Match.groupdict: Meaning of default argument?

[Loris Bennett]

Julio Di Egidio <julio at diegidio.name> writes:

> On Friday, 29 April 2022 at 09:50:08 UTC+2, Loris Bennett wrote:
>> Hi, 
>> 
>> If I do 
>> 
>> import re 
>> pattern = re.compile(r'(?P<days>\d*)(-?)(?P<hours>\d\d):(?P<minutes>\d\d):(?P<seconds>\d\d)')
>> s = '104-02:47:06' 
>> match = pattern.search(s) 
>> match_dict = match.groupdict('0') 
>> 
>> I get 
>> 
>> match_dict 
>> {'days': '104', 'hours': '02', 'minutes': '47', 'seconds': '06'} 
>> 
>> However, if the string has no initial part (corresponding to the number of 
>> days), e.g. 
>> 
>> s = '02:47:06' 
>> match = pattern.search(s) 
>> match_dict = match.groupdict('0') 
>> 
>> I get 
>> 
>> match_dict 
>> {'days': '', 'hours': '02', 'minutes': '47', 'seconds': '06'} 
>> 
>> I thought that 'days' would default to '0'. 
>> 
>> What am I doing wrong? 
>
> You tell, but it's quite obvious that you (just) run a regex on a string and captures are going to be strings: indeed, '02' is not a number either...
>
> Julio

I am not sure what you are trying to tell me.  I wasn't expecting
anything other than strings.  The problem was, as Stefan helped me to
understand, that I misunderstood what 'participating in the match'
means.

Cheers,

Loris

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