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pythoniterablegenerator-expression

Passing generator expression to all()

The built-in function all() is supposed to be equivalent to:

def all(iterable):
    for element in iterable:
        if not element:
            return False
    return True

But when a generator expression is passed to all(), the behaviors are different:

l=(1,2,3)
def all2(iterable):
    for element in iterable:
        if not element:
            return False
    return True
print all(e > 0 for e in l) # <generator object <genexpr> at 0x00000000096AB510>
print all2(e > 0 for e in l) # True

Same goes for the other similar built-ins. Is there a simple way to fix it? (Converting the generator expression to a tuple or a list is not really an option, because of the footprint.)

Solution

  • numpy has its own all function that behaves differently from the built-in all:

    >>> numpy.all(x for x in range(3))
<generator object <genexpr> at 0x0000000001FD2900>
>>> all(x for x in range(3))
False

    If for some reason all refers to numpy.all instead of __builtin__.all, perhaps due to a from numpy import * or due to automatic imports performed by the Python distribution you're using, you will get the NumPy behavior instead of what the built-in does.