Question

我有一个包含数百万字的字符串，我希望有一个正则表达式可以返回任何美元符号周围的五个单词。例如：

string = 'I have a sentence with $10.00 within it and this sentence is done. '

我希望正则表达式返回

surrounding = ['I', 'have', 'a', 'sentence', 'with', 'within', 'it', 'and', 'this', 'sentence']

我的最终目标是计算所有围绕提及'$'的单词，以便上面的列表可以完成：

final_return = [('I', 1), ('have', 1), ('a', 1), ('sentence', 2), ('with', 1), ('within', 1), ('it', 1), ('and', 1), ('this', 1)]

我到目前为止开发的正则表达式可以返回附加到货币符号的字符串，周围有5个字符。有没有办法编辑正则表达式来捕获周围的五个单词？我（如果是这样，如何）使用NLTK的标记器来实现这一目标吗？

   import re
 .....\$\s?\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{1,2})?.....

Answer 1

使用拆分来拆分单词，用isalpha删除非单词，然后计算列表中单词的频率。

string='I have a sentence with $10.00 within it and this sentence is done. '
string1=string.split()
string2=[s for s in string1 if s.isalpha()]
[[x,string2.count(x)] for x in set(string2)] 
#[['and', 1], ['within', 1], ['sentence', 2], ['it', 1], ['a', 1], ['have', 1], ['with', 1], ['this', 1], ['is', 1], ['I', 1]]

Answer 2

你可以开始使用下面的代码，我试图以更简单的方式解决它。

import re

string = 'I have a sentence with $10.00 within it and this sentence is done. '

surrounding  = re.search(r'(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*\$\d+\.?\d{2}?\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)', string, flags=0).groups()

print(surrounding )

Answer 3

我不认为正则表达式是否是这个问题的正确选择。相反，你可以提取围绕一个美元符号的所有10个单词循环，然后在找到匹配时跟踪五个先前遍历的单词。

在这种情况下，您可以使用collections.deque()这是一个具有有限数量项目的正确数据结构，以保留前五个单词。然后，您可以使用collections.Counter()对象返回阈值内的单词计数器。

from collections import deque
from collections import Counter
from itertools import chain

def my_counter(string):
    container = deque(maxlen=5)
    words = iter(string.split())
    def next_five(words):
        for _ in range(5):
            try:
                yield next(words)
            except StopIteration:
                pass

    for w in words:
        if w.startswith('$'):
            yield Counter(chain(container, next_five(words)))
        else:
            container.append(w)

演示：

In [8]: s =  ' extra1 extra2 I have a sentence with $10.00 within it and this sentence is done.asdf asdf a b c d e $5 k j n m k gg ee'

In [9]: 

In [9]: list(my_counter(s))
Out[9]: 
[Counter({'I': 1,
          'a': 1,
          'and': 1,
          'have': 1,
          'it': 1,
          'sentence': 2,
          'this': 1,
          'with': 1,
          'within': 1}),
 Counter({'a': 1,
          'b': 1,
          'c': 1,
          'd': 1,
          'e': 1,
          'j': 1,
          'k': 2,
          'm': 1,
          'n': 1})]

Answer 4

您可以将正则表达式与计数器结合使用，如下所示：

(?P<before>(?:\w+\W+){5})
\$\d+(?:\.\d+)?
(?P<after>(?:\W+\w+){5})

见a demo on regex101.com。

<小时/> 在Python：

from collections import Counter
import re

rx = re.compile(r'''
    (?P<before>(?:\w+\W+){5})
    \$\d+(?:\.\d+)?
    (?P<after>(?:\W+\w+){5})
    ''', re.VERBOSE)

sentence = 'I have a sentence with $10.00 within it and this sentence is done. '
words = [Counter(m.group('before').split() + m.group('after').split())
                    for m in rx.finditer(sentence)]
print(words)

<小时/> 这会产生（请注意，Counter已经是dict）：

[Counter({'sentence': 2, 'I': 1, 'have': 1, 'a': 1, 'with': 1, 'within': 1, 'it': 1, 'and': 1, 'this': 1})]

Regex python：返回字符周围的单词

4 个答案: