2d barcode vb.net Document 2 = 0.8246 Document 3 = 0.3271 Document 1 = 0.0801 in Java

Encoding Data Matrix 2d barcode in Java Document 2 = 0.8246 Document 3 = 0.3271 Document 1 = 0.0801

Document 2 = 0.8246 Document 3 = 0.3271 Document 1 = 0.0801
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Notice in our table of calculations that terms occurring very often in many documents do nothing to increase the score of a document. For example, of and some other words in our example actually calculate to zero. Therefore they aren t included in further calculations. This is due to the global information component, idf. In our example these terms appear in every document, which causes their idf value and therefore their weights to calculate to zero. In large repositories if they had not appeared in every document, they still would have appeared in the vast majority of them, and their weights would calculate to very small quantities and therefore have little effect on query outcomes. Common words like the, and, and but are known as stop words. Many indexing and querying schemes allow for their removal both from documents before they are put into a search repository and from queries before they are applied to a search. Is there still a problem here The three documents now are for all intents and purposes the same length. What would happen if document 1 were inordinately longer than the other two Let s say that document 1 reads, Shipment of gold damaged in a fire, gold was undamaged, gold truck was total loss, gold exchange notified. It is a natural assumption that long documents concerning a specific topic would probably contain higher term frequencies of the term they were concerned with. Witness the gold term in our new document. How can we take that into account We ll discuss that in the next section.
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12.1.2 Normalizing document length to level the playing field
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Term frequency counts by themselves are not a good measure of relevancy to a term query Q because:
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With different length documents, a longer document may be scored higher because Q could appear more often. Equal-length documents are scored higher for more occurrences of Q.
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Document ranking
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We will discuss the document-length problem first, then examine the term-count problem. Long documents tend to contain higher individual term counts, but the idf remains constant; therefore the term weight w increases proportionally. Suppose we have a document D1, which has a certain weight w1 for a given term t1. Now suppose we increase the size of D1 by appending a copy of D1 to itself. What have we accomplished The document count has not changed and neither has the df, but the term frequency count and therefore the score have doubled. How do we solve or at least minimize this problem that document length can pose We normalize the tf weights of all terms occurring in a document by the maximum tf in that document. Formally this is shown in equation 12.17.
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fi , j = tfi , j max tfi , j
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Equation 12.17 Dividing a term s frequency by the largest term frequency value yields a normalized frequency.
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fi , j = normalized frequency tfi , j = frequency of term i in document j
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max tfi , j = maximum frequency of term i in document j
Consider a document with the terms and frequencies shown in table 12.2. The truck term occurs most often, so the normalized frequencies are determined as shown in table 12.3.
Table 12.2 Example terms and frequencies Term delivery shipment silver truck Frequency 1 2 4 5 delivery shipment silver truck 1/5 = 0.20 2/5 = 0.40 4/5 = 0.80 5/5 = 1 Table 12.3 Terms with their frequencies normalized Term Normalized frequency
The weight of term i in document j is given in equation 12.18. This is the formula used in normalized frequency term vector calculations.
wi , j = tfi , j max tfi , j D * log dfi
Equation 12.18 The weight of term i in document j.
Queries also can be normalized if necessary. The formal equation for queries is shown in equation 12.19.
Scoring documents
fQ ,i = 0.5 + 0.5 *
fQ ,i tfQ ,i
tfQ ,i max tfQ ,i
Equation 12.19 The formula used in normalized frequency query vector calculations
= normalized query frequency = frequency of term i in query = maximum frequency of term i in query
max tfQ ,i
As an example, suppose you have the following query: Q = shipment silver shipment. The frequencies are
shipment 2 silver 1
The shipment term occurs most often, so the normalized frequencies are
shipment (0.5 + 0.5 * 2/2) = 1 silver (0.5 + 0.5 * 1/2) = 0.75
The weight of term i in query Q is given in equation 12.20. This formula was derived entirely from experimentation and measuring the results as the constants were changed to produce the best results.
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