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From: Stefan Behnel on 23 Jan 2010 12:34
ceciliaseidel(a)gmx.de, 23.01.2010 17:29:
> Arnaud Delobelle wrote:
>> ceciliaseidel(a)gmx.de writes:
>>
>>> As you were talking about list.pop()...
>>>
>>> Is anyone able to reproduce the following and explain why this happens
>>> by chance? (Using 3.1.1)
>>>
>>> l1 = ["ready", "steady", "go"]
>>> l2 = ["one", "two", "tree"]
>>> l3 = ["lift off"]
>>>
>>> for w in l1:
>
> Ouch... thanks Arnaud... The stable way would've been
>
> for w in l1[:]: #use copy of l1 for iteration
> print(l1.pop()) #decomposite list

If the intention is to exhaust the list during the iteration, I'd go for this:

while l1:
print(l1.pop())

Stefan
From: Steven D'Aprano on 23 Jan 2010 12:44
On Sat, 23 Jan 2010 18:29:33 +0100, Roel Schroeven wrote:

>> for w in l1[:]: #use copy of l1 for iteration
>> print(l1.pop()) #decomposite list
>
> I would prefer:
>
> while l1:
> print(l1.pop())


I would prefer:

for x in reversed(l1):
print(x)
l1[:] = []


And garbage dispose of the entire list in one go, instead of an item at a
time.



--
Steven
From: Stefan Behnel on 24 Jan 2010 07:02
Steven D'Aprano, 23.01.2010 18:44:
> On Sat, 23 Jan 2010 18:29:33 +0100, Roel Schroeven wrote:
>
>>> for w in l1[:]: #use copy of l1 for iteration
>>> print(l1.pop()) #decomposite list
>> I would prefer:
>>
>> while l1:
>> print(l1.pop())
>
>
> I would prefer:
>
> for x in reversed(l1):
> print(x)
> l1[:] = []
>
>
> And garbage dispose of the entire list in one go, instead of an item at a
> time.

IIRC, that's what CPython does anyway, so no need to make this more complex
than necessary.

Stefan
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