在Python中简单实现N-Gram，tf-idf和余弦相似性

Question

我需要比较存储在数据库中的文档，并得出0到1之间的相似性得分。

我需要使用的方法必须非常简单。 实现n-gram的vanilla版本（可以定义要使用的克数），以及tf-idf和余弦相似度的简单实现。

有没有可以做到这一点的程序？ 或者我应该从头开始写这个？

Answer 1

查看NLTK包： http ：//www.nltk.org它拥有您需要的一切

对于cosine_similarity：

def cosine_distance(u, v):
    """
    Returns the cosine of the angle between vectors v and u. This is equal to
    u.v / |u||v|.
    """
    return numpy.dot(u, v) / (math.sqrt(numpy.dot(u, u)) * math.sqrt(numpy.dot(v, v))) 

对于ngrams：

def ngrams(sequence, n, pad_left=False, pad_right=False, pad_symbol=None):
    """
    A utility that produces a sequence of ngrams from a sequence of items.
    For example:

    >>> ngrams([1,2,3,4,5], 3)
    [(1, 2, 3), (2, 3, 4), (3, 4, 5)]

    Use ingram for an iterator version of this function.  Set pad_left
    or pad_right to true in order to get additional ngrams:

    >>> ngrams([1,2,3,4,5], 2, pad_right=True)
    [(1, 2), (2, 3), (3, 4), (4, 5), (5, None)]

    @param sequence: the source data to be converted into ngrams
    @type sequence: C{sequence} or C{iterator}
    @param n: the degree of the ngrams
    @type n: C{int}
    @param pad_left: whether the ngrams should be left-padded
    @type pad_left: C{boolean}
    @param pad_right: whether the ngrams should be right-padded
    @type pad_right: C{boolean}
    @param pad_symbol: the symbol to use for padding (default is None)
    @type pad_symbol: C{any}
    @return: The ngrams
    @rtype: C{list} of C{tuple}s
    """

    if pad_left:
        sequence = chain((pad_symbol,) * (n-1), sequence)
    if pad_right:
        sequence = chain(sequence, (pad_symbol,) * (n-1))
    sequence = list(sequence)

    count = max(0, len(sequence) - n + 1)
    return [tuple(sequence[i:i+n]) for i in range(count)] 

对于tf-idf你必须首先计算分布，我使用Lucene来做到这一点，但是你很可能用NLTK做类似的事情，使用FreqDist：

http://nltk.googlecode.com/svn/trunk/doc/book/ch01.html#frequency_distribution_index_term

如果你喜欢pylucene，这将告诉你如何计算tf.idf

    # reader = lucene.IndexReader(FSDirectory.open(index_loc))
    docs = reader.numDocs()
    for i in xrange(docs):
        tfv = reader.getTermFreqVector(i, fieldname)
        if tfv:
            rec = {}
            terms = tfv.getTerms()
            frequencies = tfv.getTermFrequencies()
            for (t,f,x) in zip(terms,frequencies,xrange(maxtokensperdoc)):
                    df= searcher.docFreq(Term(fieldname, t)) # number of docs with the given term
                        tmap.setdefault(t, len(tmap))
                        rec[t] = sim.tf(f) * sim.idf(df, max_doc)  #compute TF.IDF
            # and normalize the values using cosine normalization
            if cosine_normalization:
                denom = sum([x**2 for x in rec.values()])**0.5
                for k,v in rec.items():
                    rec[k] = v / denom

Answer 2

如果你有兴趣，我已经完成了关于tf-idf和使用Scikits.learn（sklearn） Python模块的教程系列（ 第一部分和第二部分 ）。

第3部分具有余弦相似性。

Answer 3

这里只是python + numpy的答案，简而言之：

余弦 ：

def cosine_sim(u,v):
    return np.dot(u,v) / (sqrt(np.dot(u,u)) * sqrt(np.dot(v,v)))

Ngrams ：

def ngrams(sentence, n):
  return zip(*[sentence.split()[i:] for i in range(n)])

TF-IDF （它有点奇怪，但它有效）：

def tfidf(corpus, vocab):
    """
    INPUT:

    corpus = [('this is a foo bar', [1, 1, 0, 1, 1, 0, 0, 1]), 
    ('foo bar bar black sheep', [0, 2, 1, 1, 0, 0, 1, 0]), 
    ('this is a sentence', [1, 0, 0, 0, 1, 1, 0, 1])]

    vocab = ['a', 'bar', 'black', 'foo', 'is', 'sentence', 
    'sheep', 'this']

    OUTPUT:

    [[0.300, 0.300, 0.0, 0.300, 0.300, 0.0, 0.0, 0.300], 
    [0.0, 0.600, 0.600, 0.300, 0.0, 0.0, 0.600, 0.0], 
    [0.375, 0.0, 0.0, 0.0, 0.375, 0.75, 0.0, 0.375]]

    """
    def termfreq(matrix, doc, term):
        try: return matrix[doc][term] / float(sum(matrix[doc].values()))
        except ZeroDivisionError: return 0
    def inversedocfreq(matrix, term):
        try: 
            return float(len(matrix)) /sum([1 for i,_ in enumerate(matrix) if matrix[i][term] > 0])
        except ZeroDivisionError: return 0

    matrix = [{k:v for k,v in zip(vocab, i[1])} for i in corpus]
    tfidf = defaultdict(dict)
    for doc,_ in enumerate(matrix):
        for term in matrix[doc]:
            tf = termfreq(matrix,doc,term)
            idf = inversedocfreq(matrix, term)
            tfidf[doc][term] = tf*idf

    return [[tfidf[doc][term] for term in vocab] for doc,_ in enumerate(tfidf)]

以下是测试的长答案：

import numpy as np
from math import sqrt, log
from itertools import chain, product
from collections import defaultdict

def cosine_sim(u,v):
    return np.dot(u,v) / (sqrt(np.dot(u,u)) * sqrt(np.dot(v,v)))

def ngrams(sentence, n):
  return zip(*[sentence.split()[i:] for i in range(n)])

def tfidf(corpus, vocab):
    """
    INPUT:

    corpus = [('this is a foo bar', [1, 1, 0, 1, 1, 0, 0, 1]), 
    ('foo bar bar black sheep', [0, 2, 1, 1, 0, 0, 1, 0]), 
    ('this is a sentence', [1, 0, 0, 0, 1, 1, 0, 1])]

    vocab = ['a', 'bar', 'black', 'foo', 'is', 'sentence', 
    'sheep', 'this']

    OUTPUT:

    [[0.300, 0.300, 0.0, 0.300, 0.300, 0.0, 0.0, 0.300], 
    [0.0, 0.600, 0.600, 0.300, 0.0, 0.0, 0.600, 0.0], 
    [0.375, 0.0, 0.0, 0.0, 0.375, 0.75, 0.0, 0.375]]

    """
    def termfreq(matrix, doc, term):
        try: return matrix[doc][term] / float(sum(matrix[doc].values()))
        except ZeroDivisionError: return 0
    def inversedocfreq(matrix, term):
        try: 
            return float(len(matrix)) /sum([1 for i,_ in enumerate(matrix) if matrix[i][term] > 0])
        except ZeroDivisionError: return 0

    matrix = [{k:v for k,v in zip(vocab, i[1])} for i in corpus]
    tfidf = defaultdict(dict)
    for doc,_ in enumerate(matrix):
        for term in matrix[doc]:
            tf = termfreq(matrix,doc,term)
            idf = inversedocfreq(matrix, term)
            tfidf[doc][term] = tf*idf

    return [[tfidf[doc][term] for term in vocab] for doc,_ in enumerate(tfidf)]


def corpus2vectors(corpus):
    def vectorize(sentence, vocab):
        return [sentence.split().count(i) for i in vocab]
    vectorized_corpus = []
    vocab = sorted(set(chain(*[i.lower().split() for i in corpus])))
    for i in corpus:
        vectorized_corpus.append((i, vectorize(i, vocab)))
    return vectorized_corpus, vocab

def create_test_corpus():
    sent1 = "this is a foo bar"
    sent2 = "foo bar bar black sheep"
    sent3 = "this is a sentence"

    all_sents = [sent1,sent2,sent3]
    corpus, vocab = corpus2vectors(all_sents)
    return corpus, vocab

def test_cosine():
    corpus, vocab = create_test_corpus()

    for sentx, senty in product(corpus, corpus):
        print sentx[0]
        print senty[0]
        print "cosine =", cosine_sim(sentx[1], senty[1])
        print

def test_ngrams():
    corpus, vocab = create_test_corpus()
    for sentx in corpus:
        print sentx[0]
        print ngrams(sentx[0],2)
        print ngrams(sentx[0],3)
        print

def test_tfidf():
    corpus, vocab = create_test_corpus()
    print corpus
    print vocab
    print tfidf(corpus, vocab)

print "Testing cosine..."
test_cosine()
print
print "Testing ngrams..."
test_ngrams()
print
print "Testing tfidf..."
test_tfidf()
print

[OUT]：

Testing cosine...
this is a foo bar
this is a foo bar
cosine = 1.0

this is a foo bar
foo bar bar black sheep
cosine = 0.507092552837

this is a foo bar
this is a sentence
cosine = 0.67082039325

foo bar bar black sheep
this is a foo bar
cosine = 0.507092552837

foo bar bar black sheep
foo bar bar black sheep
cosine = 1.0

foo bar bar black sheep
this is a sentence
cosine = 0.0

this is a sentence
this is a foo bar
cosine = 0.67082039325

this is a sentence
foo bar bar black sheep
cosine = 0.0

this is a sentence
this is a sentence
cosine = 1.0


Testing ngrams...
this is a foo bar
[('this', 'is'), ('is', 'a'), ('a', 'foo'), ('foo', 'bar')]
[('this', 'is', 'a'), ('is', 'a', 'foo'), ('a', 'foo', 'bar')]

foo bar bar black sheep
[('foo', 'bar'), ('bar', 'bar'), ('bar', 'black'), ('black', 'sheep')]
[('foo', 'bar', 'bar'), ('bar', 'bar', 'black'), ('bar', 'black', 'sheep')]

this is a sentence
[('this', 'is'), ('is', 'a'), ('a', 'sentence')]
[('this', 'is', 'a'), ('is', 'a', 'sentence')]


Testing tfidf...
[('this is a foo bar', [1, 1, 0, 1, 1, 0, 0, 1]), ('foo bar bar black sheep', [0, 2, 1, 1, 0, 0, 1, 0]), ('this is a sentence', [1, 0, 0, 0, 1, 1, 0, 1])]
['a', 'bar', 'black', 'foo', 'is', 'sentence', 'sheep', 'this']
[[0.30000000000000004, 0.30000000000000004, 0.0, 0.30000000000000004, 0.30000000000000004, 0.0, 0.0, 0.30000000000000004], [0.0, 0.6000000000000001, 0.6000000000000001, 0.30000000000000004, 0.0, 0.0, 0.6000000000000001, 0.0], [0.375, 0.0, 0.0, 0.0, 0.375, 0.75, 0.0, 0.375]]

Answer 4

如果您仍然对这个问题感兴趣，我使用Lucene Java和Jython做了类似的事情。 这是我的代码中的一些片段。

Lucene使用所谓的分析器预处理文档和查询。 这个使用Lucene的内置n-gram过滤器：

class NGramAnalyzer(Analyzer):
    '''Analyzer that yields n-grams for minlength <= n <= maxlength'''
    def __init__(self, minlength, maxlength):
        self.minlength = minlength
        self.maxlength = maxlength
    def tokenStream(self, field, reader):
        lower = ASCIIFoldingFilter(LowerCaseTokenizer(reader))
        return NGramTokenFilter(lower, self.minlength, self.maxlength)

要将ngrams列表转换为Document ：

doc = Document()
doc.add(Field('n-grams', ' '.join(ngrams),
        Field.Store.YES, Field.Index.ANALYZED, Field.TermVector.YES))

要在索引中存储文档：

wr = IndexWriter(index_dir, NGramAnalyzer(), True,
                 IndexWriter.MaxFieldLength.LIMITED)
wr.addDocument(doc)

构建查询是比较困难的一点点Lucene的QueryParser期待与特殊运营商，报价等的查询语言，但它可以规避（如部分原因在这里 ）。

Answer 5

对于我们的信息检索课程，我们使用一些由我们的Java教授编写的代码。 对不起，没有python端口。 “它仅在GNU通用公共许可证下发布用于教育和研究目的。”

您可以查看文档http://userweb.cs.utexas.edu/~mooney/ir-course/doc/

但更具体地说： http ： //userweb.cs.utexas.edu/users/mooney/ir-course/doc/ir/vsr/HashMapVector.html

你可以下载它http://userweb.cs.utexas.edu/users/mooney/ir-course/