从包含None元素的列表中获取最大值

Question

我正在尝试使用以下代码从包含nonetype的列表对象中获取最大值:

import numpy as np

LIST = [1,2,3,4,5,None]
np.nanmax(LIST)

但我收到此错误消息

'>=' not supported between instances of 'int' and 'NoneType'

显然np.nanmax()不适用None.从包含None值的列表对象中获取最大值的替代方法是什么？

Answer 1

在Python 2中

max([i for i in LIST if i is not None])

Python 3 及以上版本很简单

max(filter(None.__ne__, LIST))

或者更详细地说

max(filter(lambda v: v is not None, LIST))

Answer 2

One approach could be -

max([i for i in LIST if i is not None])

Sample runs -

In [184]: LIST = [1,2,3,4,5,None]

In [185]: max([i for i in LIST if i is not None])
Out[185]: 5

In [186]: LIST = [1,2,3,4,5,None, 6, 9]

In [187]: max([i for i in LIST if i is not None])
Out[187]: 9

Based on comments from OP, it seems we could have an input list of all Nones and for that special case, it output should be [None, None, None]. For the otherwise case, the output would be the scalar max value. So, to solve for such a scenario, we could do -

a = [i for i in LIST if i is not None]
out = [None]*3 if len(a)==0 else max(a)

Answer 3

首先,转换为numpy数组.指定dtype=np.floatX,所有这些None将被np.nan输入到类型.

import numpy as np

lst = [1, 2, 3, 4, 5, None]

x = np.array(lst, dtype=np.float64)
print(x)
array([  1.,   2.,   3.,   4.,   5.,  nan])

现在,致电np.nanmax:

print(np.nanmax(x))
5.0

要将max作为整数返回,您可以使用.astype:

print(np.nanmax(x).astype(int)) # or int(np.nanmax(x))
5

这种方法适用于v1.13.1.