[Tutor] recursion

[Peter Otten]

Ben Finney wrote:

> Alan Gauld <alan.gauld at btinternet.com> writes:
> 
>> On 05/02/16 02:03, noopy via Tutor wrote:
>>
>> > def permutations(items):
>> >      n = len(items)
>> >      if n==0: yield []
>> >      else:
>>
>> I assume this bit is clear enough?
> 
> I think it would be clearer without the needless opaque name ‘n’.
> Better::
> 
>     def permutations(items):
>         if not items:
>             # ‘items’ is empty (or is not a container).
>             yield []
>         else:
>             for i in range(len(items)):
>                 …
> 
> Binding a name that is used exactly once should be done only if the name
> clarifies the purpose. An opaque name like ‘n’ is not helpful.

While you're at it you can also throw out the range(len(...)) construct:

 
>>> def p(items):
...     if items:
...         for i, item in enumerate(items):
...             for cc in p(items[:i] + items[i+1:]):
...                 yield [item] + cc
...     else:
...         yield []
... 
>>> list(p("xyz"))
[['x', 'y', 'z'], ['x', 'z', 'y'], ['y', 'x', 'z'], ['y', 'z', 'x'], ['z', 
'x', 'y'], ['z', 'y', 'x']]