kj wrote:
In <mailman.158.1242328059.8015.python-l...@python.org> Dave Angel
<da...@ieee.org> writes:
kj wrote:
In <mailman.113.1242254593.8015.python-l...@python.org> Terry Reedy
<tjre...@udel.edu> 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
You already had a better answer from Chris Rebert:
def foo(**kwargs):
return bar(kwargs)
kwargs at this point is exactly a dictionary of the named arguments to foo.
I disagree. If I defined foo as you show above, then there is no
error checking on the named parameters passed to foo; anything
goes.
Because if you try to do anything in this function, you'll probably be
adding more local variables. And then they'd be passed to bar as well.
That problem is easily solved: just make "x = locals()" the first
statement in the definition of foo.
kynn
Well, you caught me by surprise. The name and number of locals is
determined at compile time, not at run time. But when I run an
experiment, it seems you're right, that the dict changes as the function
executes.
In Python 2.62 at least, the following function gives the following results:
def myfunc(x, y):
print locals()
s = 42
print locals()
myfunc(y=12, x=33)
{'y': 12, 'x': 33}
{'y': 12, 'x': 33, 's': 42, 'r': {'y': 12, 'x': 33}}
{'y': 12, 'x': 33}
So, if one wanted to call bar, but not till after one had defined new
locals, you could do the following:
def foo(x=None, y=None, z=None):
args = locals().copy()
a = new stuff
dosomestuff with a, and maybe y
bar(args)
A shallow copy would be adequate in this case.
Thanks for the correction.
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