How to elegantly define a generator producing an empty sequence

Asked Oct 21 '19 at 22:27

Active Oct 21 '19 at 22:32

Viewed 33 times

Because of the yield keyword, the Python compiler understands the method g in the code below as being a generator:

class B(A):
    def g(self):
        yield 1

However, I would like the A base class to provide a default implementation for g producing an empty sequence such as:

list(B().g())
--> [1]

list(A().g())
--> []

To achieve that, I define A as:

class A:
    def g(self):
        return; yield # do NOT remove the yield keyword:
                      # it is necessary for Python to
                      # understand `g` as a generator

It works, but I don't find that very elegant since, barring the comment, it is tempting to remove the unreachable yield statement.

So, is there a better, more pythonic way of defining a generator producing an empty sequence?

asked Oct 21 '19 at 22:27

Sylvain Leroux

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`return iter(())`? – jonrsharpe Oct 21 '19 at 22:29
Maybe @jorn. You may post that as an answer and we will see how other people vote – Sylvain Leroux Oct 21 '19 at 22:31
What do you mean *"maybe"*? This is the problem with *"elegantly"*, *"pythonic"*, and even *"better"* - what do you *mean*? Why does it matter how other people vote? – jonrsharpe Oct 21 '19 at 22:31
Actually, I'm not even sure it is absolutely necessary to define `g` as a generator. Maybe `return ()` will do the trick. – Sylvain Leroux Oct 21 '19 at 22:32

How to elegantly define a generator producing an empty sequence

0 Answers0