HELP:sorting list of outline numbers

[Delaney, Timothy (Tim)]

Scott David Daniels wrote:

> Delaney, Timothy (Tim) wrote:
>> Scott David Daniels wrote:
>>> For 2.3: (using DSU -- Decorate, Sort, Undecorate)
>         ...
>>>     lst = ['1', '1.2', '1.12', '1.1', '3.1']
>>>     decorated = [(numparts(txt), txt) for txt in lst]    
>>>     decorated.sort() lst[:] = [txt for code, txt in decorated]
>> 
>> Slightly better to do::
>>      ...
>>      lst = ['1', '1.2', '1.12', '1.1', '3.1']
>>      decorated = [(numparts(txt), i, txt) for i, txt in
>>      enumerate(lst)]      decorated.sort() lst[:] = [txt[-1] for d
>> in decorated] 
> 
> In the particular case given, I don't think it matters.

True.

> In general, however, I agree the latter is a better idea.  Just
> so someone hopping in here is clear about what is better, and
> explicitly _not_ because I think Tim misunderstands the issue,

Hey! I resent that! I misunderstand lots of things! <wink>

> The latter version, in the case of a match on the "decorated"
> version compares first the "key", and then, if the keys match,
> it compares the position in the list.  In no case does it actually
> compare the original list elements.  The 2.4 version (just using the
> key= form of sort) does the same kind of thing; it avoids comparing
> the element and only compares keys.  The reason you care about this
> is that is can be arbitrarily expensive to compare elements, or the
> elements themselves may be "incomparable" (think of complex numbers).

Yep - I really should have explained the reasoning better. I guess
having been involved in the 2.4 sort discussions on python-dev, it kinda
seems obvious to me why only comparing the key is an important feature.

Tim Delaney