python regex查找带有数字的文本

Question

I have text like:

sometext...one=1290...sometext...two=12985...sometext...three=1233...

How can I find one=1290 and two=12985 but not three or four or five? There are can be from 4 to 5 digits after = . I tried this:

import re
pattern = r"(one|two)+=+(\d{4,5})+\D"
found = re.findall(pattern, sometext, flags=re.IGNORECASE)
print(found)

It gives me results like: [('one', '1290')] . If i use pattern = r"((one|two)+=+(\\d{4,5})+\\D)" it gives me [('one=1290', 'one', '1290')] . How can I get just one=1290 ?

Answer 1

You were close. You need to use a single capture group (or none for that matter):

((?:one|two)+=+\d{4,5})+

Full code:

import re

string = 'sometext...one=1290...sometext...two=12985...sometext...three=1233...'

pattern = r"((?:one|two)+=+\d{4,5})+"
found = re.findall(pattern, string, flags=re.IGNORECASE)
print(found)
# ['one=1290', 'two=12985']

Answer 2

使内部组不捕获： ((?:one|two)+=+(?:\\d{4,5})+\\D)

Answer 3

The reason that you are getting results like [('one', '1290')] rather than one=1290 is because you are using capture groups . Use:

r"(?:one|two)=(?:\d{4,5})(?=\D)"

• I have removed the additional + repeaters, as they were (I think?) unnecessary. You don't want to match things like oneonetwo===1234 , right?
• Using (?:...) rather than (...) defines a non-capture group. This prevents the result of the capture from being returned, and you instead get the whole match.
• Similarly, using (?=\\D) defines a look-ahead - so this is excluded from the match result.