Regex-python:返回字符周围的单词
我有一个包含数百万个单词的字符串,我希望有一个正则表达式,可以返回围绕任何美元符号的五个单词。例如:Regex-python:返回字符周围的单词,python,regex,python-3.x,tokenize,Python,Regex,Python 3.x,Tokenize,我有一个包含数百万个单词的字符串,我希望有一个正则表达式,可以返回围绕任何美元符号的五个单词。例如: string = 'I have a sentence with $10.00 within it and this sentence is done. ' 我想要正则表达式返回 surrounding = ['I', 'have', 'a', 'sentence', 'with', 'within', 'it', 'and', 'this', 'sentence'] 我的最终目标是对提及“
string = 'I have a sentence with $10.00 within it and this sentence is done. '
我想要正则表达式返回
surrounding = ['I', 'have', 'a', 'sentence', 'with', 'within', 'it', 'and', 'this', 'sentence']
我的最终目标是对提及“$”的所有单词进行统计,因此上面的列表将包括:
final_return = [('I', 1), ('have', 1), ('a', 1), ('sentence', 2), ('with', 1), ('within', 1), ('it', 1), ('and', 1), ('this', 1)]
到目前为止,我开发的下面的正则表达式可以返回附加到货币符号的字符串,其中包含5个字符。有没有办法编辑正则表达式以捕获周围的五个单词?我是否应该(如果是,如何)使用NLTK的标记器来实现这一点
import re
.....\$\s?\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{1,2})?.....
您可以开始使用下面的代码,我正在尝试用更简单的方法来解决它
import re
string = 'I have a sentence with $10.00 within it and this sentence is done. '
surrounding = re.search(r'(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*\$\d+\.?\d{2}?\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)\s*(\w+)', string, flags=0).groups()
print(surrounding )
我不认为正则表达式是解决这个问题的合适选择。相反,您可以提取围绕美元符号buy的所有10个单词,在这些单词上循环,并跟踪之前遍历的5个单词,以便在找到匹配项时返回 在这种情况下,您可以使用
collections.deque()
,这是一种适当的数据结构,项目数量有限,可以保留前面的五个单词。然后可以使用collections.Counter()
对象返回阈值内的单词计数器
from collections import deque
from collections import Counter
from itertools import chain
def my_counter(string):
container = deque(maxlen=5)
words = iter(string.split())
def next_five(words):
for _ in range(5):
try:
yield next(words)
except StopIteration:
pass
for w in words:
if w.startswith('$'):
yield Counter(chain(container, next_five(words)))
else:
container.append(w)
演示:
可以将正则表达式与计数器组合,如下所示:
(?P<before>(?:\w+\W+){5})
\$\d+(?:\.\d+)?
(?P<after>(?:\W+\w+){5})
这将产生(请注意,
计数器
已经是dict
):
使用split分割单词,使用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]]
您能导入
regex
模块吗?非常感谢!这真的很有帮助。无论如何,我可以按数字顺序返回这些单词吗?
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)
[Counter({'sentence': 2, 'I': 1, 'have': 1, 'a': 1, 'with': 1, 'within': 1, 'it': 1, 'and': 1, 'this': 1})]
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]]