如何从列表Python/NumPy中删除Nan

Question

我有一个列出值的列表,我得到的值之一是'nan'

countries= [nan, 'USA', 'UK', 'France']

我试图删除它,但我每次都会收到错误

cleanedList = [x for x in countries if (math.isnan(x) == True)]
TypeError: a float is required

当我尝试这个时:

cleanedList = cities[np.logical_not(np.isnan(countries))]
cleanedList = cities[~np.isnan(countries)]

TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

Answer 1

问题已经改变,所以有答案:

无法使用字符串进行测试,math.isnan因为这需要一个float参数.在countries列表中,您有浮点数和字符串.

在您的情况下,以下内容应该足够:

cleanedList = [x for x in countries if str(x) != 'nan']

老答案

在你的countries列表中,文字'nan'是一个字符串而不是Python浮点数nan,相当于:

float('NaN')

在您的情况下,以下内容应该足够:

cleanedList = [x for x in countries if x != 'nan']

Answer 2

问题来自于np.isnan()没有正确处理字符串值的事实.例如,如果你这样做:

np.isnan("A")
TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

但是,pandas版本pd.isnull()适用于数字和字符串值:

pd.isnull("A")
> False

pd.isnull(3)
> False

pd.isnull(np.nan)
> True

pd.isnull(None)
> True

Answer 3

导入numpy为np

import numpy as np

mylist = [3, 4, 5, np.nan]
l = [x for x in mylist if ~np.isnan(x)]

这应该删除所有NaN.当然,我认为它不是一个字符串,而是实际的NaN.

Answer 4

我喜欢从这样的列表中删除缺失值：

list_no_nan = [x for x in list_with_nan if pd.notnull(x)]

Answer 5

如果您检查元素类型

type(countries[1])

结果将是<class float> 这样您就可以使用以下代码：

[i for i in countries if type(i) is not float]

Answer 6

使用numpy 花式索引:

In [29]: countries=np.asarray(countries)

In [30]: countries[countries!='nan']
Out[30]: 
array(['USA', 'UK', 'France'], 
      dtype='|S6')

Answer 7

使用您的示例，其中...

countries= [nan, 'USA', 'UK', 'France']

由于nan不等于nan（nan！= nan）且country [0] = nan，因此应注意以下几点：

countries[0] == countries[0]
False

然而，

countries[1] == countries[1]
True
countries[2] == countries[2]
True
countries[3] == countries[3]
True

因此，以下应工作：

cleanedList = [x for x in countries if x == x]

Answer 8

直接去除nan值的一种方法是：

import numpy as np    
countries.remove(np.nan)