yield_all needed in Python

[Douglas Alan]

Andrew Dalke <dalke at dalkescientific.com> writes:

> On Mon, 28 Feb 2005 18:25:51 -0500, Douglas Alan wrote:

>> While writing a generator, I was just thinking how Python needs a
>> "yield_all" statement.  With the help of Google, I found a
>> pre-existing discussion on this from a while back in the
>> Lightweight Languages mailing list.  I'll repost it here in order
>> to improve the chances of this enhancement actually happening
>> someday.

> You should also have looked for the responses to that. Tim Peter's
> response is available from

>   http://aspn.activestate.com/ASPN/Mail/Message/624273

[...]

> Here is the most relevant parts.

[...]

>    BTW, Python almost never worries about worst-case behavior, and people
>    using Python dicts instead of, e.g., balanced trees, get to carry their
>    shame home with them hours earlier each day <wink> .

If you'll reread what I wrote, you'll see that I'm not concerned with
performance, but rather my concern is that I want the syntactic sugar.
I'm tired of writing code that looks like

   def foogen(arg1):

      def foogen1(arg2):
         # Some code here

      # Some code here
      for e in foogen1(arg3): yield e
      # Some code here
      for e in foogen1(arg4): yield e
      # Some code here
      for e in foogen1(arg5): yield e      
      # Some code here
      for e in foogen1(arg6): yield e      

when it would be much prettier and easier to read if it looked like:

   def foogen(arg1):

      def foogen1(arg2):
         # Some code here

      # Some code here
      yield_all foogen1(arg3)
      # Some code here
      yield_all foogen1(arg4)
      # Some code here
      yield_all foogen1(arg5)
      # Some code here
      yield_all foogen1(arg6)

|>oug