用 None 列出 Python 中的最小值？
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List minimum in Python with None?
提问by c00kiemonster
Is there any clever in-built function or something that will return 1for 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 Nonevalues really bad!)
是否有任何巧妙的内置函数或会1为min()下面的示例返回的东西？（我敢打赌它不返回任何东西是有充分理由的，但在我的特殊情况下，我需要它忽略None非常糟糕的值！）
>>> max([None, 1,2])
2
>>> min([None, 1,2])
>>>
回答by nosklo
Noneis being returned
None正在被退回
>>> print min([None, 1,2])
None
>>> None < 1
True
If you want to return 1you have to filter the Noneaway:
如果你想返回，1你必须过滤None掉：
>>> L = [None, 1, 2]
>>> min(x for x in L if x is not None)
1
回答by Adrien Plisson
using a generator expression:
使用生成器表达式：
>>> min(value for value in [None,1,2] if value is not None)
1
eventually, you may use filter:
最终，您可以使用过滤器：
>>> min(filter(lambda x: x is not None, [None,1,2]))
1
回答by gregory
Make None infinite for min():
为 min() 设置 None 无限：
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.
注意：这种方法也适用于 python 3。
