How to write a pager for Python iterators?

Why aren't you using this?

def grouper( page_size, iterable ):
    page= []
    for item in iterable:
        page.append( item )
        if len(page) == page_size:
            yield page
            page= []
    yield page

"Each page would itself be an iterator with up to page_size" items. Each page is a simple list of items, which is iterable. You could use yield iter(page) to yield the iterator instead of the object, but I don't see how that improves anything.

It throws a standard StopIteration at the end.

What more would you want?

Look at grouper(), from the itertools recipes.

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)