Python 3.8: definition with if/else condition and return/yield (always returns a generator)

Question

[Python 3.8 | Spyder]

I was trying to introduce a simple definition that would either return [1,2,3] or yield the numbers sequentially.

The minimum working code is as follows:

def example(condition):
    if condition:
        yield 1
        yield 2
        yield 3
    else:
        return [1,2,3]

If one attempts to use the def, it will always return a generator. Even if the return appears first in the if/else pair:

def example(condition):
    if not condition:
        return [1,2,3]

    else:
        yield 1
        yield 2
        yield 3

It seems python is not ignoring the presence of the yields even if condition = False. This is an unexpected behaviour.

Answer 1

You can return the generator in the form of a generator expression:

>>> def example(condition):
...     if condition:
...         return (i for i in (1, 2, 3))
...     else:
...         return [1, 2, 3]

>>> example(1)
<generator object example.<locals>.<genexpr> at 0x000000000257BBA0>

>>> example(0)
[1, 2, 3]

Or you can define the generator separately:

>>> def g():
...     yield 1
...     yield 2
...     yield 3
...
>>> def example(condition):
...     if condition:
...         return g()
...     else:
...         return [1, 2, 3]

>>> example(1)
<generator object g at 0x000000000257BBA0>

>>> example(0)
[1, 2, 3]