automated ranking of database query results
DESCRIPTION
Automated Ranking Of Database Query Results. Sanjay Agarwal - Microsoft Research Surajit Chaudhuri - Microsoft Research Gautam Das - Microsoft Research Aristides Gionis - Computer Science Dept Stanford University Presented by: Suvigya Jaiswal (Fall 10). - PowerPoint PPT PresentationTRANSCRIPT
Automated Ranking Of Database Query Results
Sanjay Agarwal - Microsoft Research Surajit Chaudhuri - Microsoft Research Gautam Das - Microsoft Research Aristides Gionis - Computer Science Dept Stanford University
Presented by: Suvigya Jaiswal (Fall 10)
Ranking
As the name suggests ‘Ranking’ is the process of ordering a set of values (or data items) based on some parameter that is of high relevance to the user of ranking process.
Ranking and returning the most relevant results of user’s query is a popular paradigm in information retrieval.
Database Ranking Example
IntroductionAutomated Ranking is used in Information
Retrieval(IR).Database Systems do not support
Automated Ranking but support only Boolean Query model.
Following scenarios not handled well by SQL Systems
1. Empty Answers(Query too specific)2. Many Answers(Query not specific)
IntroductionHow to adapt ranking functions from IR to
handle Database ranking problem?1.When each of the attribute in the relation is a categorical attribute , mimic the IR solution by applying the TD-IDF idea of frequency of values2. When attributes are also numerical ,extend the TD-IDF concepts to numerical domains.
In some cases the TF-IDF idea does not produce results with desired accuracy, in these cases we use workload information to arrive at better results.
Contributions of Paper
IDF Similarity QF similarityQFIDF SimilarityIndex Based Threshold Algorithm:
IDF Similarity: Intro
Given: A Table R
Attributes {A1,….,Am)Tuples {T1,….,Tn}Valuek
Query’s WHERE clause is of the form:“Where C1 AND C2 AND ….. AND Cm”
AttributesNumerical Attribute
CategoricalAttribute
SNO MFR PRICE COLOR MODEL TYPE1 AUDI 2000.00 RED Q5 SUV
2 BMW 3000.50 RED Z4
3 TOYOTA 3000.00 BLUE CAMRY SEDAN
4 HONDA 2000.00 GREEN ACCORD SEDAN
5 NISSAN 4000.00 WHITE 350Z CONVERTIBLE
Tuples
Cosine Similarity
Cosine Similarity from IR can be applied when the database has only categorical attributes.
Tuple and Query are considered a small document. A documents is an m-dimensional vector with m
words
ith element in the vector represents the TF of the word.
Cosine Similarity:
Cosine Similarity
IDF used to further refine Cosine Similarity IDF(w)= log(N/F(w)) N is number of documents F(w) is the number of documents in which w appears.
Idea behind using IDF?More often occurring words convey information than rarely occurring words.
IDF Similarity
For every value t in the domain of Ak , IDFk(t) is defined as
IDFk(t)=log(n/Fk(t)) n=# of tuples , Fk(t)) is the frequency of
tuples Ak=tT=<t1,……tm> Q=<q1,…...qm> Condition is of the form “WHERE A1=q1 AND A2=q2 ,….., AND
Am=qm “
Sk (u,v) = IDFk(u) if u=vOtherwise,Sk (u,v)=0
),(),(1
qtS kk
m
kk
QTSIM
Uses
As an Example say we want to find all convertibles made by Nissan.
The System will return the following:1. All Convertibles made by Nissan.2. All the Convertibles made by other
manufacturers.3. All Nissan Cars which may not be convertibles.
Why so?Convertible is a rarer car type than other Nissan
cars.
IDF Similarity for Numerical Data
Why the IDF Similarity for categorical data cannot be used for numeric data?
SELECT *FROM RWHERE PRICE=300K AND BEDROOM = 10;
S(u,v) will incorrectly evaluate to zero.Since 315K and 305K are close to 300K (resp. 9 & 10) but not
equal.
ID PRICE BEDROOM
CITY
1 315K 9 DALLAS2 300K 10 FTW3 305K 10 ARLINGTO
N
IDF Similarity for Numerical Data
Solution:{t1,t2…..tn} be the values of attribute A. For every value t
The denominator represents the sum of contributions to t from every other ti
Further t is from ti , lesser is the contribution from ti
IDF Similarity for Numerical Data
Similarity is defined as density at ‘t’ of a Gaussian Distribution centered at q.
Suppose there are n1 tuples that have the same value and the remaining
n-n1 tuples have a value far from t. 1. If q belongs to n-n1 then S(t,q) almost 02. If q belongs to n1 then S(t,q) = log (n/nt)
QF Similarity
Why is IDF Similarity not sufficient ?Examples:1. In a database , more homes are built in
recent years than in the pervious years (1980’s). But IDF of the homes built in recent years will be less. Yet the demand for newer homes is more.
2. In a bookstore DB, the demand of a particular author’s work might be more even if he has written many books. But the IDF of that author will be low.
QF Similarity
The Idea behind QF Similarity is that the importance of attribute values is related to the frequency of their occurrence in the query string in the workload
In previous example it is reasonable to assume that the queries for newer homes appears more often than queries for older homes
Also the query for a particular author might appear more often than the other authors if his books are more popular in spite of him having many books
QF Similarity
We define query frequency QF as
QF(q) = RQF(q)/ RQFMax
RQF(q) raw frequency of occurrence of value q of attribute A in query strings of workload
RQFMax raw frequency of most frequently occurring value in workload
S(t,q)=QF(q) if q=t else 0
Similarity between different attributes
If we use IDF or QF Similarity to measure any of the following we get 0 as the answerS(Toyota, Honda)=0S(Accord, Camry)=0
1.But we know that Honda and Toyota makecars that are directed toward the same market segment.
2.Accord and Camry are the same type of Cars of comparable quality
Similarity between different attributes
To solve this problem we apply the intuition that if certain pair of values(t<>u) often occur together in the work load then they are similar.
For example if we receive many queries which has C-Conditions of the form“MFR IN {Toyota, Honda, Nissan}”
It suggest that Toyota, Honda and Nissan are more similar to each other than they are to Ferrari or Mercedes
Hence we can say that by using this metric,S(Toyota, Honda)=0.8S(Ferrari, Toyota)=o.1
Similarity between different attributes
Let W(t) be the subset of Queries in workload W in
which the categorical value t appears(in our example
say Toyota) in an IN clause.
Jaccard Coefficient measures similarity b/w W(t) and W(q)
Similarity coefficient is then defined as:
QFIDF Similarity
QF Similarity can be unreliable in certain situations.This happens because QF Similarity is purely
workload based. It doesn’t take data values into account.
To tackle this we define QFIDF Similarity:
S(t,q)=QF(q) *IDF(q) when t=q 0, otherwise
where QF(q)=(RQF(q)+1)/(RQFMax+1).
1 is added to the numerator and denominator so that QF is never zero.
Many Answer Problem.
IDF Similarity and QF Similarity may sometimes run into problem: many tuples may tie for the same similarity score and thus get ordered arbitrarily.
Approach is to determine weights of missing attribute values that reflect their “global importance” for ranking purposes
IF we seek homes with four bedrooms in DB, we can examine attributes other than number of bedrooms to rank the result set. If we knew that “Dallas” is a more important location than “Fort-Worth” in a global sense, we would rank four bedroom homes in Dallas higher than four bedroom homes in Fort-Worth.
We use workload information to determine global importance of missing attribute values.
We define the global importance of missing attribute value tk as log(QFk(tk))
Extend QF Similarity to use the quantity Sum(log(QFk(tk))) to break ties in each equivalence
class (larger this quantity1, higher the rank of the tuple) where the summation is over missing attributes.
An alternative strategy is to rank tied tuples higher if their missing attribute values have small IDF, i.e. occur more frequently in the database.
Implementation
Two Phases: Pre-processing component Query processing component
Pre-processing component
Compute IDF(t) (resp. QF(t)) for all categorical values t involves scanning the database (resp. scanning/parsing the workload) to compute frequency of occurrences of values in the database (resp. workload), and store the results in auxiliary tables.
We cannot pre-compute IDF(q) (resp. QF(q)) for numerical attributes; thus we have to store an approximate representation of the smooth function IDF( ) (resp. QF( )) so that the function value at any q can be retrieved at runtime.
Query processing component
main task of the query processing component is, given a query Q and an integer K, to efficiently retrieve the Top-K tuples from the database using one of the ranking functions.
A simpler query processing problem
Inputs: (a) a database table R with m categorical columns, clustered on key
column TID, where standard database indexes exist on a subset of columns,
(b) A query expressed as a conjunction of m single-valued conditions of the form Ak = qk., and
(c) an integer K.Similarity function: Overlap SimilarityOutput: The Top-K tuples of R most similar to Q.
An index-based Top-K implementation:
monotonic property: if T and U are two tuples such that for all k, Sk(tk,qk)< Sk(uk,qk) then
SIM(T,Q) <=SIM(U, Q).adapt Fagin’s Threshold Algorithm (TA)Two types of access methods required1. Sorted Access2. Random Accessuse of an early stopping condition, by which the
algorithm can detect that the final Top-K tuples have been retrieved before all tuples have been processed.
Read all grades of an object once seen from a sorted access• No need to wait until the lists give k common objects
Do sorted access (and corresponding random accesses) until you have seen the top k answers.
• How do we know that grades of seen objects are higher than the grades of unseen objects ?
• Predict maximum possible grade unseen objects:
a: 0.9b: 0.8c: 0.72
.
.
.
.
L1L2
d: 0.9a: 0.85b: 0.7
c: 0.2
.
.
.
.f: 0.65
d: 0.6
f: 0.6
Seen
Possibly unseen
Threshold value
Threshold Algorithm
T = min(0.72, 0.7) = 0.7
ID A1 A2 Min(A1,A2)
Step 1: - parallel sorted access to each list
(a, 0.9)
(b, 0.8)
(c, 0.72)
(d, 0.6)
.
.
.
.
L1 L2
(d, 0.9)
(a, 0.85)
(b, 0.7)
(c, 0.2)
.
.
.
.
a
d
0.9
0.9
0.85 0.85
0.6 0.6
For each object seen: - get all grades by random access - determine Min(A1,A2) - amongst 2 highest seen ? keep in
buffer
Example – Threshold Algorithm
ID A1 A2 Min(A1,A2)a: 0.9
b: 0.8
c: 0.72
d: 0.6
.
.
.
.
L1 L2
d: 0.9
a: 0.85
b: 0.7
c: 0.2
.
.
.
.
Step 2: - Determine threshold value based on objects currently seen under sorted access. T = min(L1, L2)
a
d
0.9
0.9
0.85 0.85
0.6 0.6
T = min(0.9, 0.9) = 0.9
- 2 objects with overall grade ≥ threshold value ? stop else go to next entry position in sorted list and repeat step 1
Example – Threshold Algorithm
ID A1 A2 Min(A1,A2)
Step 1 (Again): - parallel sorted access to each list
(a, 0.9)
(b, 0.8)
(c, 0.72)
(d, 0.6)
.
.
.
.
L1 L2
(d, 0.9)
(a, 0.85)
(b, 0.7)
(c, 0.2)
.
.
.
.
a
d
0.9
0.9
0.85 0.85
0.6 0.6
For each object seen: - get all grades by random access - determine Min(A1,A2) - amongst 2 highest seen ? keep in buffer
b 0.8 0.7 0.7
Example – Threshold Algorithm
ID A1 A2 Min(A1,A2)a: 0.9
b: 0.8
c: 0.72
d: 0.6
.
.
.
.
L1 L2
d: 0.9
a: 0.85
b: 0.7
c: 0.2
.
.
.
.
Step 2 (Again): - Determine threshold value based on objects currently seen. T = min(L1, L2)
a
b
0.9
0.7
0.85 0.85
0.8 0.7
T = min(0.8, 0.85) = 0.8
- 2 objects with overall grade ≥ threshold value ? stop else go to next entry position in sorted list and repeat step 1
Example – Threshold Algorithm
ID A1 A2 Min(A1,A2)a: 0.9
b: 0.8
c: 0.72
d: 0.6
.
.
.
.
L1 L2
d: 0.9
a: 0.85
b: 0.7
c: 0.2
.
.
.
.
Situation at stopping condition
a
b
0.9
0.7
0.85 0.85
0.8 0.7
T = min(0.72, 0.7) = 0.7
Example – Threshold Algorithm
Algorithm
Experiment Results
Quality resultsFor queries with empty answers, QFIDF produced the best
rankings, followed by QF, then IDF, and finally Overlap. For queries with empty answers, the ranking quality of QF
improves with increasing workload size. For queries with numerous answers, QF produced better
rankings than IDF.Performance results The preprocessing time and space requirements of all
techniques scale linearly with data size.When all indexes are present, ITA is more efficient than SQL
Server Top-K for all similarity functions.Even when a subset of indexes is present, ITA can perform well
References
http://www.emeraldinsight.com/journals.htm?articleid=1563479
Ppt Slides by Ramya Soumri(Fall 09)[14] R. Fagin. Fuzzy Queries in Multimedia
Database Systems. PODS 1998.
Thank You