itertools.ilen?

[Raymond Hettinger]

"Jeremy Fincher"
> Sometimes I find myself simply wanting the length of an iterator.  For
> example, to collect some (somewhat useless ;)) statistics about a program
> of mine, I've got code like this:
>
>         objs = gc.get_objects()
>         classes = len([obj for obj in objs if inspect.isclass(obj)])
>         functions = len([obj for obj in objs if inspect.isroutine(obj)])
>         modules = len([obj for obj in objs if inspect.ismodule(obj)])
>         dicts = len([obj for obj in objs if type(obj) == types.DictType])
>         lists = len([obj for obj in objs if type(obj) == types.ListType])
>         tuples = len([obj for obj in objs if type(obj) == types.TupleType])
>
> Now, obviously I can (and will, now that 2.3 is officially released :))
> replace the list comprehensions with itertools.ifilter, but I need an
> itertools.ilen to find the length of such iterators.
>
> I can imagine such a need arises in more useful situations than this, but
> this is the particular case that brought the need to mind.
>
> The Python code is simple, obviously:
>
> def ilen(iterator):
>     i = 0
>     for _ in iterator:
>         i += 1
>     return i
>
> But it's a pity to use itertools' super-fast iterators and have to use slow,
> raw Python to determine their length :)


For your application, it is not hard to build a itertools version:

>>> import itertools
>>> def countif(predicate, seqn):
...     return sum(itertools.imap(predicate, seqn))

>>> def isEven(x):
...     return x&1 == 0

>>> countif(isEven, xrange(1000000))
500000

>>> def isTuple(x):
...     return type(x) == types.TupleType

>>> tuples = countif(isTuple, objs)


Raymond Hettinger