如何从NumPy数组中删除NaN值?
[1, 2, NaN, 4, NaN, 8] ⟶ [1, 2, 4, 8]
只需填写
x = numpy.array([
[0.99929941, 0.84724713, -0.1500044],
[-0.79709026, numpy.NaN, -0.4406645],
[-0.3599013, -0.63565744, -0.70251352]])
x[numpy.isnan(x)] = .555
print(x)
# [[ 0.99929941 0.84724713 -0.1500044 ]
# [-0.79709026 0.555 -0.4406645 ]
# [-0.3599013 -0.63565744 -0.70251352]]
pandas引入了一种选项,可以将所有数据类型转换为缺失值。
np.isnan()
函数不支持所有数据类型,例如:
>>> import numpy as np
>>> values = [np.nan, "x", "y"]
>>> np.isnan(values)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
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''
pd.isna()
和pd.notna()
函数与许多数据类型兼容,而pandas引入了一个pd.NA
值:
>>> import numpy as np
>>> import pandas as pd
>>> values = pd.Series([np.nan, "x", "y"])
>>> values
0 NaN
1 x
2 y
dtype: object
>>> values.loc[pd.isna(values)]
0 NaN
dtype: object
>>> values.loc[pd.isna(values)] = pd.NA
>>> values.loc[pd.isna(values)]
0 <NA>
dtype: object
>>> values
0 <NA>
1 x
2 y
dtype: object
#
# using map with lambda, or a list comprehension
#
>>> values = [np.nan, "x", "y"]
>>> list(map(lambda x: pd.NA if pd.isna(x) else x, values))
[<NA>, 'x', 'y']
>>> [pd.NA if pd.isna(x) else x for x in values]
[<NA>, 'x', 'y']
最简单的方法是:
numpy.nan_to_num(x)
文档:https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html