question on namedtuple

hetchkay hetchkay at
Thu Apr 1 20:35:14 CEST 2010

For purposes I don't want to go into here, I have the following code:
def handleObj(obj):
  if isinstance(obj, MyType):
     return obj.handle()
  elif isinstance(obj, (list, tuple, set)):
     return obj.__class__(map (handleObj, obj))
  elif isinstance(obj, dict):
     return obj.__class__((handleObj(k), handleObj(v)) for k, v in
     return obj

This works fine except if obj is a namedtuple. A namedtuple object has
different constructor signature from tuple:
>>> tuple([1,2])
>>> collections.namedtuple("sample", "a, b")([1, 2])
Traceback (most recent call last):
  File "CommandConsole", line 1, in <module>
TypeError: __new__() takes exactly 3 arguments (2 given)
>>> collections.namedtuple("sample", "a, b")(1, 2)
sample(a=1, b=2)

Is there any easy way of knowing that the obj is a namedtuple and not
a plain tuple [so that I could use obj.__class__(*map(handleObj, obj))
instead of obj.__class__(map(handleObj, obj)) ].

Thanks in advance for your help.


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