1. Information Retrieval and Data Mining (AT71. 07) Comp. Sc. and Inf
Information Retrieval and Data Mining (AT71.07) Comp. Sc. and Inf. Mgmt. Asian Institute of Technology
Instructor: Prof. Sumanta Guha Slide Sources: Introduction to Information Retrieval book slides from Stanford University, adapted and supplemented Chapter 6: Scoring, term weighting, and the vector space model

2. CS276: Information Retrieval and Web Search
Christopher Manning and Prabhakar Raghavan
Lecture 6: Scoring, term weighting, and the vector space model

3. This lecture; IIR Sections 6.2-6.4.3
Ranked retrieval
Scoring documents
Term frequency
Collection statistics
Weighting schemes
Vector space scoring

4. Ranked retrieval Thus far, our queries have all been Boolean.
Ch. 6
Ranked retrieval
Thus far, our queries have all been Boolean.
Documents either match or don’t.
Good for expert users with precise understanding of their needs and the collection.
Also good for applications: Applications can easily consume 1000s of results.
Not good for the majority of users.
Most users incapable of writing Boolean queries (or they are, but they think it’s too much work).
Most users don’t want to wade through 1000s of results.
This is particularly true of web search.

5. Problem with Boolean search: feast or famine
Boolean queries often result in either too few (=0) or too many (1000s) results.
Query 1: “standard user dlink 650” → 200,000 hits
Query 2: “standard user dlink 650 no card found”: 0 hits
It takes a lot of skill to come up with a query that produces a manageable number of hits.
AND gives too few; OR gives too many
Cf. our discussion of how Westlaw Boolean queries didn’t actually outperform free text querying

6. Ranked retrieval models
Rather than a set of documents satisfying a query expression, in ranked retrieval models, the system returns an ordering over the (top) documents in the collection with respect to a query
Free text queries: Rather than a query language of operators and expressions, the user’s query is just one or more words in a human language
In principle, there are two separate choices here, but in practice, ranked retrieval models have normally been associated with free text queries and vice versa

7. Feast or famine: not a problem in ranked retrieval
Ch. 6
Feast or famine: not a problem in ranked retrieval
When a system produces a ranked result set, large result sets are not an issue
Indeed, the size of the result set is not an issue
We just show the top k ( ≈ 10) results
We don’t overwhelm the user
Premise: the ranking algorithm works

8. Scoring as the basis of ranked retrieval
Ch. 6
Scoring as the basis of ranked retrieval
We wish to return in order the documents most likely to be useful to the searcher
How can we rank-order the documents in the collection with respect to a query?
Assign a score – say in [0, 1] – to each document
This score measures how well document and query “match”.

9. Parametric and zone indexes
Consider query: “find docs authored by Shakespeare in 1601 containing the phrase alas poor Yorick”
Fields can have well-defined set of values, e.g., numeric or character strings of fixed max length.
date field
author field
Parametric search interface to enter parameter values (=field values)

10. Parametric and zone indexes
Zone are similar to fields, except contents of a zone can be arbitrary free text. E.g., title zone, abstract zone, body zone, …
Indexes for fields and zones can be
Separate (drawback larger dictionary):
Combined (drawback larger postings, but dictionary is not enlarged – preferable to get a compact dictionary, e.g., to fit in main):
william.author 11
177
244
255
william.title 11
134
244
255
william.body 4
134
213
255
william
4.body
11.author.title
134.title.body

11. Weighted zone scoring
Given a Boolean query q and doc d, weighted zone* scoring assigns each pair (q, d) a score between 0 and 1 by computing a linear combination of zone scores.
Suppose docs each have l zones.
Let g1, g2, …, gl be the respective zone weights, s.t., ∑i=1..l gi = 1.
Let s1, s2, …, sl be a Boolean score for the respective zones, being 1/0 according as q occurs/does not occur in the i th zone of doc d.
Then, weighted zone score is
∑i=1..l gisi
E.g., for indexes of previous slide, if zone weights of author, title and body are, resp., g1 = 0.2, g2 = 0.3, g3 = 0.5, then
WEIGHTEDZONE(“william”, 11) = 1* * *0.5 = 0.5
*Here and in the following, by zone we mean zone and fields.

12. Weighted zone scoring
Algorithm to compute weighted zone score from two postings lists given an AND query.
ZONESCORE(q1, q2) // Score for query “q1 AND q2”
// Boolean score 1 if both queries present in zone; 0, otherwise
1 float scores[N] = [0]
2 constant g[ℓ]
3 p1 ← postings(q1)
4 p2 ← postings(q2)
5 // scores[] is an array with a score entry for each document, initialized to zero.
6 //p1 and p2 are initialized to point to the beginning of their respective postings.
7 //Assume g[] is initialized to the respective zone weights.
8 while p1 ≠ NIL AND p2 ≠ NIL
9 do if docID(p1) == docID(p2)
then scores[docID(p1)] ← WEIGHTEDZONE(p1, p2, g)
p1 ← next(p1)
12 p2 ← next(p2)
13 else if docID(p1) < docID(p2)
then p1 ← next(p1)
15 else p2 ← next(p2)
16 return scores
∑i=1..l gisi, where si is 1 if both queries
present in zone i; 0, otherwise.

13. Learning weights (simple machine learning)
Assume only two zones title (T) and body (B) with zone weights g and 1 – g, respectively.
Example DocID d Query sT sB Judgment (human expert) r (quantized judgment)
φ linux Relevant
φ penguin Non-relevant
φ system Relevant
φ penguin Non-relevant
φ kernel Relevant
φ driver Relevant
φ driver Non-relevant
Training examples
sT sB Score
– g g
Four possible combinations of sT and sB and the corresponding
score(d, q) = g * sT(d, q) + (1 – g) * sB(d, q)

14. Learning weights (simple machine learning)
Squared error of the scoring function with weight g on example φ is
ε(g, φ) = ( r(d, q) – score(d, q) )2
Example d Query sT sB Score r ε ε assuming g = 0.4
φ linux
φ penguin – g (1 – g)
φ system – g g
φ penguin
φ kernel
φ driver – g g
φ driver g g
Tot_ ε Tot_ ε = 0.84
sT sB Score Score assuming g = 0.4
– g g

15. Learning weights (simple machine learning)
Total error of a set of training examples Tot_ ε = ∑j ε(g, φj) = ∑j ( r(dj, q) – score(dj, q) )2
Goal is to choose g to minimize the total error.
Note: Our example has only two zones with weights g and 1 – g, respectively! Generally, there will be l zones with weights g1, …, gl. Same principles!
sT sB Score r No ε
n00n
n00r
– g n01n (1 – g)2
– g n01r g2
g n10n g2
g n10r (1 – g)2
n11n
n11r
Total error Tot_ ε : (n01r + n10n )g2 + (n10r + n01n)(1 – g)2 + n00r + n11n

16. Learning weights (simple machine learning)
Want to minimize total error Tot_ ε = (n01r + n10n )g2 + (n10r + n01n)(1 – g) n00r + n11n
Differentiating w.r.t. g: d(Tot_ ε)/dg
= 2(n01r + n10n )g – 2(n10r + n01n)(1 – g)
Find minimum by solving:
2(n01r + n10n )g – 2(n10r + n01n)(1 – g) = 0
→ (n10r + n10n + n01r + n01n )g = n10r + n01n
→ g = (n10r + n01n )/ (n10r + n10n + n01r + n01n )

17. Exercises
Exercise 6.1: When using weighted scoring, is it necessary for all zones to use the same match function?
Exercise 6.2: If author, title and body zones have weights g1 = 0.2, g2 = 0.31 and g3 = 0.49, what are all the distinct score values a doc may get?
Exercise 6.3: Rewrite the algorithm in Fig. 6.4 to the case of more than two queries, viz., q1 AND q2 AND … AND qm.
Exercise 6.4: Write pseudocode for the function WeightedZone for the case of two postings lists in Fig. 6.4.
Exercise 6.5: Apply Eq. 6.6 to the sample training set in Fig. 6.5 to estimate the best value of g for this example.
Exercise 6.6: For the value of g estimated in Ex. 6.5, compute the weighted zone score of each (query, doc) example. How do these scores relate to the relevance judgments of Fig. 6.4 (quantized to 0/1)?
Exercise 6.7: Why does the expression for g in Eq. 6.6 not involve the training examples in which sT(d, q) and sB(d, q) have the same value?

18. Query-document matching scores
We need a way of assigning a score to a query/document pair
Let’s start with a one-term query
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 (should be)
We will look at a number of alternatives for this.

19. Take 1: Jaccard coefficient
Ch. 6
Take 1: Jaccard coefficient
A commonly used measure of overlap of two sets A and B:
jaccard(A,B) = |A ∩ B| / |A ∪ B|
jaccard(A, A) = 1
jaccard(A, B) = 0 if A ∩ B = 0
jaccard(B, A) = jaccard(A, B)
A and B don’t have to be the same size.
Always assigns a number between 0 and 1.
Exercise: A = {1, 2, 3, 4}, B = {1, 2, 4}, C = {1, 2, 4, 5}.
Calculate jaccard(A, B), jaccard(B, C), jaccard(A, C).

20. Jaccard coefficient: Scoring example
Ch. 6
Jaccard coefficient: Scoring example
Exercise: What is the query-document match score that the Jaccard coefficient computes for each of the two documents below?
Query: ides of march
Document 1: caesar died in march
Document 2: the long march

21. Issues with Jaccard for scoring
Ch. 6
Issues with Jaccard for scoring
It doesn’t consider term frequency (how many times a term occurs in a document)
Rare terms in a collection are more informative than frequent terms. Jaccard doesn’t consider this information
We need a more sophisticated way of normalizing for length

22. Recall (Lecture 1): Binary term-document incidence matrix
Sec. 6.2
Recall (Lecture 1): Binary term-document incidence matrix
Each document is represented by a binary vector ∈ {0,1}|V| !

23. Term-document count matrices
Sec. 6.2
Term-document count matrices
Consider the number of occurrences of a term in a document*:
Each document is represented by a count vector ∈ ℕ|V| !
*Recall from Ch. 1 that term frequency can be stored with a document in the inverted index.

24. Bag of words model
Vector representation doesn’t consider the ordering of words in a document
John is quicker than Mary and Mary is quicker than John have the same vectors
This is called the bag of words model.
In a sense, this is a step back: The positional index was able to distinguish these two documents.
We will look at “recovering” positional information later in this course.
For now: bag of words model

25. Term frequency tf
The term frequency tft,d of term t in document d is defined as the number of times that t occurs in d.
We want to use tf when computing query-document match scores. But how?
Raw term frequency is not what we want:
A document with 10 occurrences of the term is more relevant than a document with 1 occurrence of the term.
But not 10 times more relevant!
Relevance does not increase proportionally with term frequency.

26. Log-frequency weighting
Sec. 6.2
Log-frequency weighting
The log frequency weight of term t in d is
0 → 0, 1 → 1, 2 → 1.3, 10 → 2, 1000 → 4, etc.
Score for a document-query pair: sum over terms t in both q and d:
score
The score is 0 if none of the query terms is present in the document.

27. Document frequency Rare terms are more informative than frequent terms
Sec
Document frequency
Rare terms are more informative than frequent terms
Recall stop words
Consider a term in the query that is rare in the collection (e.g., capricious) vs. a term that is frequent (e.g., person)
A document containing this term “capricious” is very likely to be relevant to a query containing “capricious”
→ We want a high weight for rare terms like capricious.

28. Document frequency, continued
Sec
Document frequency, continued
Frequent terms are less informative than rare terms
Consider a query term that is frequent in the collection (e.g., high, increase, line)
A document containing such a term is more likely to be relevant than a document that doesn’t
But it’s not a sure indicator of relevance.
→ For frequent terms, we want high positive weights for words like high, increase, and line
But lower weights than for rare terms.
We will use document frequency (df) to capture this.

29. Sec
idf weight
dft is the document frequency of t: the number of documents that contain t
dft is an inverse measure of the informativeness of t
dft  N
We define the idf (inverse document frequency) of t by
We use log (N/dft) instead of N/dft to “dampen” the effect of idf.
Will turn out the base of the log is immaterial.

30. idf example, suppose N = 1 million
Sec
idf example, suppose N = 1 million
term
dft
idft
calpurnia 1 6
animal
100 4
sunday
1,000 3
fly
10,000 2
under
100,000
the
1,000,000
There is one idf value for each term t in a collection.

31. Effect of idf on ranking
Question: Does idf have an effect on ranking for one-term queries, like
iPhone
idf has no effect on ranking one term queries
idf affects the ranking of documents for queries with at least two terms
For the query capricious person, idf weighting makes occurrences of capricious count for much more in the final document ranking than occurrences of person.

32. Collection vs. Document frequency
Sec
Collection vs. Document frequency
The collection frequency of t is the number of occurrences of t in the collection, counting multiple occurrences.
Example:
Which word is a better search term (and should get a higher weight)?
Word
Collection frequency
Document frequency
insurance
10440
3997
try
10422
8760
Why do you get these numbers?
Suggests df is better.

33. Sec
tf-idf weighting
The tf-idf weight of a term is the product of its tf weight and its idf weight.
Best known weighting scheme in information retrieval
Note: the “-” in tf-idf is a hyphen, not a minus sign!
Alternative names: tf.idf, tf x idf
Increases with the number of occurrences within a document
Increases with the rarity of the term in the collection

34. Final ranking of documents for a query
Sec
Final ranking of documents for a query

35. Binary → count → weight matrix
Sec. 6.3
Binary → count → weight matrix
Each document is now represented by a real-valued vector of tf-idf weights ∈ R|V|

36. Exercise 6.8: Why is the idf of a term always finite?
Exercise 6.9: What is the idf of a term that occurs in every document? Compare this with the use of stop word lists.
Exercise 6.10: Consider the table of term frequencies for 3 docs:
term dft idft Doc Doc Doc3
car ,
auto ,
insurance ,
best ,
Compute the tf-idf weights for the terms car, auto, insurance and best for each doc using the given idf values.
Exercise 6.11: Can the tf-idf weight of a term in a doc exceed 1?
Exercise 6.12: How does the base of the logarithm in affect the score calculation by ? How does it affect the relative scores of two docs on a given query?
Exercise 6.13: If the logarithm in is computed base 2, suggest a simple approximation of the idf of a term.

37. Documents as vectors So we have a |V|-dimensional vector space
Sec. 6.3
Documents as vectors
So we have a |V|-dimensional vector space
Terms are axes of the space
Documents are points or vectors in this space
Very high-dimensional: tens of millions of dimensions when you apply this to a web search engine
These are very sparse vectors – most entries are zero.

38. Sec. 6.3
Queries as vectors
Key idea 1: Do the same for queries: represent them as vectors in the space
Key idea 2: Rank documents according to their proximity to the query in this space
proximity = similarity of vectors
proximity ≈ inverse of distance
Recall: We do this because we want to get away from the you’re-either-in-or-out Boolean model.
Instead: rank more relevant documents higher than less relevant documents

39. Formalizing vector space proximity
Sec. 6.3
Formalizing vector space proximity
First cut: distance between two points
( = distance between the end points of the two vectors)
Euclidean distance?
Euclidean distance is a bad idea . . .
. . . because Euclidean distance is large for vectors of different lengths.

40. Why distance is a bad idea
Sec. 6.3
Why distance is a bad idea
The Euclidean distance between q
and d2 is large even though the
distribution of terms in the query q and the distribution of
terms in the document d2 are
very similar.

41. Use angle instead of distance
Sec. 6.3
Use angle instead of distance
Thought experiment: take a document d and append it to itself. Call this document d′.
“Semantically” d and d′ have the same content
The Euclidean distance between the two documents can be quite large
The angle between the two documents is 0, corresponding to maximal similarity.
Key idea: Rank documents according to angle with query.

42. From angles to cosines The following two notions are equivalent.
Sec. 6.3
From angles to cosines
The following two notions are equivalent.
Rank documents in decreasing order of the angle between query and document
Rank documents in increasing order of cosine(query,document)
Cosine is a monotonically decreasing function for the interval [0o, 180o]

43. Sec. 6.3
From angles to cosines
But how – and why – should we be computing cosines?

44. Sec. 6.3
Length normalization
A vector can be (length-) normalized by dividing each of its components by its length – for this we use the L2 norm:
Dividing a vector by its L2 norm makes it a unit (length) vector (on surface of unit hypersphere)
Effect on the two documents d and d′ (d appended to itself) from earlier slide: they have identical vectors after length-normalization.
Long and short documents now have comparable weights

45. cosine(query,document)
Sec. 6.3
cosine(query,document)
Dot product
Unit vectors
qi is the tf-idf weight of term i in the query
di is the tf-idf weight of term i in the document
cos(q,d) is the cosine similarity of q and d … or,
equivalently, the cosine of the angle between q and d.
See Law of Cosines (Cosine Rule) wikipedia page

46. Cosine for length-normalized vectors
For length-normalized vectors, cosine similarity is simply the dot product (or scalar product):
for q, d length-normalized.

47. Cosine similarity illustrated

48. Cosine similarity amongst 3 documents
Sec. 6.3
Cosine similarity amongst 3 documents
How similar are
the novels
SaS: Sense and
Sensibility
PaP: Pride and
Prejudice, and
WH: Wuthering
Heights?
term
SaS
PaP WH
affection
115 58 20
jealous 10 7 11
gossip 2 6
wuthering 38
Term frequencies (counts)
Note: To simplify this example, we don’t do idf weighting.

49. 3 documents example contd.
Sec. 6.3
3 documents example contd.
Log frequency weighting
After length normalization
term
SaS
PaP WH
affection
3.06
2.76
2.30
jealous
2.00
1.85
2.04
gossip
1.30
1.78
wuthering
2.58
term
SaS
PaP WH
affection
0.789
0.832
0.524
jealous
0.515
0.555
0.465
gossip
0.335
0.405
wuthering
0.588
cos(SaS,PaP) ≈
0.789 × × × × 0.0
≈ 0.94
cos(SaS,WH) ≈ 0.79
cos(PaP,WH) ≈ 0.69
Why do we have cos(SaS,PaP) > cos(SAS,WH)?
SaS and PaP were both written by Jane Austen, WH by Emily Brontë.

50. Computing cosines
Suppose the terms in the query q are t1, t2, …, tn. E.g., if q = “cat dog cat rat”, then n =3 and t1 = cat, t2 = dog, t3 = rat. Suppose the document is d.
Now, q = (0, …, 0, wt1,q, 0, …, 0, wt2,q, …, 0, …, 0, wtn,q , 0, …, 0)
d = (x, …, x, wt1,d, x, …, x, wt2,d, …, x, …, x, wtn,d , x, …, x)
where 0’s are terms not in q and x’s are corresponding values in d.
Now, q d = wt1,q x wt1,d + wt2,q x wt2,d + … + wtn,q x wtn,d
and
cos(q, d) = wt1,q x wt1,d + wt2,q x wt2,d + … + wtn,q x wtn,d
|q| x |d|
where |q| = Length(q) = (w2t1,q+ w2t2,q+ …+ w2tn,q ) and
|d| = Length(d) = (Σt runs over all terms in d w2t,d)

51. Computing cosines example
E.g., if q = “cat dog cat rat”, then n =3 and t1 = cat, t2 = dog, t3 = rat. Suppose the document is d = “cat cat cat dog goat dog horse”.
Then,
cos(q, d) = wcat,q x wcat,d + wdog,q x wdog,d + wrat,q x wrat,d
|q| x |d|
where |q| = Length(q) = (w2cat,q + w2dog,q + w2rat,q ) and
|d| = Length(d) = (w2cat,d + w2dog,d + w2goat,d + w2horse,d)

52. Computing cosine scores
Sec. 6.3
Computing cosine scores
Can traverse posting lists one
term at time – which is called
term-at-a-time scoring. Or can
traverse them concurrently
as in the INTERSECT algorithm
of Ch. 1 – which is called
document-at-a-time scoring
No need to divide by Length[q] as it is the same for all docs!
No need to store these per doc per posting list. Can be computed on-the-fly from the dft value at the head of the postings list and the tft,d value in the doc.
Priority queue=heap!

53. tf-idf weighting has many variants
Sec. 6.4
tf-idf weighting has many variants
n default is just term frequency
ltc is best known form of weighting
Why is the base of the log in idf immaterial?

54. Weighting may differ in queries vs documents
Sec. 6.4
Weighting may differ in queries vs documents
Many search engines allow for different weightings for queries vs. documents
SMART Notation: denotes the combination in use in an engine, with the notation ddd.qqq, using the acronyms from the previous table
A very standard weighting scheme is: lnc.ltc
Document: logarithmic tf (l first character), no idf (n second character), cosine normalization…
Query: logarithmic tf (l first character), idf (t second character), cosine normalization …
Leaving off idf weighting on documents is good for both efficiency and system effectiveness reasons.
A bad idea?

55. tf-idf example: lnc.ltc
Sec. 6.4
tf-idf example: lnc.ltc
Document: car insurance auto insurance
Query: best car insurance
Term
Query
Document
Prod
tf-raw
tf-wt df
idf wt
n’lize
auto
5000
2.3 1
0.52
best
50000
1.3
0.34
car
10000
2.0
0.27
insurance
1000
3.0
0.78 2
0.68
0.53
N, the number of docs = 1,000,000
Doc length =
Score = = 0.8

56. |x – y| = sqrt( ∑i=1..m (xi – yi)2 )
Exercise 6.14: If we were to stem jealous and jealousy to a common stem before setting up the vector space, detail how the definitions of tf and idf should be modified.
Exercise 6.15: Recall the tf-idf weights computed in Exercise Compute the Euclidean normalized document vectors for each of the docs, where each has four components, one for each of the four terms.
Exercise 6.16: Verify that the sum of the squares of the components of each of the document vectors in Exercise 6.15 is 1 (to within rounding error). Why?
Exercise 6.17: With term weights as computed in Exercise 6.15, rank the three documents by computed score for the query car insurance for each of the following cases of term weight in the query:
The weight of the term is 1 if present in the query, 0 otherwise.
Euclidean normalized idf.
Exercise 6.18: One measure of the similarity of two vectors x and y is the Euclidean (or L2) distance between them:
|x – y| = sqrt( ∑i=1..m (xi – yi)2 )
Given a query q and docs d1, d2, …, we may rank the docs di in order of increasing distance from q. Show that if q and the di are all normalized to unit vectors, then the rank ordering produced by Euclidean distance is identical to that produced by cosine similarity.

57. Exercise 6.19: Compare the vector space similarity between the query “digital cameras” and the document “digital cameras and video cameras” by filling out the empty columns in the table below.
query document
word tf wf df idf qi=wf-idf tf wf di=normalized wf qi · di
digital ,000
video ,000
cameras ,000
Assume N = 10,000,000, logarithmic term weighting (wf columns) for query and doc, idf weighting for the query only and cosine normalization for the doc only. Treat and as a stop word. Enter term counts in the tf columns. What is the final similarity score?

58. Exercise 6.20: Show that for the query affection, the relative ordering of the scores of the three docs in the table below is the reverse of the ordering for the query jealous gossip.
term SaS PaP WH
affection
jealous
gossip
Exercise 6.21: In turning a query into a unit vector in the table above, we assigned equal weights to each of the query terms. What other principled approaches are plausible?
Exercise 6.22: Consider the case of a query term that is not in the set of M indexed terms.; thus, our standard construction of the query vector results in V(q) not being in the vector space created from the collection. How would one adapt the vector space construction to handle this case?
Exercise 6.23: Refer to the tf and idf values for the four terms and three docs in Exercise Compute the two top-scoring docs on the query best car insurance for each of the following weighting schemes (i) nnn.atc (ii) ntc.atc.
Exercise 6.24: Suppose the word coyote does not occur in the collection used in Exercises 6.10 and How would one compute ntc.atc scores for the query coyote insurance?

59. Summary – vector space ranking
Represent the query as a weighted tf-idf vector
Represent each document as a weighted tf-idf vector
Compute the cosine similarity score for the query vector and each document vector
Rank documents with respect to the query by score
Return the top K (e.g., K = 10) to the user

60. Resources for today’s lecture
Ch. 6
Resources for today’s lecture
IIR 6.2 – 6.4.3
Term weighting and cosine similarity tutorial for SEO folk!