>>> match = re.findall('a.*?a', 'a 1 a 2 a 3 a 4 a')
>>> match
['a 1 a', 'a 3 a']
How do I get it to print
我如何打印它
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
Thank you!
谢谢！
3 个解决方案
#1
6
I think using a positive lookahead assertion should do the trick:
我认为使用积极的先行断言应该可以解决这个问题：
>>> re.findall('(?=(a.*?a))', 'a 1 a 2 a 3 a 4 a')
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
re.findall returns all the groups in the regex - including those in look-aheads. This works because the look-ahead assertion doesn't consume any of the string.
re.findall返回正则表达式中的所有组 - 包括那些在前瞻中的组。这是有效的，因为前瞻断言不会消耗任何字符串。
#2
5
You may use alternative regex module which allows overlapping matches:
您可以使用允许重叠匹配的替代正则表达式模块：
>>> regex.findall('a.*?a', 'a 1 a 2 a 3 a 4 a', overlapped = True)
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
#3
4
r = re.compile('a.*?a') # as we use it multiple times
matches = [r.match(s[i:]) for i in range(len(s))] # all matches, if found or not
matches = [m.group(0) for m in matches if m] # matching string if match is not None
print matches
gives
给
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
I don't know if it is the best solution, but here I test every substring reaching to the end of the string for matching with the given pattern.
我不知道它是否是最好的解决方案，但在这里我测试每个到达字符串末尾的子字符串以匹配给定的模式。
#1
6
I think using a positive lookahead assertion should do the trick:
我认为使用积极的先行断言应该可以解决这个问题：
>>> re.findall('(?=(a.*?a))', 'a 1 a 2 a 3 a 4 a')
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
re.findall returns all the groups in the regex - including those in look-aheads. This works because the look-ahead assertion doesn't consume any of the string.
re.findall返回正则表达式中的所有组 - 包括那些在前瞻中的组。这是有效的，因为前瞻断言不会消耗任何字符串。
#2
5
You may use alternative regex module which allows overlapping matches:
您可以使用允许重叠匹配的替代正则表达式模块：
>>> regex.findall('a.*?a', 'a 1 a 2 a 3 a 4 a', overlapped = True)
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
#3
4
r = re.compile('a.*?a') # as we use it multiple times
matches = [r.match(s[i:]) for i in range(len(s))] # all matches, if found or not
matches = [m.group(0) for m in matches if m] # matching string if match is not None
print matches
gives
给
['a 1 a', 'a 2 a', 'a 3 a', 'a 4 a']
I don't know if it is the best solution, but here I test every substring reaching to the end of the string for matching with the given pattern.
我不知道它是否是最好的解决方案，但在这里我测试每个到达字符串末尾的子字符串以匹配给定的模式。