Python re模块 - 使用'或'时重复的符号限制

Question

I have a question. Lets say there are such patterns:

>>> import re
>>> pt ='^a{1,2}$'
>>> re.search(pt, 'aa') # looks ok
<_sre.SRE_Match object at 0x020B2288>
>>> re.search(pt, 'aaa') # ok too
>>>

Now lets try to look for match with another pattern:

>>> pt = '^a{1,2}|x$'
>>> re.search(pt, 'a') # this one looks ok
<_sre.SRE_Match object at 0x020B25D0>
>>> re.search(pt, 'aaax') # (1) Now this one?
<_sre.SRE_Match object at 0x020B2288>
>>> re.search(pt, 'aaaaaax') # (2) and this one?
<_sre.SRE_Match object at 0x020B25D0>
>>> re.search(pt, 'aaa') # (3) and this one?
<_sre.SRE_Match object at 0x020B25D0>

(1)(2)(3) To me it looks like it should match string that starts with one or two 'a' or one 'x' or both combinations and ends between these letters, but nothing else. Or I don't get it something? Is it should be like that? Like when you use '|', it ignores what limit is put inside {}? Can someone explain me this?

Answer 1

The $ is affected by the grouping. Your regex is interpreted as (^a{1,2})|(x$) , which matches either "one or two as at the beginning of the string" OR "an x at the end of the string". If you want to have the | apply only to the as and xs, you need to group them:

pt = '^(a{1,2}|x)$'

Or, if you don't want to capture the group, use a noncapturing group:

pt = '^(?:a{1,2}|x)$'

Edit: I'm not sure I understand what you're trying to match, but perhaps try:

pt = '^(a{1,2}x?|x)$"

Answer 2

^foo|bar$匹配^foo或bar$ 。