Ask for help on using re
Jach Feng
jfong at ms4.hinet.net
Thu Aug 5 20:57:14 EDT 2021
ast 在 2021年8月5日 星期四下午11:29:15 [UTC+8] 的信中寫道:
> Le 05/08/2021 à 17:11, ast a écrit :
> > Le 05/08/2021 à 11:40, Jach Feng a écrit :
> >> I want to distinguish between numbers with/without a dot attached:
> >>
> >>>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>>>> re.compile(r'ch \d{1,}[.]').findall(text)
> >> ['ch 1.', 'ch 23.']
> >>>>> re.compile(r'ch \d{1,}[^.]').findall(text)
> >> ['ch 23', 'ch 4 ', 'ch 56 ']
> >>
> >> I can guess why the 'ch 23' appears in the second list. But how to get
> >> rid of it?
> >>
> >> --Jach
> >>
> >
> > >>> import re
> >
> > >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >
> > >>> re.findall(r'ch \d+\.', text)
> > ['ch 1.', 'ch 23.']
> >
> > >>> re.findall(r'ch \d+(?!\.)', text) # (?!\.) for negated look ahead
> > ['ch 2', 'ch 4', 'ch 56']
> import regex
>
> # regex is more powerful that re
> >>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
> >>> regex.findall(r'ch \d++(?!\.)', text)
>
> ['ch 4', 'ch 56']
>
> ## ++ means "possessive", no backtrack is allowed
Can someone explain how the difference appear? I just can't figure it out:-(
>>> text = 'ch 1. is\nch 23. is\nch 4 is\nch 56 is\n'
>>> re.compile(r'ch \d+[^.]').findall(text)
['ch 23', 'ch 4 ', 'ch 56 ']
>>> re.compile(r'ch \d+[^.0-9]').findall(text)
['ch 4 ', 'ch 56 ']
--Jach
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