Is there any clever in-built function or 开发者_StackOverflow社区something that will return 1
for the min()
example below? (I bet there is a solid reason for it not to return anything, but in my particular case I need it to disregard None
values really bad!)
>>> max([None, 1,2])
2
>>> min([None, 1,2])
>>>
None
is being returned
>>> print min([None, 1,2])
None
>>> None < 1
True
If you want to return 1
you have to filter the None
away:
>>> L = [None, 1, 2]
>>> min(x for x in L if x is not None)
1
using a generator expression:
>>> min(value for value in [None,1,2] if value is not None)
1
eventually, you may use filter:
>>> min(filter(lambda x: x is not None, [None,1,2]))
1
Make None infinite for min():
def noneIsInfinite(value):
if value is None:
return float("inf")
else:
return value
>>> print min([1,2,None], key=noneIsInfinite)
1
Note: this approach works for python 3 as well.
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