如何从句子中提取字符ngram？-蟒蛇

Question

以下word2ngrams函数从一个单词中提取字符3gram：

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

这篇文章显示了单个单词的字符ngram提取，使用python快速实现字符n-gram。

但是，如果我有句子并且想提取字符ngram，该word2ngram()怎么办呢？

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

我试过了：

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

[出]：

0.0205280780792
0.0271739959717
True

已编辑

regex方法看起来很优雅，但比迭代调用的执行速度慢word2ngram()：

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

[出]：

0.0211708545685
0.0284190177917
0.0303599834442
True

Answer 1

为什么不只是(?=(...))

edit 相同的内容，但不是空格(?=(\S\S\S))
edit2您也可以使用您想要的内容。前任。仅使用字母(?=([^\W_]{3}))

使用前瞻捕获 3 个字符。然后引擎在每场比赛中将排名提升 1 次
。然后捕获下 3 个。

结果foobar是
foo
oob
oba
bar

 # 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