Information Retrieval and Data Mining (AT71.07) Comp. Sc. and Inf. Mgmt. Asian Institute of Technology

1
Information Retrieval and Data Mining
(AT71.07)Comp. Sc. and Inf. Mgmt.Asian
Institute of Technology

• Instructor Dr. 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

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

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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
Ch. 6

• Thus far, our queries have all been Boolean.
• Documents either match or dont.
• 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 its too much work).
• Most users dont want to wade through 1000s of
  results.
• This is particularly true of web search.

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Problem with Boolean searchfeast or famine
Ch. 6

• 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

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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 users 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

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Feast or famine not a problem in ranked retrieval
Ch. 6

• 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 dont overwhelm the user
• Premise the ranking algorithm works

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Scoring as the basis of ranked retrieval
Ch. 6

• 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.

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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)
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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
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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., ?i1..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
• ?i1..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) 10.2 10.3
  00.5 0.5

Here and in the following, by zone we mean zone
and fields.
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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 scoresN 0
• 2 constant gl
• 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)
• 10 then scoresdocID(p1) ? WEIGHTEDZONE(p1,
  p2, g)
• 11 p1 ? next(p1)
• 12 p2 ? next(p2)
• 13 else if docID(p1) lt docID(p2)
• 14 then p1 ? next(p1)
• 15 else p2 ? next(p2)
• 16 return scores

?i1..l gisi, where si is 1 if both
queries present in zone i 0, otherwise.
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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)
• f1 37 linux 1
  1 Relevant
  1
• f2 37 penguin 0
  1 Non-relevant
  0
• f3 238 system 0
  1 Relevant
  1
• f4 238 penguin 0
  0 Non-relevant
  0
• f5 1741 kernel 1
  1 Relevant
  1
• f6 2094 driver 0
  1 Relevant
  1
• f7 3191 driver 1
  0 Non-relevant
  0
• Training examples
• sT sB Score
• 0 0 0
• 0 1 1 g
• 1 0 g
• 1 1 1
• Four possible combinations of sT and sB and the
  corresponding
• score(d, q) g sT(d, q) (1 g)
  sB(d, q)

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Learning weights (simple machine learning)

• Squared error of the scoring function with weight
  g on example f is
• e(g, f) ( r(d, q) score(d, q) )2
• Example d Query sT sB
  Score r e
  e assuming g 0.4
• f1 37 linux 1
  1 1 1 0
  0
• f2 37 penguin 0 1
  1 g 0 (1 g)2
  0.36
• f3 238 system 0 1
  1 g 1 g2
  0.16
• f4 238 penguin 0 0
  0 0 0
  0
• f5 1741 kernel 1 1
  1 1 0
  0
• f6 2094 driver 0 1
  1 g 1 g2
  0.16
• f7 3191 driver 1 0
  g 0 g2
  
  0.16
• 
  
  Tot_ e Tot_ e 0.84
• sT sB Score Score
  assuming g 0.4
• 0 0 0
  0
• 0 1 1 g
  0.6
• 1 0 g
  0.4
• 1 1 1
  1

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Learning weights (simple machine learning)

• Total error of a set of training examples Tot_ e
  ?j e(g, fj) ?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.
  e
• 0 0 0 0
  n00n 0
• 0 0 0 1
  n00r 1
• 0 1 1 g 0 n01n
  (1 g)2
• 0 1 1 g 1 n01r
  g2
• 1 0 g 0
  n10n g2
• 1 0 g 1
  n10r (1 g)2
• 1 1 1 0
  n11n 1
• 1 1 1 1
  n11r 0
• 
  Total error Tot_ e (n01r n10n )g2
  (n10r n01n)(1 g)2 n00r n11n

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Learning weights (simple machine learning)

• Want to minimize total error Tot_ e (n01r
  n10n )g2 (n10r n01n)(1 g)2
  n00r n11n
• Differentiating w.r.t. g d(Tot_ e)/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 )

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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?

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Query-document matching scores
Ch. 6

• We need a way of assigning a score to a
  query/document pair
• Lets 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.

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Take 1 Jaccard coefficient
Ch. 6

• A commonly used measure of overlap of two sets A
  and B
• jaccard(A,B) A n B / A ? B
• jaccard(A, A) 1
• jaccard(A, B) 0 if A n B 0
• jaccard(B, A) jaccard(A, B)
• A and B dont 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).

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Jaccard coefficient Scoring example
Ch. 6

• 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

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Issues with Jaccard for scoring
Ch. 6

• It doesnt consider term frequency (how many
  times a term occurs in a document)
• Rare terms in a collection are more informative
  than frequent terms. Jaccard doesnt consider
  this information
• We need a more sophisticated way of normalizing
  for length

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Recall (Lecture 1) Binary term-document
incidence matrix
Sec. 6.2
Each document is represented by a binary vector ?
0,1V!
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Term-document count matrices
Sec. 6.2

• Consider the number of occurrences of a term in a
  document

Each document is represented by a count vector ?
NV!
Recall from Ch. 1 that term frequency can be
stored with a document in the inverted index.
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Bag of words model

• Vector representation doesnt 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

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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.

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Log-frequency weighting
Sec. 6.2

• 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.

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Document frequency
Sec. 6.2.1

• 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.

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Document frequency, continued
Sec. 6.2.1

• 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 doesnt
• But its 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.

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idf weight
Sec. 6.2.1

• 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.
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idf example, suppose N 1 million
Sec. 6.2.1
term dft idft
calpurnia 1 6
animal 100 4
sunday 1,000 3
fly 10,000 2
under 100,000 1
the 1,000,000 0
There is one idf value for each term t in a
collection.
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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.

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Collection vs. Document frequency
Sec. 6.2.1

• 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
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tf-idf weighting
Sec. 6.2.2

• 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

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Final ranking of documents for a query
Sec. 6.2.2
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Binary ? count ? weight matrix
Sec. 6.3
Each document is now represented by a real-valued
vector of tf-idf weights ? RV
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• 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 Doc1 Doc2 Doc3
• car 18,165 1.65
  27 4 24
• auto 6,723 2.08
  3 33 0
• insurance 19,241 1.62
  0 33 29
• best 25,235 1.50
  14 0 17
• 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.

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Documents as vectors
Sec. 6.3

• 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.

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Queries as vectors
Sec. 6.3

• 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 youre-either-in-or-out Boolean model.
• Instead rank more relevant documents higher than
  less relevant documents

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Formalizing vector space proximity
Sec. 6.3

• 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.

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Why distance is a bad idea
Sec. 6.3

• 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.

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Use angle instead of distance
Sec. 6.3

• 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.

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From angles to cosines
Sec. 6.3

• 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

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From angles to cosines
Sec. 6.3

• But how and why should we be computing
  cosines?

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Length normalization
Sec. 6.3

• 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

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cosine(query,document)
Sec. 6.3
Dot product
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.
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Cosine for length-normalized vectors

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

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Cosine similarity illustrated
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Cosine similarity amongst 3 documents
Sec. 6.3

• 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 0 6
wuthering 0 0 38
Term frequencies (counts)
Note To simplify this example, we dont do idf
weighting.
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3 documents example contd.
Sec. 6.3

• 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 0 1.78
wuthering 0 0 2.58
term SaS PaP WH
affection 0.789 0.832 0.524
jealous 0.515 0.555 0.465
gossip 0.335 0 0.405
wuthering 0 0 0.588
cos(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) gt cos(SAS,WH)? SaS
and PaP were both written by Jane Austen, WH by
Emily Brontë.
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Computing cosine scores
Sec. 6.3
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 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 queueheap!
51
tf-idf weighting has many variants
Sec. 6.4
Why is the base of the log in idf immaterial?
52
Weighting may differ in queries vs documents
Sec. 6.4

• 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

A bad idea?
53
tf-idf example lnc.ltc
Sec. 6.4
Document car insurance auto insurance Query
best car insurance
Term Query Query Query Query Query Query Document Document Document Document Prod
tf-raw tf-wt df idf wt nlize tf-raw tf-wt wt nlize
auto 0 0 5000 2.3 0 0 1 1 1 0.52 0
best 1 1 50000 1.3 1.3 0.34 0 0 0 0 0
car 1 1 10000 2.0 2.0 0.52 1 1 1 0.52 0.27
insurance 1 1 1000 3.0 3.0 0.78 2 1.3 1.3 0.68 0.53
N, the number of docs 1,000,000
Score 000.270.53 0.8
54

• 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 6.10. 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( ?i1..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.

55

• 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 qiwf-idf tf wf
  dinormalized wf qi di
• digital 10,000
• video 100,000
• cameras 50,000
• 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?

56

• 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 0.996 0.993
  0.847
• jealous 0.087
  0.120 0.466
• gossip 0.017 0
  0.254
• 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 6.10.
  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 6.23. How would one compute ntc.atc scores
  for the query coyote insurance?

57
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

58
Resources for todays lecture
Ch. 6

• IIR 6.2 6.4.3
• http//www.miislita.com/information-retrieval-tuto
  rial/cosine-similarity-tutorial.html
• Term weighting and cosine similarity tutorial for
  SEO folk!