pandas 在 python 中删除 NaN 值的列表的中位数
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Median of a list with NaN values removed, in python
提问by wolfsatthedoor
Is it possible to calculate the median of a list without explicitly removing the NaN's, but rather, ignoring them?
是否可以在不明确删除 NaN 的情况下计算列表的中位数，而是忽略它们？
I want median([1,2,3,NaN,NaN,NaN,NaN,NaN,NaN])to be 2, not NaN.
我想median([1,2,3,NaN,NaN,NaN,NaN,NaN,NaN])成为 2，而不是 NaN。
回答by Warren Weckesser
numpy 1.9.0 has the function nanmedian:
numpy 1.9.0 具有以下功能nanmedian：
nanmedian(a, axis=None, out=None, overwrite_input=False, keepdims=False)
Compute the median along the specified axis, while ignoring NaNs.
Returns the median of the array elements.
.. versionadded:: 1.9.0
E.g.
例如
>>> from numpy import nanmedian, NaN
>>> nanmedian([1,2,3,NaN,NaN,NaN,NaN,NaN,NaN])
2.0
If you can't use version 1.9.0 of numpy, something like @Parker's answer will work; e.g.
如果您不能使用 numpy 的 1.9.0 版，则可以使用@Parker 的回答之类的方法；例如
>>> import numpy as np
>>> x = np.array([1,2,3,NaN,NaN,NaN,NaN,NaN,NaN])
>>> np.median(x[~np.isnan(x)])
2.0
or
或者
>>> np.median(x[np.isfinite(x)])
2.0
(When applied to a boolean array, ~is the unary operator notation for not.)
（当应用于布尔数组时，~是 的一元运算符符号not。）
回答by Parker
I would clean the list of all NaN's, and then get the median of the cleaned list. There're two ways that come to mind. If you're using the numpy library, you can do:
我会清理所有 NaN 的列表，然后获取清理列表的中位数。我想到了两种方法。如果您使用的是 numpy 库，则可以执行以下操作：
x = x[numpy.logical_not(numpy.isnan(x))]where xis the list you want to get the median of
x = x[numpy.logical_not(numpy.isnan(x))]x您想要获得中位数的列表
在哪里
Or, if you just want to use the included libraries you can do:
或者，如果您只想使用包含的库，您可以执行以下操作：
import math
x = [value for value in x if not math.isnan(value)]
Then to get the median just use the cleaned list: `median(x)``
然后要获得中位数，只需使用清理过的列表：`median(x)``
