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# Lecture 1 : Term Weighting and VSM - PowerPoint Presentation

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Slide1

Lecture 1 : Term Weighting and VSM

1Slide2

Basics to Informational Retrieval

2Slide3

3

Definition of

information

retrieval

Information retrieval (IR) is

finding

material (usually documents) ofan unstructured nature (usually text) that satisfies an informationneed from within large collections (usually stored on computers).IR is the foundation to Text miningInfo Retrieval & Extraction → Text Mining → Knowledge Discovery先能處理大量資訊，再將處理層次提升Ex. 資訊檢索 → 相似度計算 → 分類分群→ 摘要 → 主題偵測及追蹤 → 情緒及意見分析 → 實體辨識及語意網路 → 自然語言對話 → 找出答案 ….

3Slide4

4

Unstructured data

in 1650

Which plays of Shakespeare contain the words

B

RUTUS AND

CAESAR, but not CALPURNIA ?One could scan all of Shakespeare’s plays for BRUTUS and CAESAR, then strip out lines containing CALPURNIA Why is scan not the solution?Slow (for large collections)Advanced operations not feasible (e.g., find the word ROMANS near COUNTRYMAN )4Slide5

5

Term-document

incidence

matrix

Entry is 1 if term occurs. Example: CALPURNIA occurs in

Julius

Caesar. Entry is 0 if term doesn’t occur. Example: CALPURNIAdoesn’t occur in The tempest.5Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCY

WORSER. . .111

0

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

1

1

0

1

1

0

0

1

1

0

0

1

0

0

1

1

1

0

1

0

0

1

0Slide6

6

Incidence vectors

So we have a 0/1 vector for each term.

B

RUTUS AND

CAESAR AND NOT CALPURNIA:Take the vectors for BRUTUS, CAESAR AND NOT CALPURNIA Complement the vector of CALPURNIA Do a (bitwise) and on the three vectors110100 AND 110111 AND 101111 = 1001006Slide7

7

0/1 vector

for

B

RUTUS

7

Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .1110111111100

00

0

0

0

0

1

1

0

1

1

0

0

1

1

0

0

1

0

0

1

1

1

0

10010result:100100

Antony and Cleopatra

HamletSlide8

8

Too big to build the incidence matrix

M

= 500,000 × 10

6

= half a trillion 0s and 1s. (5000

)But the matrix has no more than 109 1s.Matrix is extremely sparse. (only 10/5000 has values)What is a better representations?We only record the 1s.8Consider N = 106 documents, each with about 1000 tokens ⇒ total of 109 tokens (10億)Assume there are M = 500,000 distinct terms in the collectionSlide9

9

Inverted Index

For each term

t

, we store a

list

of all documents that contain

t.9dictionary (sorted) postings Slide10

Ranked Retrieval

10Slide11

11

Problem with Boolean search

Boolean retrieval return documents

either match or

don't.

Boolean queries often result in either too few (=0) or too many (1000s) results.

Example

query : [standard user dlink 650] → 200,000 hits Example query : [standard user dlink 650 no card found] → 0 hitsGood for expert users with precise understanding of their needs and of the collection. Not good for the majority of users11Slide12

12

Ranked retrieval

With ranking, large result sets are not an issue.

More relevant results are ranked higher than less relevant results

.

The user may decide how many results he/she wants.

12Slide13

13

Scoring as the basis of ranked retrieval

Assign a score to each query-document pair, say in [0, 1], to measure how well document and query “match”.

If the query term does not occur in the document: score should be 0.

The more frequent the query term in the document, the higher the score

13Slide14

Term Frequency

14Slide15

15

Binary incidence

matrix

Each document is represented as a binary vector ∈ {0, 1}

|V|.

15

Anthony

and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .111011111110000000011

01100

1

1

0

0

1

0

0

1

1

1

0

1

0

0

1

0Slide16

16

Count matrix

Each document is now represented as a count vector ∈ N

|

V

|.

16

Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .157423205722731572271000000000310

220081

0

0

1

0

0

5

1

1

0

0

0

0

8

5Slide17

17

Here is Bag of words model

Do not consider the

order

of words in a document.

John is quicker than Mary , and

Mary is quicker than John

are represented the same way.17Slide18

18

Term frequency

tf

The term frequency

tf

t,d

of term

t in document d is defined as the number of times that t occurs in d.Use tf when computing query-document match scores.But Relevance does not increase proportionally with term frequency.Example A document with tf = 10 occurrences of the term is more relevant than a document with tf = 1 occurrence of the term, but not 10 times more relevant.18Slide19

19

Log frequency weighting

The log frequency weight of term t in d is defined as follows

tf

t,d

→ w

t,d

: 0 → 0, 1 → 1, 2 → 1.3, 10 → 2, 1000 → 4, etc.Why use log ? 在數量少時, 差1即差很多； 但隨著數量越多，差1的影響變得越小tf-matching-score(q, d) = t∈q∩d (1 + log tft,d )

19Slide20

20

Exercise

Compute the

tf

matching score for the following query-document pairs.

q: [information on cars] d: “all you have ever wanted to know

tf = 1+log1q: [information on cars] d: “information on trucks, information on planes, information on trains” tf = (1+log3) + (1+log3)20Slide21

TF-IDF Weighting

21Slide22

22

Desired weight for frequent terms

Frequent terms are less informative than rare terms.

Consider a term in the query that is

frequent

in the collection

(e.g.,

GOOD, INCREASE, LINE). → common term or 無鑑別力的詞22Slide23

23

Desired weight for rare terms

Consider a term in the query that is

rare

in the collection

(e.g.,

ARACHNOCENTRIC).A document containing this term is very likely to be relevant.→ We want high weights for rare terms like ARACHNOCENTRIC.23Slide24

24

Document frequency

We want

high weights for rare terms

like

ARACHNOCENTRIC

.

We want low (still positive) weights for frequent words like GOOD, INCREASE and LINE.We will use document frequency to factor this into computing the matching score.The document frequency is the number of documents in the collection that the term occurs in.24Slide25

25

idf weight

df

t

is the document frequency, the number of documents that

t

occurs in.dft is an inverse measure of the informativeness of term t.We define the idf weight of term t as follows: (N is the number of documents in the collection.)idft is a measure of the informativeness of the term.[log N/dft ] instead of [N/dft ] to balance the effect of idf

(i.e. use log for both tf and

df

)

25Slide26

26

Examples for

idf

Compute

idf

t

using the formula:26termdftidftcalpurniaanimalsundayflyunderthe

1

100

1000

10,000

100,000

1,000,000

6

4

3

2

1

0Slide27

27

Collection frequency

vs.

Document

frequency

Collection frequency of

t: number of tokens of t in the collectionDocument frequency of t: number of documents t occurs inDocument/collection frequency weighting is computed from known collection, or estimatedWhich word is a more informative ?27wordcollection frequencydocument frequencyINSURANCETRY1044010422

39978760Slide28

Example

cf 出現總次數 與 df 文件數。差異範例如下： Word

cf

df

29

tf-idf weighting

The

tf-idf

weight of a term is the

product of its

tf

weight and its idf weight.tf-weightidf-weightBest known weighting scheme in information retrievalNote: the “-” in tf-idf is a hyphen, not a minus signAlternative names: tf.idf , tf x idf29Slide30

30

Summary: tf-idf

Assign a

tf-idf

weight for each term t in each document

d

:

The tf-idf weight . . .. . . increases with the number of occurrences within a document. (term frequency). . . increases with the rarity of the term in the collection. (inverse document frequency)30Slide31

31

Exercise: Term, collection and document frequency

Relationship between

df

and

cf

?

Relationship between tf and cf?Relationship between tf and df?31QuantitySymbolDefinitionterm frequency document frequency collection frequency

tft,d

df

t

cf

t

number of occurrences of

t

in

d

number of documents in the

collection that

t

occurs in

total number of occurrences of

t

in

the

collectionSlide32

Vector Space Model

32Slide33

33

Binary incidence

matrix

Each document is represented as a binary vector ∈ {0, 1}

|

V

|.

33Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .11101111111000000

00110

1

1

0

0

1

1

0

0

1

0

0

1

1

1

0

1

0

0

1

0Slide34

34

Count matrix

Each document is now represented as a count vector ∈ N

|

V

|.

34

Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .1574232057227315722710000000003

10220

0

8

1

0

0

1

0

0

5

1

1

0

0

0

0

8

5Slide35

35

Binary → count →

weight

matrix

Each document is now represented as a real-valued vector of

tf

idf weights ∈ R|V|.35Anthony and CleopatraJulius Caesar The TempestHamlet Othello Macbeth . . .ANTHONYBRUTUS CAESARCALPURNIACLEOPATRAMERCYWORSER. . .5.251.218.590.02.85

1.511.373.186.102.54

1.54

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

1.90

0.11

0.0

1.0

1.51

0.0

0.0

0.12

4.15

0.0

0.0

0.25

0.0

0.05.250.250.350.00.00.00.00.881.95Slide36

36

Documents as

vectors

Each document is now represented as a real-valued vector of

tf-idf

weights ∈ R|V|.So we have a |V|-dimensional real-valued vector space.Terms are axes of the space.Documents are points or vectors in this space.Each vector is very sparse - most entries are zero.Very high-dimensional: tens of millions of dimensions when apply this to web (i.e. too many different terms on web)36Slide37

Vector Space Model

Vector Space Model

13.1x11.4 + 3.0 x 8.3

Applications of Vector Space Model

：挑選最相近的類別

Centroid

Issues about Vector Space Model (1)

，故稱完全不相似

D1=(1,0) D2=(0,1)

，其內積為0，但兩文件是否真的不相似？當詞為彼此有相依性 (dependence)挑出正交（不相依）的詞將維度進行數學轉換（找出正交軸）Slide41

Issues about Vector Space Model (2)

feature selection

the dimensionality is 7Slide42

42

Queries as

vectors

Do the same for queries: represent them as vectors in the high-dimensional space

Rank documents according to their proximity to

the

queryproximity = similarity ≈ negative distanceRank relevant documents higher than nonrelevant documents42Slide43

43

Rank documents according to angle with query

For example : take a document d and append it to itself. Call this document

d′

.

d′

is twice as long as d.“Semantically” d and d′ have the same content.The angle between the two documents is 0, corresponding to maximal similarity . . .. . . even though the Euclidean distance between the two documents can be quite large.43Slide44

44

From angles

to

cosines

The following two notions are equivalent.

Rank documents according to the angle between query and document in decreasing orderRank documents according to cosine(query,document) in increasing order44Slide45

45

Length normalization

A vector can be (length-) normalized by dividing each of its components by its length – here we use the

L

2

norm:

This maps vectors onto the unit sphere . . .

. . . since after normalization: As a result, longer documents and shorter documents have weights of the same order of magnitude.Effect on the two documents d and d′ (d appended to itself) : they have identical vectors after length-normalization.45Slide46

46

Cosine similarity between query and document

q

i

is the

tf-idf

weight of term

i in the query.di is the tf-idf weight of term i in the document.| | and | | are the lengths of and This is the cosine similarity of and . . . . . . or, equivalently, the cosine of the angle between and 46Slide47

47

Cosine similarity

illustrated

47Slide48

48

Cosine: Example

term frequencies (counts)

48

term

SaS

PaP

WHAFFECTIONJEALOUSGOSSIPWUTHERING1151020587002011638How similar are these novels? SaS: Sense and Sensibility 理性與感性 PaP:Pride and Prejudice

WH: Wuthering Heights

49

Cosine: Example

term

frequencies

(

counts) log frequency weighting (To simplify this example, we don't do idf weighting.)49termSaSPaPWHAFFECTIONJEALOUSGOSSIPWUTHERING3.062.01.300

2.761.850

0

2.30

2.04

1.78

2.58

term

SaS

PaP

WH

AFFECTION

JEALOUS

GOSSIP

WUTHERING

115

10

2

0

58

7

0

0

20

11638Slide50

50

Cosine: Example

log

frequency

weighting

log

frequency weighting & cosine normalization 50termSaSPaPWHAFFECTIONJEALOUSGOSSIPWUTHERING3.062.01.30

02.761.85

0

0

2.30

2.04

1.78

2.58

term

SaS

PaP

WH

AFFECTION

JEALOUS

GOSSIP

WUTHERING

0.789

0.515

0.335

0.0

0.832

0.555

0.0

0.0

0.5240.4650.4050.588cos(SaS,PaP

) ≈ 0.789 ∗ 0.832 + 0.515 ∗ 0.555 + 0.335 ∗ 0.0 + 0.0 ∗ 0.0 ≈ 0.94.

cos(

SaS,WH

) ≈ 0.79

cos(

PaP,WH

) ≈ 0.69

Why do we have

cos

(

SaS,PaP

) >

cos

(

SaS,WH

)?Slide51

51

Ranked retrieval in the Vector Space Model

Represent the query as a weighted

tf-idf

vector

Represent each document as a weighted

tf-idf

vectorCompute the cosine similarity between the query vector and each document vectorRank documents with respect to the queryReturn the top K (e.g., K = 10) to the user51Slide52

Conclusion

Ranking search results is important (compared with unordered Boolean results)Term frequencytf-idf ranking: best known traditional ranking schemeVector space model: One of the most important formal models for information retrieval (along with Boolean and probabilistic models)52