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Python正则表达式找到所有重叠的匹配项？

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问题：Python正则表达式找到所有重叠的匹配项？
回答 0
回答 1
回答 2

问题：Python正则表达式找到所有重叠的匹配项？

我正在尝试在Python 2.6中使用re查找更大系列的数字中的每10位数字系列。

我很容易就能抓住不重叠的比赛，但我希望数字系列中的每场比赛。例如。

在“ 123456789123456789”中

我应该得到以下列表：

[1234567891,2345678912,3456789123,4567891234,5678912345,6789123456,7891234567,8912345678,9123456789]

我已经找到了对“超前”的引用，但是我所看到的示例仅显示了成对的数字，而不是更大的分组，而且我无法将其转换为两位数以外的数字。

点击查看英文原文

I’m trying to find every 10 digit series of numbers within a larger series of numbers using re in Python 2.6.

I’m easily able to grab no overlapping matches, but I want every match in the number series. Eg.

in “123456789123456789”

I should get the following list:

[1234567891,2345678912,3456789123,4567891234,5678912345,6789123456,7891234567,8912345678,9123456789]

I’ve found references to a “lookahead”, but the examples I’ve seen only show pairs of numbers rather than larger groupings and I haven’t been able to convert them beyond the two digits.

回答 0

在前瞻范围内使用捕获组。前瞻捕捉您感兴趣的文本，但是实际匹配在技术上是前瞻之前的零宽度子字符串，因此匹配在技术上是不重叠的：

import re 
s = "123456789123456789"
matches = re.finditer(r'(?=(\d{10}))',s)
results = [int(match.group(1)) for match in matches]
# results: 
# [1234567891,
#  2345678912,
#  3456789123,
#  4567891234,
#  5678912345,
#  6789123456,
#  7891234567,
#  8912345678,
#  9123456789]
点击查看英文原文

Use a capturing group inside a lookahead. The lookahead captures the text you’re interested in, but the actual match is technically the zero-width substring before the lookahead, so the matches are technically non-overlapping:

import re 
s = "123456789123456789"
matches = re.finditer(r'(?=(\d{10}))',s)
results = [int(match.group(1)) for match in matches]
# results: 
# [1234567891,
#  2345678912,
#  3456789123,
#  4567891234,
#  5678912345,
#  6789123456,
#  7891234567,
#  8912345678,
#  9123456789]

回答 1

您也可以尝试使用支持重叠匹配的第三方regex模块（不是re）。

>>> import regex as re
>>> s = "123456789123456789"
>>> matches = re.findall(r'\d{10}', s, overlapped=True)
>>> for match in matches: print match
...
1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789
点击查看英文原文

You can also try using the third-party regex module (not re), which supports overlapping matches.

>>> import regex as re
>>> s = "123456789123456789"
>>> matches = re.findall(r'\d{10}', s, overlapped=True)
>>> for match in matches: print(match)  # print match
...
1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789

回答 2

我喜欢正则表达式，但是这里不需要它们。

只是

s =  "123456789123456789"

n = 10
li = [ s[i:i+n] for i in xrange(len(s)-n+1) ]
print '\n'.join(li)

结果

1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789
点击查看英文原文

I’m fond of regexes, but they are not needed here.

Simply

s =  "123456789123456789"

n = 10
li = [ s[i:i+n] for i in xrange(len(s)-n+1) ]
print '\n'.join(li)

result

1234567891
2345678912
3456789123
4567891234
5678912345
6789123456
7891234567
8912345678
9123456789