python - 如何从句子中提取字符ngram？ - Python

Question

以下 word2ngrams 函数从一个单词中提取字符 3grams:

>>> x = 'foobar'
>>> n = 3
>>> [x[i:i+n] for i in range(len(x)-n+1)]
['foo', 'oob', 'oba', 'bar']

这篇文章展示了单个单词 Quick implementation of character n-grams using python 的字符 ngrams 提取。

但是如果我有句子并且我想提取字符 ngrams 怎么办，有没有比迭代调用 word2ngram() 更快的方法？

实现相同 word2ngram 和 sent2ngram 输出的正则表达式版本是什么？会更快吗？

我试过:

import string, random, time
from itertools import chain

def word2ngrams(text, n=3):
  """ Convert word into character ngrams. """
  return [text[i:i+n] for i in range(len(text)-n+1)]

def sent2ngrams(text, n=3):
    return list(chain(*[word2ngrams(i,n) for i in text.lower().split()]))

def sent2ngrams_simple(text, n=3):
    text = text.lower()
    return [text[i:i+n] for i in range(len(text)-n+1) if not " " in text[i:i+n]]

# Generate 10000 random strings of length 100.
sents = [" ".join([''.join(random.choice(string.ascii_uppercase) for j in range(10)) for i in range(100)]) for k in range(100)]

start = time.time()
x = [sent2ngrams(i) for i in sents]
print time.time() - start        

start = time.time()
y = [sent2ngrams_simple(i) for i in sents]
print time.time() - start        

print x==y

[输出]:

0.0205280780792
0.0271739959717
True

已编辑

正则表达式方法看起来很优雅，但它的执行速度比迭代调用 word2ngram() 慢:

import string, random, time, re
from itertools import chain

def word2ngrams(text, n=3):
  """ Convert word into character ngrams. """
  return [text[i:i+n] for i in range(len(text)-n+1)]

def sent2ngrams(text, n=3):
    return list(chain(*[word2ngrams(i,n) for i in text.lower().split()]))

def sent2ngrams_simple(text, n=3):
    text = text.lower()
    return [text[i:i+n] for i in range(len(text)-n+1) if not " " in text[i:i+n]]

def sent2ngrams_regex(text, n=3):
    rgx = '(?=('+'\S'*n+'))'
    return re.findall(rgx,text)

# Generate 10000 random strings of length 100.
sents = [" ".join([''.join(random.choice(string.ascii_uppercase) for j in range(10)) for i in range(100)]) for k in range(100)]

start = time.time()
x = [sent2ngrams(i) for i in sents]
print time.time() - start        

start = time.time()
y = [sent2ngrams_simple(i) for i in sents]
print time.time() - start        

start = time.time()
z = [sent2ngrams_regex(i) for i in sents]
print time.time() - start  

print x==y==z

[输出]:

0.0211708545685
0.0284190177917
0.0303599834442
True

Answer 1

为什么不只是 (?=(...))

编辑 同样的东西，但不是空格 (?=(\S\S\S))
edit2 您也可以随意使用。前任。仅使用字母数字 (?=([^\W_]{3}))

使用前瞻来捕获 3 个字符。然后引擎每次将位置向上颠簸1次
匹配。然后捕获下一个 3。

foobar 的结果是
富
oob
奥巴马
酒吧

 # Compressed regex
 #  (?=(...))

 # Expanded regex
 (?=                   # Start Lookahead assertion
      (                     # Capture group 1 start
           .                     # dot - metachar, matches any character except newline
           .                     # dot - metachar
           .                     # dot - metachar
      )                     # Capture group 1 end
 )                     # End Lookahead assertion