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

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

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

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.

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

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

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

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”.

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)

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

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*0.2 + 1*0.3 + 0*0.5 = 0.5 *Here and in the following, by zone we mean zone and fields.

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) 10 then scores[docID(p1)] ← WEIGHTEDZONE(p1, p2, g) 11 p1 ← next(p1) 12 p2 ← next(p2) 13 else if docID(p1) < docID(p2) 14 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.

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) φ1 37 linux 1 1 Relevant 1 φ2 37 penguin 0 1 Non-relevant 0 φ3 238 system 0 1 Relevant 1 φ4 238 penguin 0 0 Non-relevant 0 φ5 1741 kernel 1 1 Relevant 1 φ6 2094 driver 0 1 Relevant 1 φ7 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)

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 φ1 37 linux 1 1 1 1 0 0 φ2 37 penguin 0 1 1 – g 0 (1 – g)2 0.36 φ3 238 system 0 1 1 – g 1 g2 0.16 φ4 238 penguin 0 0 0 0 0 0 φ5 1741 kernel 1 1 1 1 0 0 φ6 2094 driver 0 1 1 – g 1 g2 0.16 φ7 3191 driver 1 0 g 0 g2 0.16 Tot_ ε Tot_ ε = 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

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 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_ ε : (n01r + n10n )g2 + (n10r + n01n)(1 – g)2 + n00r + n11n

Learning weights (simple machine learning) Want to minimize total error Tot_ ε = (n01r + n10n )g2 + (n10r + n01n)(1 – g)2 + 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 )

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?

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.

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).

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

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

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| !

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.

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

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.

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.

Document frequency Rare terms are more informative than frequent terms Sec. 6.2.1 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.

Document frequency, continued Sec. 6.2.1 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.

Sec. 6.2.1 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.

idf example, suppose N = 1 million Sec. 6.2.1 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 6 4 3 2 1 0 There is one idf value for each term t in a collection.

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.

Collection vs. Document frequency Sec. 6.2.1 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.

Sec. 6.2.2 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

Final ranking of documents for a query Sec. 6.2.2 Final ranking of documents for a query

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|

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.

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.

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

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.

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.

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.

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]

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

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

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

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

Cosine similarity illustrated

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.

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 × 0.832 + 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)? SaS and PaP were both written by Jane Austen, WH by Emily Brontë.

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)

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)

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!

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?

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?

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+0+0.27+0.53 = 0.8

|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 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( ∑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.

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 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?

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?

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

Resources for today’s lecture Ch. 6 Resources for today’s lecture IIR 6.2 – 6.4.3 http://www.miislita.com/information-retrieval-tutorial/cosine-similarity-tutorial.html Term weighting and cosine similarity tutorial for SEO folk!