Python - Generator case where nothing to return

Question

I have a generator like so:

def iterate_my_objects_if_something(self):
    for x in self.my_objects:
        if x.something:
            yield x

Which I call like so:

for x in self.iterate_my_objects_if_something():
    pass

In the case where there is nothing to return, this tries to iterate over NoneType and throws an exception.

How do I return an empty generator instead?

Why is `my_objects` `None` when you don't have any objects? Why isn't it a list of no objects? — user2357112, Jul 18 '13 at 01:49
I think you might be misdiagnosing the problem. You should only get this error if `self.my_objects` is `None`, not if there are no objects yielded. — DSM, Jul 18 '13 at 01:49
you should probably post the stack trace, my bet is that the exception is being thrown _within_ the generator. All you need to do is skip the loop if you can't iterate over `self.my_objects`, or catch the exception within the generator. — Shep, Jul 18 '13 at 01:56
Turns out I was chaining this generator in a different generator and it was throwing an error there because this one returned none. Stack trace was confusing. — whats canasta, Jul 18 '13 at 02:00

score 8 · Accepted Answer · answered Jul 18 '13 at 01:47

8

Just do a simple check:

def iterate_my_objects_if_something(self):
    if self.my_objects:
        for x in self.my_objects:
            if x.something:
                yield x

answered Jul 18 '13 at 01:47

Wolph

78,177
11
137
148

score 5 · Answer 2 · answered Jul 18 '13 at 02:03

It is important to know, which iteration causes the error. That is certainly pointed in traceback, but in this case traceback is not necessary (keep reading).

Is iteration over generator an issue?

After you take a look at that, it is obvious, but worth clarifying that:

empty generator is not of NoneType, so iterating through it will not cause such issue:

>>> def test_generator():
    for i in []:
        yield i


>>> list(test_generator())  # proof it is empty
[]
>>> for x in test_generator():
    pass

>>>

generator is recognized by Python during definition (I am simplifying) and trying to mix generators and simple functions (eg. by using conditional, as below) will be a syntax error:

>>> def test_generator_2(sth):
    if sth:
        for i in []:
            yield i
    else:
        return []

SyntaxError: 'return' with argument inside generator (<pyshell#73>, line 6)

Is the iteration inside generator an issue?

Based on the above the conclusion is that the error is not about iterating through iterator, but what happens when it is created (the code within generator):

def iterate_my_objects_if_something(self):
    for x in self.my_objects:  # <-- only iteration inside generator
        if x.something:
            yield x

So seemingly in some cases self.my_objects becomes None.

Solution

To fix that issue either:

guarantee that self.my_objects is always an iterable (eg. empty list []), or

check it before iteration:

def iterate_my_objects_if_something(self):
    # checks, if value is None, otherwise assumes iterable:
    if self.my_objects is not None:
        for x in self.my_objects:
            if x.something:
                yield x

Óscar López · Answer 3 · 2013-07-18T01:50:38.860

3

Check before iterating:

if self.my_objects:
    for x in self.my_objects:
        if x.something:
          yield x

edited Jul 18 '13 at 01:50

answered Jul 18 '13 at 01:45

Óscar López

232,561
37
312
386

3

Don't return `[]`. It's a generator; just drop off the end without yielding anything. This won't even compile. – user2357112 Jul 18 '13 at 01:45
@user2357112 If I do that, the same error will occur at the call site: `'NoneType' object is not iterable` – Óscar López Jul 18 '13 at 01:47
1

@ÓscarLópez: your code is broken, you cannot have a `return` (except the empty one) in a generator. – Wolph Jul 18 '13 at 01:50
@user2357112 ok, fixed it then. – Óscar López Jul 18 '13 at 01:51

score 0 · Answer 4 · edited Jun 19 '20 at 22:10

0

If there is nothing to iterate throw StopIteration exception. See the examples below.

def iterate_my_objects_if_something(self):
    if not self.my_objects:
        raise StopIteration
    for x in self.my_objects:
        if x.something:
            yield x

I'd go even further:

def iterate_my_objects_if_something(self):
    if not self.my_objects:
        raise StopIteration
    yield from (x for x in self.my_objects if x.something)

edited Jun 19 '20 at 22:10

jonchar

6,183
1
14
19

answered Apr 22 '20 at 19:46

Albert Tugushev

1,504
13
25

1

This is decrepated: https://stackoverflow.com/a/30217723/4986615 – guhur Nov 10 '20 at 09:01

score 0 · Answer 5 · answered Nov 10 '20 at 09:06

This answer was answered in another question: What is the difference between raise StopIteration and a return statement in generators?.

Basically, you just need to return None to end your generator. Raising StopIteration is now deprecated.

So your code can simply be:

def iterate_my_objects_if_something(self):
    if self.my_objects == []:
        return
    for x in self.my_objects:
        if x.something:
            yield x

This is very close to the answer of this question, but it is just more explicit.

Python - Generator case where nothing to return

5 Answers5

Is iteration over generator an issue?

Is the iteration inside generator an issue?

Solution