Passing generator expressions to any() and all()

Question

I was just messing around in the Python interpreter and I came across some unexpected behavior.

>>> bools = (True, True, True, False)
>>> all(bools)
False
>>> any(bools)
True

Ok, so far nothing out of the ordinary...

>>> bools = (b for b in (True, True, True, False))
>>> all(bools)
False
>>> any(bools)
False

Here's where things start getting spooky. I figure this happens because the all function iterates over the generator expression, calling its __next__ method and using up the values until it encounters one that is False. Here's some evidence to back that theory up:

>>> bools = (b for b in (True, False, True, True))
>>> all(bools)
False
>>> any(bools)
True

I think the result is different because the False is not at the end, so there are still some unused values left in the generator. If you type

>>> bools = (b for b in (True, False, True, True))
>>> all(bools)
False
>>> list(bools)
[True, True]

It seems like there are only 2 remaining values.

So, why exactly does this really happen? I'm sure there are many details that I'm missing.

Answer 1

The problem that you are having is that you are using the generator after it has produced all the values.

You can verify this by running the following code:

>>> bools = (b for b in (True, False, True, True))
>>> all(bools) # once the False is found it will stop producing values
True
>>> next(bools) # next value after False which is True
True
>>> next(bools) # next value after True which is True
True
>>> next(bools)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

This will work:

>>> bools = (b for b in (True, False, True, True))
>>> all(bools)
False
>>> bools = (b for b in (True, False, True, True))
>>> any(bools)
True