[martin_k]

As i said, f() returns a generator object in this case. So you're calling list on a generator that yields None once, the result of which is a list that contains None. If you change the `yield` to `yield "foobar"`, list(f()) will get you `["foobar"]`.

Edit: Perhaps, to clear things up more, about where the value 7 went- your snippet can be expressed without using lambdas as

  >>> def g(x):
  ...  return 7
  ... 
  >>> def f():
  ...  v = (yield)
  ...  g(v)
  ... 
  >>> list(f())
  [None]

Note that there's no return before g(v) because generators (at least in 2.7) can only yield, not return a value.

[pyre]

I'm thinking that the parent is expecting 'yield' to return 'None' into g(), and then for g() to return 7 since it doesn't do anything with the 'x' parameter, and is confused why that isn't the case.

[martin_k]

Yes, but I think this is cleared up if you consider that calling a generator function does not execute its body, but return generator object. It doesn't really matter if it's a lambda or a regular named function.

[cool-RR]

Sure, I understand all of that. I'm just saying it's unusual to see a lambda function throwing away the value like that.

[martin_k]

The value of the lambda is the generator object. How else could you call list() on it if it was thrown away?

[cool-RR]

You're just using the word "value" in a different way. I meant "value" as in the 7.

[martin_k]

You're right. I am still referring to that particular part of your blog post:

  >> So that’s the only case I can think of where Python
  >> completely throws away the value of a lambda function.