  | OH    |   4.0 |          162 |
	| Koko Bakery | OutdoorSeating             | 0     | OH    |   4.0 |          162 |
	| Koko Bakery | BusinessAcceptsCreditCards | 1     | OH    |   4.0 |          162 |
	| Koko Bakery | WheelchairAccessible       | 1     | OH    |   4.0 |          162 |
	| Starbucks   | WiFi                       | free  | AZ    |   3.0 |           52 |
	| Starbucks   | OutdoorSeating             | 1     | AZ    |   3.0 |           52 |
	| Starbucks   | BusinessAcceptsCreditCards | 1     | AZ    |   3.0 |           52 |
	+-------------+----------------------------+-------+-------+-------+--------------+
	
iv. Provide the SQL code you used to create your final dataset:

SELECT b.name AS business
	,att.name AS attribute
	,att.value
	,b.STATE
	,b.stars
	,b.review_count
FROM business b
INNER JOIN category c ON c.business_id = b.id
INNER JOIN attribute att ON att.business_id = b.id
WHERE (
		att.name LIKE 'WiFi'
		OR att.name LIKE 'WheelchairAccessible'
		OR att.name LIKE 'OutdoorSeating'
		OR att.name LIKE 'DogsAllowed'
		OR att.name LIKE 'BusinessAcceptsCreditCards'
		)
	AND category = 'Coffee & Tea'
ORDER BY stars DESC