Compute one BM25 relevance score per tokenized document. Repeated query terms are counted once.
idf(t)=log(df(t)+0.5N−df(t)+0.5+1) score(D,Q)=t∈Q∑idf(t)tf(t,D)+k1(1−b+bavgdl∣D∣)tf(t,D)(k1+1)Here, N is the document count, df(t) counts documents containing term t, tf(t,D) counts the term in document D, ∣D∣ is document length, and avgdl is average document length. Return a NumPy array whose entries follow the original document order.
Input: query_tokens = ["machine", "learning"], docs = [["introduction", "to", "machine", "learning"], ["deep", "learning", "basics"], ["cooking", "pasta", "guide"]], k1 = 1.2, b = 0.75
Output: [1.341106, 0.490052, 0.0]
Explanation: The first document matches both query terms, the second matches one, and the third matches neither.
Input: query_tokens = ["data"], docs = [["data", "data", "mining"], ["applied", "data", "science", "science", "science"], ["sports", "news"]], k1 = 1.2, b = 0.75
Output: [0.664957, 0.390192, 0.0]
Use one Counter per document for term frequencies.
Use Counter.update(set(document)) to count document frequencies.
Build a NumPy vector of one term's frequency across all documents before applying the formula.
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Compute one BM25 relevance score per tokenized document. Repeated query terms are counted once.
idf(t)=log(df(t)+0.5N−df(t)+0.5+1) score(D,Q)=t∈Q∑idf(t)tf(t,D)+k1(1−b+bavgdl∣D∣)tf(t,D)(k1+1)Here, N is the document count, df(t) counts documents containing term t, tf(t,D) counts the term in document D, ∣D∣ is document length, and avgdl is average document length. Return a NumPy array whose entries follow the original document order.
Input: query_tokens = ["machine", "learning"], docs = [["introduction", "to", "machine", "learning"], ["deep", "learning", "basics"], ["cooking", "pasta", "guide"]], k1 = 1.2, b = 0.75
Output: [1.341106, 0.490052, 0.0]
Explanation: The first document matches both query terms, the second matches one, and the third matches neither.
Input: query_tokens = ["data"], docs = [["data", "data", "mining"], ["applied", "data", "science", "science", "science"], ["sports", "news"]], k1 = 1.2, b = 0.75
Output: [0.664957, 0.390192, 0.0]
Use one Counter per document for term frequencies.
Use Counter.update(set(document)) to count document frequencies.
Build a NumPy vector of one term's frequency across all documents before applying the formula.
Sign in to take notes on this problem
Accepts: array
Accepts: array
Accepts: number
Accepts: number