Is there an all(map(...)) optimization in Python?

Question

I'd like to use all(map(func, iterables)), because it is clear, but I'm very interested in whether this approach is optimized? For example, if any calculated result of map() is not True mapping should stop.

Example from my project:

for item in candidate_menu:
    if not item.is_max_meals_amount_ok(daily_menus):
        return False
return True

I prefer to use functional-like style:

all(map(operator.methodcaller('is_max_meals_amount_ok', daily_menus), candidate_menu)

I there any optimization for all(map(...)) or any(map(...)) in Python?

Edit: Python 2.7 on board.

Answer 1

In addition to all(func(i) for i in iterable) suggested by others, you can also use itertools.imap - all(imap(func, iterable)). Compared to map, imap is an iterator thus if all stops consuming from it, it won't advance any further.

Answer 2

If you want to get lazy evaluation in Python 2.x, try:

all(func(i) for i in iterables)

This will not build the whole list then evaluate, unlike map. In Python 3.x, map returns an iterator so will be lazy by default.

Answer 3

Edit: this answer isn't True for Python 3, where map is a generator!

No. all and any are short-circuited, but map isn't: it can't know that it was called from a bit of code that does short circuiting, and always just returns the full list.

The solution is to use generator expressions:

all(func(i) for i in iterables)

That does short-circuit.

Answer 4

"all/any" do use short circuits.

but it wouldn't propagate across the map() command ,

to do that , change map to a generator

all(func(x) for x in iterable)

should work