Newbie question...

[Quinn Dunkan]

On 28 Sep 2000 07:28:18 GMT, Mike 'Cat' Perkonigg <blablu at gmx.net> wrote:
>ivnowa at hvision.nl (Hans Nowak) wrote in <39D1FC71.6768 at hvision.nl>:
>
>>y must have an initial value... try inserting  y = 0  before the for.
>>Aside from that, you probably want to use args.values() rather than
>>args.keys().
>>
>>This won't work for your second line (good="a", etc) though, because it
>>uses strings for values, and y is initialized as an integer.
>>
>>If you want a more generic function, you could try:
>>
>>def adder3(**args):
>>    return reduce(lambda x, y, a=args: x+y, args.values())
>>
>>>>> print adder3(good=1, bad=2, ugly=3)
>>6
>>>>> print adder3(good="a", bad="b", ugly="c")
>>"abc"
>
>Why do you add 'a=args' to the lambda parameter list? args is in local 
>namespace and is not a parameter of lambda but a parameter of reduce.

Try taking it out, and see what happens :)

'args' is *not* in the local namespace of the lambda.  You're probably coming
from a lexically scoped language like scheme, but in python there are only two
namespaces: local and global (well, and __builtins__).  lambda won't see
'args' because it's not local or global to the lambda.  python's scoping rules
are a bit unusual, so it's a faq item.

a = 1
# we can see 'a' (local and global) from here
def f():
    b = 2
    # we can see 'b' (local) and 'a' (global) from here
    def g():
        c = 3
        # we can see 'c' (local) and 'a' (global) from here (but not 'b'!)

Passing things in via default args is the usual python trick to emulate
lexical scoping.