Python正则表达式中的非贪婪 [英] Non-greedy in Python Regex

问题描述

我尝试了解python中的非贪婪正则表达式，但我不明白为什么以下示例具有此结果:

I try to understand the non-greedy regex in python, but I don't understand why the following examples have this results:

print(re.search('a??b','aaab').group())
ab
print(re.search('a*?b','aaab').group())
aaab

我认为第一个为'b'，第二个为'ab'. 谁能解释一下?

I thought it would be 'b' for the first and 'ab' for the second. Can anyone explain that?

推荐答案

之所以会发生这种情况，是因为您要求匹配的是之后.如果您尝试遵循a??b的匹配从左到右的方式，您将看到类似这样的内容:

This happens because the matches you are asking match afterwards. If you try to follow how the matching for a??b happens from left to right you'll see something like this:

• 尝试0 a加b vs aaab:无匹配项(b != a)
• 尝试1个a加b vs aaab:不匹配(ab != aa)
• 尝试0 a加b vs aab:无匹配项(b != a)(匹配位置向右移动一位)
• 尝试1个a加b vs aab:不匹配(ab != aa)
• 尝试0 a加b vs ab:无匹配项(b != a)(匹配位置向右移动一位)
• 尝试1个a加b与ab:匹配(ab == ab)

• Try 0 a plus b vs aaab: no match (b != a)
• Try 1 a plus b vs aaab : no match (ab != aa)
• Try 0 a plus b vs aab: no match (b != a) (match position moved to the right by one)
• Try 1 a plus b vs aab : no match (ab != aa)
• Try 0 a plus b vs ab: no match (b != a) (match position moved to the right by one)
• Try 1 a plus b vs ab : match (ab == ab)

与*?类似.

事实是search函数返回最左匹配项.使用??和*?只会更改行为，以偏爱最左边最短的匹配，但是 not 不会返回较短的匹配，该匹配从已找到的匹配的右边开始

The fact is that the search function returns the leftmost match. Using ?? and *? changes only the behaviour to prefer the shortest leftmost match but it will not return a shorter match that starts at the right of an already found match.

还请注意，re模块不会返回重叠的匹配项，因此即使使用findall或finditer，您也将找不到要查找的两个匹配项.

Also note that the re module doesn't return overlapping matches, so even using findall or finditer you will not be able to find the two matches you are looking for.

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