Useful for Data Analysis

Author: ayushi-b

SQL Data Cleaning/Lesson 5 - SQL Data Cleaning.sql

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+/*
+In the accounts table, there is a column holding the website for each 
+company. The last three digits specify what type of web address they 
+are using. Pull these extensions and provide how many of each website 
+type exist in the accounts table.
+*/
+
+SELECT RIGHT(website, 3),
+	COUNT(*)
+FROM accounts
+GROUP BY 1;
+
+
+
+
+
+/*
+Use the accounts table to pull the first letter of each company name 
+to see the distribution of company names that begin with each letter 
+(or number). 
+*/
+
+SELECT LEFT(name, 1),
+	COUNT(*)
+FROM accounts
+GROUP BY 1
+ORDER BY 2 DESC;
+
+
+
+
+
+/*
+Use the accounts table and a CASE statement to create two groups: one 
+group of company names that start with a number and a second group of 
+those company names that start with a letter. What proportion of 
+company names start with a letter?
+*/
+
+SELECT CASE WHEN LEFT(name, 1) IN ('0','1','2','3','4','5','6','7','8','9')
+		    THEN 'number'
+		    ELSE 'letter'
+			END
+		AS type,
+		COUNT(*)
+FROM accounts
+GROUP BY 1
+ORDER BY 2 DESC;
+
+
+
+
+
+/*
+Consider vowels as a, e, i, o, and u. What proportion of company names 
+start with a vowel, and what percent start with anything else?
+*/
+
+SELECT CASE WHEN LEFT(UPPER(name), 1) IN ('A','E','I','O','U')
+		    THEN 'vowel'
+		    ELSE 'else'
+			END
+		AS type,
+		COUNT(*)
+FROM accounts
+GROUP BY 1
+ORDER BY 2 DESC;
+
+
+
+
+
+/*
+Use the accounts table to create first and last name columns that hold the 
+first and last names for the primary_poc. 
+*/
+
+SELECT LEFT(primary_poc, STRPOS(primary_poc, ' ')-1) AS firstname,
+	RIGHT(primary_poc, LENGTH(primary_poc) - STRPOS(primary_poc, ' ')) AS lastname
+FROM accounts;
+
+
+
+
+
+/*
+Now see if you can do the same thing for every rep name in the sales_reps 
+table. Again provide first and last name columns.
+*/
+
+SELECT LEFT(name, STRPOS(name, ' ')-1) AS firstname,
+RIGHT(name, LENGTH(name) - STRPOS(name, ' ')) AS lname
+FROM sales_reps;
+
+
+
+
+
+/*
+Each company in the accounts table wants to create an email address for each 
+primary_poc. The email address should be the first name of the primary_poc . 
+last name primary_poc @ company name .com.
+*/
+
+SELECT primary_poc, name,
+	LOWER(
+		LEFT(primary_poc, STRPOS(primary_poc, ' ')-1) || 
+		'.' || 
+		RIGHT(primary_poc, LENGTH(primary_poc) - STRPOS(primary_poc, ' ')) || 
+		'@' || 
+		name || 
+		'.com') 
+	AS email
+FROM accounts;
+
+
+
+
+
+
+
+/*
+You may have noticed that in the previous solution some of the company names 
+include spaces, which will certainly not work in an email address. See if 
+you can create an email address that will work by removing all of the spaces 
+in the account name, but otherwise your solution should be just as in 
+question above.
+*/
+
+SELECT primary_poc, name,
+	LOWER(
+		LEFT(primary_poc, STRPOS(primary_poc, ' ')-1) || 
+		'.' || 
+		RIGHT(primary_poc, LENGTH(primary_poc) - STRPOS(primary_poc, ' ')) || 
+		'@' || 
+		TRANSLATE(name, ' ', '') || 
+		'.com') 
+	AS email
+FROM accounts;
+
+
+
+
+
+/*
+We would also like to create an initial password, which they will change after 
+their first log in. The first password will be the first letter of the 
+primary_poc's first name (lowercase), then the last letter of their first name 
+(lowercase), the first letter of their last name (lowercase), the last letter of 
+their last name (lowercase), the number of letters in their first name, the number 
+of letters in their last name, and then the name of the company they are working 
+with, all capitalized with no spaces.
+*/
+
+WITH sub AS (
+  	SELECT LEFT(primary_poc, STRPOS(primary_poc, ' ')-1) AS firstname,
+		RIGHT(primary_poc, LENGTH(primary_poc) - STRPOS(primary_poc, ' ')) AS lastname,
+		id
+	FROM accounts)
+          
+SELECT LEFT(LOWER(sub.firstname), 1) ||
+       RIGHT(LOWER(sub.firstname), 1) ||
+       LEFT(LOWER(sub.lastname), 1) ||
+       RIGHT(LOWER(sub.lastname), 1) ||
+       LENGTH(sub.firstname) ||
+       LENGTH(sub.lastname) ||
+       TRANSLATE(UPPER(a.name), ' ', '')
+       AS pwd, 
+       a.primary_poc,
+	   a.name
+FROM accounts a
+JOIN sub
+ON sub.id = a.id;
+
+