localizing a sort

[Ricardo Aráoz]

Alex Martelli wrote:
> Ricardo Aráoz <ricaraoz at gmail.com> wrote:
>> Peter Otten wrote:
>    ...
>>>>>>> print ''.join(sorted(a, cmp=lambda x,y: locale.strcoll(x,y)))
>>>> aeiouàáäèéëìíïòóöùúü
>>> The lambda is superfluous. Just write cmp=locale.strcoll instead.
>> No it is not :
>>>>> print ''.join(sorted(a, cmp=locale.strcoll(x,y)))
>> Traceback (most recent call last):
>>   File "<input>", line 1, in <module>
>> TypeError: strcoll expected 2 arguments, got 0
>>
>> You need the lambda to assign both arguments.
> 
> No, your mistake is that you're CALLING locale.strcoll, while as Peter
> suggested you should just PASS it as the cmp argument.  I.e.,
> 
>     ''.join(sorted('ciao', cmp=locale.strcoll))
> 
> Using key=locale.strxfrm should be faster (at least when you're sorting
> long-enough lists of strings), which is why strxfrm (and key=...:-)
> exist in the first place, but cmp=locale.strcoll, while usually slower,
> is entirely correct.  That lambda _IS_ superfluous, as Peter said.
> 
> 
> Alex

Got it! And it is MUCH more elegant than my code. Thanks.