On 2012-11-08, Nils Bruin <nbr...@sfu.ca> wrote:
> On Nov 8, 1:17 pm, Simon King <simon.k...@uni-jena.de> wrote:
>> Hi Volker,
>>
>> On 2012-11-08, Volker Braun <vbraun.n...@gmail.com> wrote:
>>
>> > But then you'd always have to write a lot of
>> > boilerplate to return an iterator. Its much easier to use yield, but then
>> > you can't return any additional information about finiteness.
>>
>> No? What's wrong with __len__?
>
> There are iterators where you do know the result is finite but where
> it is hard to predict how many items will be returned, e.g.
>
> ( x for x in range(1000) if expensive_predicate(x) ).
>
> It would be silly to run through this iterable twice: once to get the
> length and establish it's finite and once to actually sum the entries.
> Would you just test for the existence of __len__ ? [On this example it
> actually fails because a generator object doesn't have a __len__
> attribute.
My suggestion was that the sum() method (and the prod() method as well)
do things like
def sum(self, L):
try:
n = len(L)
except:
raise ValueError, "the given input is not a finite iterable"
if n not in ZZ:
raise ValueError, "the given input is not a finite iterable"
<do the usual summation and return the result>
For V = QQ^3, an error would be raised on V.sum(V), simply because len(V)
raises an error. Similarly for V = ZZ.
For V = GF(5)^3, V.sum(V) would return the sum of all elements, because
len(V) is finite and V is iterable.
And from the implementation point of view, I guess if P is an iterable
parent (i.e., someone was able to implement __iter__), then usually it
would also be possible to tell the cardinality of P. There are parents
that have a method called cardinality() - why not make __len__ an alias
for the cardinality?
Best regards,
Simon
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