Information Retrieval 02

Introduction to Information RetrievalIntroduction to Information Retrieval
Introduction to
Information Retrieval
Scoring, Term Weighting and the Vector Space Model
Slides from Christopher D. Manning, Prabhakar Raghavan and Hinrich Schütze
1Department of CSE,TSSOT,AUS, SILCHAR11/28/17

Outlines
 Ranked retrieval
 Scoring documents
 Term frequency
 Collection statistics
 Weighting schemes
 Vector space scoring

Boolean Retrieval - AND/OR/NOT
A B
All documents
C
3
Department of CSE,TSSOT,AUS, SILCHAR11/28/17

Boolean Retrieval
 Weights assigned to terms are either “0” or “1”
 “0” represents “absence”: term isn’t in the document
 “1” represents “presence”: term is in the document
 Build queries by combining terms with Boolean
operators
 AND, OR, NOT
 The system returns all documents that satisfy the
query

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

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
Today’s question: Given a large list of documents that
match a query, how to order them according to their
relevance?
Answer: Scoring the documents
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”.
Let us say we have a one-term query – How will you
Compute the score ?

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)
What are the two issues with the proposed approach

Ranking based on raw term frequency - issues
 All terms have equal weight
 Larger/Bigger documents have more terms and thus
the score is larger
 N times more frequency/occurrence does not mean
N times more relevant

Jaccard coefficient
 A commonly used measure of overlap of two sets A
and B
 What is jaccard(A,B) =
 jaccard(A,B) = |A ∩ B| / |A ∪ B|
 jaccard(A,A) = 1
 jaccard(A,B) = 0 if A B =∩ 0
 A and B don’t have to be the same size.
Always assigns a number between X and Y and why?
Always assigns a number between 0 and 1 because
cause at most the intersection can be as large as the union.

Jaccard coefficient: Scoring example
 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

Jaccard Example
 Query: ides of march
 Document 1: caesar died in march
 Document 2: the long march
 A = {ides, of, march}
 B1 = {caesar, died, in, march}
 B2 = {the, long, march}
6/11/1
},,,,,{1
}{1
=
=
=
BABA
marchofinidesdiedcaesarBA
marchBA



AI B2 ={march}
AUB2 ={ides,long,march,of ,the}
AI B2/ AUB2 =1/5

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.
 Biased towards shorter documents.

14
The problem with the first example was that document
2 “won” because it was shorter, not because it was a
better match. We need a way to take into account
document length so that longer documents are not
penalized in calculating the match score.

Binary term-document incidence matrix
Antony and Cleopatra Julius Caesar The Tempest Hamlet Othello Macbeth
Antony 1 1 0 0 0 1
Brutus 1 1 0 1 0 0
Caesar 1 1 0 1 1 1
Calpurnia 0 1 0 0 0 0
Cleopatra 1 0 0 0 0 0
mercy 1 0 1 1 1 1
worser 1 0 1 1 1 0
Each document is represented by a binary vector {0,1}∈ |V

Term-document count matrices
 Consider the number of occurrences of a term in a
document:
 Each document is a count vector in ℕv
: a column below
Antony 157 73 0 0 0 0
Brutus 4 157 0 1 0 0
Caesar 232 227 0 2 1 1
Calpurnia 0 10 0 0 0 0
Cleopatra 57 0 0 0 0 0
mercy 2 0 3 5 5 1
worser 2 0 1 1 1 0

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.

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
 The log frequency weight of term t in d is
 0 → 0, 1 → 1, 2 → 1.3, 10 → 2, 1000 → 4, etc.
 How will you compute 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.


 >+
=
otherwise0,
0tfif,tflog1 10 t,dt,d
t,dw
∑ ∩∈
+= dqt dt )tflog(1 ,

Document frequency
 Rare terms are more informative than frequent terms
 stop words (a, and, the…..) are not informative
 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 to the query arachnocentric
 → We want a high weight for rare terms like
arachnocentric.

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.

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
 What can be the max value of dft
 dft ≤ N
 We define the idf (inverse document frequency) of t
by
 Why We use log (N/dft) instead of N/dft
 to “dampen” the effect of idf.
)/df(logidf 10 tt N=
Will turn out the base of the log is immaterial.

23
Mathematically the base of the log function does not
matter and constitutes a constant multiplicative factor
towards the overall result.

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

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 1
the 1,000,000 0
There is one idf value for each term t in a collection.

Effect of idf on ranking
 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.

27
So for a term one query, you're going to have one of these terms of N
over the document frequency, and it'll be worked out. But it's going to
be just a scaling factor. Which, since there's only one IDF value for each
term will be applied to every document, and therefore, it won't affect the
ranking in any way.
You only get an effect from IDF when you have multiple-terms in a
query. So, for example, if we have the query, capricious person, well,
now, we're in a situation where capricious is a much rarer word. And so
IDF will say. Pay much more attention to documents that contain the
word capricious, than to documents that contain just the word person in
ranking your retrieval results.

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
What does it mean

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
)df/(log)tf1log(w 10,, tdt Ndt
×+=

Score for a document given a query
 There are many variants
 How “tf” is computed (with/without logs)
 Whether the terms in the query are also weighted
 …
30
Score(q,d)= tf.idft,dt∈q∩d
∑

Binary → count → weight matrix
Antony 5.25 3.18 0 0 0 0.35
Brutus 1.21 6.1 0 1 0 0
Caesar 8.59 2.54 0 1.51 0.25 0
Calpurnia 0 1.54 0 0 0 0
Cleopatra 2.85 0 0 0 0 0
mercy 1.51 0 1.9 0.12 5.25 0.88
worser 1.37 0 0.11 4.15 0.25 1.95
Each document is now represented by a real-valued
vector of tf-idf weights ∈ R|V|

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.

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
 First cut: distance between two points
 ( = distance between the end points of the two vectors)
 Euclidean distance?
 Euclidean distance is a bad idea . . .Why?
 . . . because Euclidean distance is large for vectors of
different lengths.

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
 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
.
 Rank documents in decreasing order of the angle between
query and document – A minor issue?
 Rank documents in increasing order of
cosine(query,document) – Why Cosine?
Cosine is a monotonically decreasing function for the
interval [0o
, 180o
]

From angles to cosines
 But how – and why – should we be computing cosines?

cosine(query,document)
∑∑
∑
==
=
=•=
•
=
V
i i
V
i i
V
i ii
dq
dq
d
d
q
q
dq
dq
dq
1
2
1
2
1
),cos( 





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.

Cosine similarity illustrated

Cosine similarity exercises
 Exercise: Rank the following by decreasing cosine
similarity:
 Two docs that have only frequent words (the, a, an, of) in
common.
 Two docs that have no words in common.
 Two docs that have many rare words in common
(wingspan, tailfin).

Which documents are similar according to VSM
auto
engine
bonnet
tyres
lorry
boot
car
emissions
hood
make
model
trunk
make
hidden
Markov
model
emissions
normalize
D1 D2 D3
Which documents are semantically similar

Synonymy and Polysemy
auto
engine
bonnet
tyres
lorry
boot
car
emissions
hood
make
model
trunk
make
hidden
Markov
model
emissions
normalize
Synonymy
Will have small
cosine but are
related
Polysemy
Will have large
cosine but not truly
related

Information Retrieval 02

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