How do I dynamically create functions without lambda?

[Steven Bethard]

Russell wrote:
> I want my code to be Python 3000 compliant, and hear
> that lambda is being eliminated. The problem is that I
> want to partially bind an existing function with a value
> "foo" that isn't known until run-time:
> 
>    someobject.newfunc = lambda x: f(foo, x)
> 
> The reason a nested function doesn't work for this is
> that it is, well, dynamic. I don't know how many times
> or with what foo's this will be done.

I don't understand this argument here.  The code above is almost exactly 
equivalent to:

     def newfunc(x):
         return f(foo, x)
     someobject.newfunc = newfunc

Observe:

 >>> class C(object):
...     pass
...
 >>> someobject = C()
 >>> someobject.newfunc = lambda x: f(foo, x)
 >>> import dis
 >>> dis.dis(someobject.newfunc)
   1           0 LOAD_GLOBAL              0 (f)
               3 LOAD_GLOBAL              1 (foo)
               6 LOAD_FAST                0 (x)
               9 CALL_FUNCTION            2
              12 RETURN_VALUE
 >>> def newfunc(x):
...     return f(foo, x)
...
 >>> someobject.newfunc = newfunc
 >>> dis.dis(someobject.newfunc)
   2           0 LOAD_GLOBAL              0 (f)
               3 LOAD_GLOBAL              1 (foo)
               6 LOAD_FAST                0 (x)
               9 CALL_FUNCTION            2
              12 RETURN_VALUE

Note that both the lambda and the function have exactly the same 
byte-code.  The only real difference is that if you use a def-statement 
instead of a lambda, your function will get a real name, "newfunc", 
instead of <lambda>.

STeVe