ignoring keywords on func. call

[Brano Zarnovican]

Hi !

If I define 'f' like this

def f(a):
  print a

then, the call with keywords

f(1, optional=2)

fails. I have to change 'f' to

def f(a, **kwrds):
  print a

to ignore optional parameters.

BUT..

Q: Can you call 'f' with keywords that will be
  ignored, without changing 'f's definition ?

I would like to avoid code like this:
k = {}
k['optional'] = 2
try:
  f(1, **k)
except TypeError:
  f(1)

Also, the problem is that I don't know if the TypeError
was caused by calling 'f' with keywords or somewhere
"inside" f.

You can also say that I need to specify optional parameters
on caller side (not called side).

BranoZ