Steven D'Aprano:
> since g is not an arbitrary iterator, one can easily do this:
> print [(h,len(list(g))) for h,g in groupby(s)]
> No need for a special function.
If you look at my first post you can see that I have shown that
solution too, but it creates a list that may be long, that may use a
l
On Fri, 19 Jan 2007 05:04:01 -0800, bearophileHUGS wrote:
> Steven D'Aprano:
>> > s = "aaabaabb"
>> > from itertools import groupby
>> > print [(h,leniter(g)) for h,g in groupby(s)]
>>
>> s isn't an iterator. It's a sequence, a string, and an iterable, but not
>> an iterator.
>
> If you l
Steven D'Aprano:
> > s = "aaabaabb"
> > from itertools import groupby
> > print [(h,leniter(g)) for h,g in groupby(s)]
>
> s isn't an iterator. It's a sequence, a string, and an iterable, but not
> an iterator.
If you look better you can see that I use the leniter() on g, not on s.
g is th
On Thu, 18 Jan 2007 16:55:39 -0800, bearophileHUGS wrote:
> What's your point? Maybe you mean that it consumes the given iterator?
> I am aware of that, it's written in the function docstring too. But
> sometimes you don't need the elements of a given iterator, you just
> need to know how many ele
At Thursday 18/1/2007 20:26, [EMAIL PROTECTED] wrote:
def leniter(iterator):
"""leniter(iterator): return the length of an iterator,
consuming it."""
if hasattr(iterator, "__len__"):
return len(iterator)
nelements = 0
for _ in iterator:
nelements += 1
retu
[EMAIL PROTECTED] writes:
> But sometimes you don't need the elements of a given iterator, you
> just need to know how many elements it has.
AFAIK, the iterator protocol doesn't allow for that.
Bear in mind, too, that there's no way to tell from outside that an
iterater even has a finite length;
George Sakkis:
> Is this a rhetorical question ? If not, try this:
It wasn't a rhetorical question.
> >>> x = (i for i in xrange(100) if i&1)
> >>> if leniter(x): print x.next()
What's your point? Maybe you mean that it consumes the given iterator?
I am aware of that, it's written in the functi
[EMAIL PROTECTED] wrote:
> Often I need to tell the len of an iterator, this is a stupid example:
>
> >>> l = (i for i in xrange(100) if i&1)
>
> len isn't able to tell it:
>
> >>> len(l)
> Traceback (most recent call last):
> File "", line 1, in
> TypeError: object of type 'generator' has no l
Often I need to tell the len of an iterator, this is a stupid example:
>>> l = (i for i in xrange(100) if i&1)
len isn't able to tell it:
>>> len(l)
Traceback (most recent call last):
File "", line 1, in
TypeError: object of type 'generator' has no len()
This is a bad solution, it may need t
