[Tutor] Puzzled by print lst.sort()

[Liam Clarke]

Dick Moores wrote:
> At 03:22 AM 9/30/2006, Liam Clarke wrote:
>> Dick Moores wrote:
>>>
>>>  >>> lst = [5,3,7,6,2]
>>>  >>> lst.sort()
>>>  >>> lst
>>> [2, 3, 5, 6, 7]
>>>  >>> lst = [5,3,7,6,2]
>>>  >>> print lst.sort()
>>> None
>>>  >>> lst
>>> [2, 3, 5, 6, 7]
>>>
>>> I'm wondering why "print lst.sort()" doesn't print the newly
>>> sorted 
>>> list, but instead prints "None". In fact, the sorting has taken
>>> place 
>>> because of "print lst.sort()". Is this behavior a Good Thing in
>>> Python?
>>>
>>> Dick
>>>
>>> _______________________________________________
>>> Tutor maillist  - 
>>> Tutor at python.org <mailto:Tutor at python.org>
>>>
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>>>
>>>   
>> Hi Dick,
>>
>> A Python list sort is destructive, as you can see - it has modified 
>> lst. So, to emphasise that it is destructive, it returns None. You'll 
>> find this in most destructive methods and functions in Python.
>
> OK, but returning the new list would seem to make more sense. Is the 
> reason sort() doesn't, really only that it is better to emphasize that 
> it is destructive?
>
>> However, as of Python 2.4, there's a new built-in function that has 
>> the functionality you want:
>>
>> >>> x = [3,1,2]
>> >>> y = sorted(x)
>> >>> print y
>> [1, 2, 3]
>> >>> print x
>> [3, 1, 2]
>>
>> You'll note that sorted() is *not *destructive - that is, x is not 
>> modified. 
>
> Didn't know about sorted(). Thanks, Liam.
>
> Dick
>
>
> ------------------------------------------------------------------------
>
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
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>   

Oh, and PS - you'll find that list.reverse() also returns None.