检查NLTK包:
http://www.nltk.org它有一切你需要的
对于余弦相似性:
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ngram:
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,你将不得不首先计算分布,我使用Lucene做到这一点,但你可能很好地做类似于NLTK的东西,使用FreqDist:
如果你喜欢pylucene,这将告诉你如何喜欢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