using regexp

[Shane Geiger]

import re
line = '123456789123456789'
print re.findall('([0-9]{3})', line)



Shane Geiger wrote:
> You don't even need regex.
>
> def 
> split_seq(seq,size):                                                                                                               
>
>    # this is sort of the inverse of 
> flatten                                                                                           
>
>    # Source: 
> http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/425044                                                           
>
>    return [seq[i:i+size] for i in range(0, len(seq), 
> size)]                                                                           
>
>                                                                                                                                       
>
> line = 
> '123456789123456789'                                                                                                            
>
>                                                                                                                                       
>
> print 
> split_seq(line,3)                                                                                                                
>
>
>
> Will that work for you?
>
>
>
> s99999999s2003 at yahoo.com wrote:
>> hi
>> how can i use regexp to group these digits into groups of 3?
>>
>> eg
>> line 123456789123456789
>>
>> i have :
>>
>> pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)
>>
>> but this only gives the first 3. I also tried
>>
>> "line\s+(\d{3})+"
>> but also not working.
>> I need output to be ['123' ,'456','789', '123','456','789', .....]
>> thanks.
>>
>>   
>

-- 
Shane Geiger
IT Director
National Council on Economic Education
sgeiger at ncee.net  |  402-438-8958  |  http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

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