1. Computational Biology
Dr. Jens Allmer
Lecture Slides Week 7

2. MBG404 Overview
Processing
Pipelining
Generation
Data
Storage
Mining

3. A Relation is a Table Beers name manf Winterbrew Bud Lite Pete’s
Attributes
(column
headers)
Beers
name
manf
Winterbrew
Bud Lite
Pete’s
Anheuser-Busch
Tuples
(rows)
Domain
All possible values
Contains data -> Instance

4. A DB is a Collection of Relations

5. Decompositions and Normal Forms
So you stored your data in a high normal form and decomposed your
Relations because of that.
How do you now get any information from your DB?

6. Why SQL? SQL is a very-high-level language.
Say “what to do” rather than “how to do it.”
Avoid a lot of data-manipulation details needed in procedural languages like C++ or Java.
Database management system figures out “best” way to execute query.
Called “query optimization.” 6

7. Select-From-Where Statements
SELECT desired attributes
FROM one or more tables
WHERE condition about tuples of
the tables
ORDER BY desired attributes 7

8. Our Running Example
All our SQL queries will be based on the following database schema.
Underline indicates key attributes.
Beers(name, manf)
Bars(name, addr, license)
Drinkers(name, addr, phone)
Likes(drinker, beer)
Sells(bar, beer, price)
Frequents(drinker, bar) 8

9. Example
Using Beers(name, manf), what beers are made by Anheuser-Busch?
SELECT name
FROM Beers
WHERE manf = ’Anheuser-Busch’;
Notice SQL uses single-quotes for strings.
SQL is case-insensitive, except inside strings. 9

10. Result of Query The answer is a relation with a single attribute,
name
Bud
Bud Lite
Michelob
. . .
The answer is a relation with a single attribute,
name, and tuples with the name of each beer
by Anheuser-Busch, such as Bud. 10

11. Meaning of Single-Relation Query
Begin with the relation in the FROM clause.
Apply the selection indicated by the WHERE clause.
Apply the extended projection indicated by the SELECT clause. 11

12. Operational Semantics
name
manf tv
Bud
Anheuser-Busch
Include tv.name
in the result
Check if
Anheuser-Busch 12

13. * In SELECT clauses
When there is one relation in the FROM clause, * in the SELECT clause stands for “all attributes of this relation.”
Example using Beers(name, manf):
SELECT *
FROM Beers
WHERE manf = ’Anheuser-Busch’; 13

14. Result of Query: Now, the result has each of the attributes of Beers.
name manf
Bud Anheuser-Busch
Bud Lite Anheuser-Busch
Michelob Anheuser-Busch
Now, the result has each of the attributes
of Beers. 14

15. Renaming Attributes
If you want the result to have different attribute names, use “AS <new name>” to rename an attribute.
Example based on Beers(name, manf):
SELECT name AS beer, manf
FROM Beers
WHERE manf = ’Anheuser-Busch’ 15

16. Result of Query: beer manf Bud Anheuser-Busch Bud Lite Anheuser-Busch
Michelob Anheuser-Busch 16

17. Expressions in SELECT Clauses
Any mathematical expression that makes sense can appear as an element of a SELECT clause.
Example: from Sells(bar, beer, price):
SELECT bar, beer,
price * 114 AS priceInYen
FROM Sells; 17

18. Result of Query bar beer priceInYen Joe’s Bud 285 Sue’s Miller 342
… … … 18

19. Another Example: Constant Expressions
From Likes(drinker, beer) :
SELECT drinker,
’likes Bud’ AS whoLikesBud
FROM Likes
WHERE beer = ’Bud’; 19

20. Result of Query drinker whoLikesBud Sally likes Bud Fred likes Bud … …
… … 20

21. Complex Conditions in WHERE Clause
From Sells(bar, beer, price), find the price Joe’s Bar charges for Bud:
SELECT price
FROM Sells
WHERE bar = ’Joe’’s Bar’ AND
beer = ’Bud’;
Notice how we
get a single-quote
in strings. 21

22. Patterns
WHERE clauses can have conditions in which a string is compared with a pattern, to see if it matches.
General form:
<Attribute> LIKE <pattern>
<Attribute> NOT LIKE <pattern>
Pattern is a quoted string with
% = “any string”;
_ = “any character.” 22

23. Example
From Drinkers(name, addr, phone) find the drinkers with area code 555:
SELECT name
FROM Drinkers
WHERE phone LIKE ’%555-_ _ _ _’; 23

24. NULL Values
Tuples in SQL relations can have NULL as a value for one or more components.
Meaning depends on context. Two common cases:
Missing value : e.g., we know Joe’s Bar has some address, but we don’t know what it is.
Inapplicable : e.g., the value of attribute spouse for an unmarried person. 24

25. Comparing NULL’s to Values
The logic of conditions in SQL is really 3-valued logic: TRUE, FALSE, UNKNOWN.
When any value is compared with NULL, the truth value is UNKNOWN.
But a query only produces a tuple in the answer if its truth value for the WHERE clause is TRUE (not FALSE or UNKNOWN). 25

26. Surprising Example From the following Sells relation: bar beer price
Joe’s Bar Bud NULL
SELECT bar
FROM Sells
WHERE price < OR price >= 2.00;
UNKNOWN UNKNOWN
UNKNOWN 26

27. End Theory I
5 min mindmapping
10 min break

28. Practice I

29. Where is my DB?
Try to remember where you stored the Gene Ontology terms we used last week
You can also try the Recent part in the File menu (e.g.: you are on a different PC)
Select the DB and open it up for our queries

30. Query 1 How many enteries are there in the top categories
Biological process
Cellular component
Molecular function
To find out use the query wizard
At the end we want:
SELECT term.namespace, Count(term.name) AS CountOfname
FROM term
GROUP BY term.namespace;
Follow along as I use the query wizard

31. Query 2 Find all terms that contain the term ‘mito’
At the end we want:
SELECT term.namespace, Count(term.name) AS CountOfname
FROM term
GROUP BY term.namespace;
Follow along as I use the query editor

32. Query 3 Now you are on your own
Find all definitions related to catalysis and ethanol
At the end we want:
SELECT def.defstr
FROM def
WHERE (((def.defstr) Like '*catalysis*') AND ((def.defstr) Like '*ethanol*'));

33. Constant

34. End Practice I
15 min break

35. Theory II

36. Multirelation Queries
Interesting queries often combine data from more than one relation.
We can address several relations in one query by listing them all in the FROM clause.
Distinguish attributes of the same name by “<relation>.<attribute>” 36

37. Example
Using relations Likes(drinker, beer) and Frequents(drinker, bar), find the beers liked by at least one person who frequents Joe’s Bar.
SELECT beer
FROM Likes, Frequents
WHERE bar = ’Joe’’s Bar’ AND
Frequents.drinker = Likes.drinker;
Different then inner join
A simple cross product of
all combination of rows 37

38. Formal Semantics Almost the same as for single-relation queries:
Start with the product of all the relations in the FROM clause.
Apply the selection condition from the WHERE clause.
Project onto the list of attributes and expressions in the SELECT clause. 38

39. Operational Semantics
Imagine one tuple-variable for each relation in the FROM clause.
These tuple-variables visit each combination of tuples, one from each relation.
If the tuple-variables are pointing to tuples that satisfy the WHERE clause, send these tuples to the SELECT clause. 39

40. Example drinker bar drinker beer tv1 tv2 Sally Bud Sally Joe’s Likes
Frequents
to output
check these
are equal
check
for Joe 40

41. Controlling Duplicate Elimination
Force the result to be a set by SELECT DISTINCT . . .
Force the result to be a bag (i.e., don’t eliminate duplicates) by ALL, as in UNION ALL . . . 41

42. Example: DISTINCT
From Sells(bar, beer, price), find all the different prices charged for beers:
SELECT DISTINCT price
FROM Sells;
Notice that without DISTINCT, each price would be listed as many times as there were bar/beer pairs at that price. 42

43. Products and Natural Joins
R NATURAL JOIN S;
Product:
R CROSS JOIN S;
Example:
Likes NATURAL JOIN Serves;
Relations can be parenthesized subqueries, as well.
R OUTER JOIN S is the core of an outerjoin expression. It is modified by:
Optional NATURAL in front of OUTER.
Optional ON <condition> after JOIN.
Optional LEFT, RIGHT, or FULL before OUTER.
LEFT = pad dangling tuples of R only.
RIGHT = pad dangling tuples of S only.
FULL = pad both; this choice is the default. 43

44. Outerjoins
R OUTER JOIN S is the core of an outerjoin expression. It is modified by:
Optional NATURAL in front of OUTER.
Optional ON <condition> after JOIN.
Optional LEFT, RIGHT, or FULL before OUTER.
LEFT = pad dangling tuples of R only.
RIGHT = pad dangling tuples of S only.
FULL = pad both; this choice is the default. 44

45. Overview SELECT (π) AS (σ) FROM (Table) AS (σ) WHERE (Condition)
SELECT Table.Attribute AS newName ...
FROM (Table1 INNER JOIN Table2) ON Table1.Attribute = Table2.Attribute ...
GROUP BY Table.Attribute, ...
HAVING (requires GROUP BY) (COUNT|AVG|MAX|MIN|SUM) (>|=|<) (ConstValue|SubQuery)
WHERE Table1.Attribute (>|=|<|IS|IS NOT|LIKE) (ConstValue|Table2.Attribute|SubQuery)
ORDER BY Table.Attribute, ... 45

46. End Theory II
Mindmap 5 min
Break 10 min

47. Practice II

48. Join Tables Join term and relationship Any problems?
Term.id = relationship.to
SELECT term.id, term.name, term.namespace, relationship.type, relationship.to FROM relationship INNER JOIN term ON relationship.to = term.id;
Any problems?

49. Term Interrelationships
Join term with itself
Select those terms wher id matches to is_a
SELECT term.id, term.name, term.namespace, term.is_a, term_1.id, term_1.is_a FROM term INNER JOIN term AS term_1 ON term.id = term_1.is_a;

50. Term Types Join term and relationship tables
Find the number of types that are associated with each term
Order by highest count first
Final query:
SELECT term.id, Count(relationship.type) AS CountOftype
FROM term INNER JOIN relationship ON term.id = relationship.to
GROUP BY term.id
ORDER BY Count(relationship.type) DESC;
Now change to only count distinct types
How?
SELECT dt.id, Count(dt.type) AS CountOftype
FROM (SELECT DISTINCT id, type FROM term INNER JOIN relationship ON term.id = relationship.to) AS dt
GROUP BY dt.id;

51. Beneficial Things to do
Find data worth putting into a database
Even simple anlyses you do/did in the lab may be worth
Check online for interesting datasets if you cannot find anything
Model the data in MS Access
Make at least three tables
Store the data in your database