如何使用python正则表达式finditer查找所有匹配项

Question

I am trying to find a pattern. I have written the below code:

string = '000,001,100,001'
pattern = '(.*)00(.*),(.*)00(.*)'

for m in re.finditer(pattern, string):
    print(m.groups())

The code above returns ('000,001,1', '', '', '1') where as it misses the match with groups ('', '0', '', '1,100,001')

I am trying to workout if the characters before and after the '00' on consecutive lines are the same. The code i wrote matches '000,001,1 00 , 00 1'. How to match ' 00 0, 00 1,100,001'.

How the obtain match groups for the later?

Answer 1

Comments : for the string '2295051,2238451,2235301,1950522,2238451,3530333'
... You see that the groups have the same number of digits before the occur which is 2 digits and after they occur which is 1 digit

string = '2295051,2238451,2235301,1950522,2238451,3530333'  

_Step 1_  
pattern = '(\d+)'
Output: ('2295051',) ('2238451',) ('2235301',) ('1950522',) ('2238451',) ('3530333',)  

_Step 2_
pattern = '((\d\d)\d+)'  
Output: ('2295051', '22') ('2238451', '22') ('2235301', '22')  
        ('1950522', '19') ('2238451', '22') ('3530333', '35')  

_Step 3_
pattern = '((\d\d)\d+(\d))'
Output: ('2295051', '22', '1') ('2238451', '22', '1') ('2235301', '22', '1')  
('1950522', '19', '2') ('2238451', '22', '1') ('3530333', '35', '3')  

Read about the meaning of '+' in the Docs re.html#module-re .

Comment : ...what i don't understand is how it does it and how i can make use of it...

The pattern = '((\\d\\d)\\d+(\\d))' search for a substring, starting with 2 Digits \\d\\d , followed by any number of Digits, at least one, and one Digit \\d at the End. This pattern is generalized, matches any substring, at least with len=4, of Digits.

---

Try this pattern:

string = '000,001,100,001'
pattern = '((\d)00|00(\d))'

for m in re.finditer(pattern, string):
    print(m.groups())

Output :

('000', '0', None)
('001', None, '1')
('100', '1', None)
('001', None, '1')

The first item 000 have both, before and after .

Tested with Python:3.4.2 - re:2.2.1
Come back and Flag your Question as answered if this is working for you or comment why not.