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如何在findex + Python的正则表达式模式中使用{}

[英]How to use {} in regex pattern with findall + Python

I'm creating a regex as below: 我正在创建一个正则表达式,如下所示:

import re
asd = re.compile(r"(blah){2}")
mo = asd.search("blahblahblahblahblahblah ll2l 21HeHeHeHeHeHe lllo")
mo1 = asd.findall("blahblahblahblahblahblah")
print(mo.group())
print("findall output: ", mo1)

This returns output blahblah findall output: ['blah', 'blah', 'blah'] 这将返回输出blahbla findall输出:['blah','blah','blah']

-Why findall output matches 'blah' three times, when its specified {2} times only in the pattern? -为什么当findall输出仅在模式中指定{2}次时,findall输出与“ blah”匹配3次?

If I change to {4}, then findall matches: 如果我更改为{4},则findall匹配:

asd = re.compile(r"(blah){4}")
findall output:  ['blah']

-How is {m} treated with re.search and re.findall ? -如何使用re.search和re.findall处理{m}?

Thanks a lot. 非常感谢。

If you want to catch the (blah){2} (the 2 blah you have there) you should wrap it: 如果您想抓住 (blah){2} (那里有2个blah ),则应该将其包装起来:

asd = re.compile(r"((?:blah){2})")

Note that I made sure not to catch the inside blah (using ?: ) 请注意,我确保不要抓住里面的东西(使用?: blah

>>>asd = re.compile(r"((?:blah){2})")
>>>mo = asd.search("blahblahblahblahblahblah ll2l 21HeHeHeHeHeHe lllo")
>>>mo1 = asd.findall("blahblahblahblahblahblah")
>>>print(mo.group())
blahblah
>>>print("findall output: ", mo1)
findall output:  ['blahblah', 'blahblah', 'blahblah']

Exactly the same goes with the {4} you have there. 那里的{4}完全一样。 The regex will find it, but will not catch it. regex可以找到它,但不会捕获它。 if you want to catch it you should wrap it. 如果您想抓住它,则应该将其包裹起来。

(blah){2} captures and exhausts the string blahblah but only returns the last blah in blahblah . (blah){2}捕获和排气字符串blahblah但仅返回最后 blahblahblah Since you have three blahblah s in your string, it will output ['blah', 'blah', 'blah'] 由于您的字符串中包含三个blahblah ,因此它将输出['blah', 'blah', 'blah']

(blah){4} can only match once so it gives you ['blah'] (blah){4}只能匹配一次,因此可以给您['blah']

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