In <[email protected]> Terry Reedy
<[email protected]> writes:
>kj wrote:
>>
>> Suppose I have the following:
>>
>> def foo(x=None, y=None, z=None):
>> d = {"x": x, "y": y, "z": z}
>> return bar(d)
>>
>> I.e. foo takes a whole bunch of named arguments and ends up calling
>> a function bar that takes a single dictionary as argument, and this
>> dictionary has the same keys as in foo's signature, so to speak.
>>
>> Is there some builtin variable that would be the same as the variable
>> d, and would thus obviate the need to explicitly bind d?
>Use the built-in function locals()
> >>> def f(a,b):
> x=locals()
> print(x)
> >>> f(1,2)
>{'a': 1, 'b': 2}
That's *exactly* what I was looking for. Thanks!
kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded.
--
http://mail.python.org/mailman/listinfo/python-list
