hetchkay wrote:
> 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
> obj.items())
> else:
> return obj
>
> This works fine except if obj is a namedtuple. A namedtuple object has
> different constructor signature from tuple:
>>>> tuple([1,2])
> (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)) ].
I don't think there is a safe way to do this short of a comprehensive list
of namedtuple classes. In practice it may be sufficient to check the duck
type of the tuple subclass, e. g.:
elif isinstance(obj, (list, set)):
return obj.__class__(map(handleObj, obj))
elif isinstance(obj, tuple):
try:
make = obj._make
except AttributeError:
make = obj.__class__
return make(handleObj(item) for item in obj)
Peter
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