yield_all needed in Python

[Terry Reedy]

"Douglas Alan" <nessus at mit.edu> wrote in message 
news:lck6osnu68.fsf at gaffa.mit.edu...
>     We can shorten the code--and make it run in O(N) time--by adding a 
> new
>     keyword to replace the "for v in ...: yield v" pattern:

Maybe.  Until you define the semantics of yield_all and at least outline an 
implementation, I am not convinced of 'run in o(n) time'.  There was once a 
several-post discussion of a related idea of having yield somehow, 
magically, skip intermediate generators that only yielded value on up, 
without tranformation.  But it was never clear how to do this practically 
without negatively impacting all generators.  Cetainly, if <yield_all 
iterator> == <for i in iterator: yield i>, I don't see how anything is 
gained except for a few keystrokes.  If <yield_all iterator> == <yield 
list(i for i in iterator)> then the replacement is a semantic change.

>       def in_order(self):
>           if self.left is not None:
>               yield_all self.left.in_order():
>           yield self.value
>           if self.right is not None:
>               yield_all self.right.in_order():

If and when I write a text-based double-recursion to iteration transformer, 
a pseudokeyword might be be an idea for indicating that stacked yields are 
identify functions and therefore bypassable.

Terry J. Reedy