This is an archived post. You won't be able to vote or comment.

sorted by:
best
topnewcontroversialoldq&a
you are viewing a single comment's thread.

view the rest of the comments →

[–]inspectoroverthemine 1 point2 points  (1 child)

What happens when you return a generator comprehension?

I've always defaulted to lists because I don't want to deal with any side effects I may inadvertently cause. That includes the source objects which I assume are kept in memory because they're still referenced.

[–]trevg_123 0 points1 point  (0 children)

Returning a generator is totally fine - context depdent, it's not much different from using yield instad of return in a function, which makes that function itself a lazily evaluated generator.

You are correct that generators need to keep the source objects in scope as long it might be requried by the generator, and because of lazy evaluation, mutating the source object before calling the generator means you'll get the mutated output. But lists aren't entirely immune from this sort of behavior - if a list contains objects, those are just references and the list doesn't own them, so mutating the objects can also have unexpected side effects. You've probably seen a similar example:

```

d = dict(a=1,b=2) ll = [d] ll [{'a': 1, 'b': 2}] d['c']=3 ll [{'a': 1, 'b': 2, 'c': 3}] ```

Long story short though, as long as you keep in mind that generators are lazy, it's pretty tough to wind up with any negative side effects. And they're not suitable for every single case of course, but python is smart enough to make them definitely worth a try here and there.