走到这一步,我们利用Google的一元语言模型进行分词的程序基本上已经完成了,先看一下已完成的segment.py程序吧:
[cc lang="python"]
import operator

def segment( text ):
"""Return a list of words that is the best segmentation of text."""
if not text : return []
candidates = ( [first] + segment( rem ) for first, rem in splits( text ) )
return max( candidates, key = Pwords )

def splits( text, L = 20 ):
"""Return a list of all possible ( first, rem ) pairs, len( first ) <=L""" return [ ( text[:i+1], text[i+1:] ) for i in range( min(len(text), L ) ) ] def Pwords( words ): """The Naive Bayes probability of a sequence of words.""" return product( Pw(w) for w in words ) def product( nums ): """Return the product of a sequence of numbers.""" return reduce( operator.mul, nums, 1 ) class Pdist( dict ): """A probability distribution estimated from counts in datafile.""" def __init__( self, data, N = None, missingfn = None ): for key, count in data: self[key] = self.get( key, 0 ) + int( count ) self.N = float( N or sum( self.itervalues() ) ) self.missingfn = missingfn or ( lambda k, N: 1./N ) def __call__( self, key ): if key in self: return self[key] / self.N else: return self.missingfn( key, self.N ) def datafile( name, sep = '\t' ): """Read key, value pairs from file.""" for line in file( name ): yield line.split( sep ) def avoid_long_words( word, N ):   """Estimate the probability of an unknown word.""" return 10./( N * 10**len( word ) ) N = 1024908267229 ## Number of tokens in corpus Pw = Pdist( datafile( 'count_1w.txt' ), N, avoid_long_words ) [/cc]   其中Pwords和segment两个函数还没有测试,先利用Python解释器测一下。先测count_1w.txt里前三个单词the,of,and,其中:   Count(the) = 23135851162   Count(of) = 13151942776   Count(and) = 12997637966   >>> segment.N
  1024908267229L
  >>> segment.Pw( "the" )
  0.022573582340740989
  >>> 23135851162. / segment.N
  0.022573582340740989
  >>> segment.Pw( "of" )
  0.012832312116632971
  >>> 13151942776. / segment.N
  0.012832312116632971
  >>> segment.Pw( "and" )
  0.012681757364628494
  >>> 12997637966. / segment.N
  0.012681757364628494
  测一个造出来的未登录词“theofand”:
  >>> segment.Pw( "theofand" )
  9.7569707648437317e-20
  >>> 10. / ( segment.N * 10 ** len( "theofand" ) )
  9.7569707648437317e-20

  再来看Pwords函数,它返回“product( Pw(w) for w in words )”:
  >>> segment.Pwords( ["the", "of", "and"] )
  3.6735405611059254e-06
  也就是等于:
  >>> segment.Pw( "the" ) * segment.Pw( "of" ) * segment.Pw( "and" )
  3.6735405611059254e-06

  最后我们来看segment函数了,先测一下分词第一节所举的两个例子:
  >>> segment.segment( "choosespain" )
  ['choose', 'spain']
  >>> segment.segment( "insufficientnumbers" )
  ['insufficient', 'numbers']
  不错,全分正确了,再试一下“自然语言处理”的英文单词:
  >>> segment.segment( "naturallanguageprocessing" )
  ...
  陷入了长时间的停顿,看看你机器的CPU,总有一个在100%的全力运行, 但不管怎么说,还是等到了正确的分词结果:
  ['natural', 'language', 'processing']
  如果读者愿意,可以继续试一下,譬如“我爱自然语言处理”:
  >>> segment.segment( "Ilovenaturallanguageprocessing" )
  ...
  这一次,我实在等不下去了。问题出在了什么地方?再回顾一下segment函数的代码吧:
[cc lang="python"]
def segment( text ):
"""Return a list of words that is the best segmentation of text."""
if not text : return []
candidates = ( [first] + segment( rem ) for first, rem in splits( text ) )
return max( candidates, key = Pwords )
[/cc]
  关于“max(candidates,key=Pwords)”这里引用前一节评论中热心读者navygong的解释:

candidates实际上是个generator(生成器),你提到的这两行代码就是计算每种候选分词方式的概率,并从中取概率最大的那种。如”wheninthecourse”可能的分词方式有
['w', 'henin', 'the', 'course']
['wh', 'en', 'in', 'the', 'course']
['whe', 'n', 'in', 'the', 'course']

['wheninthecour', 'se']
['wheninthecours', 'e']
['wheninthecourse']。
以['wh', 'en', 'in', 'the', 'course']为例,Pwords函数作用到这个列表上后得到的是各个词出现的概率的乘积。然后用max函数取出最大乘积的那种候选分词方式。

未完待续:分词8

注:原创文章,转载请注明出处“我爱自然语言处理”:www.52nlp.cn

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作者 52nlp

《Beautiful Data-统计语言模型的应用三:分词7》有4条评论
  1. 看了您的这些博文,学习颇多,对统计语言模型有种醍醐灌顶的感觉,支持您,多多加油!!期待后续工作。

    [回复]

    52nlp 回复:

    其实很惭愧,最近几乎成了自然语言处理的旁观者,连这个博客都照顾的不周了!

    [回复]

    qingyuanxingsi 回复:

    对于英文分词问题,是否可以采用类似于中文分词的方法粗略分词得到待选项然后使用概率模型进行选择,这样效率是不是高点啊?

    [回复]

    52nlp 回复:

    英文几乎不存在分词的问题,所以没必要这样做了,当然要是遇到了这样的问题,可以考虑一些中文分词的方法

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