您可以使用以下列表理解:

You could use the following list comprehension:

import numpy as np

[[0 if np.isnan(j) else j for j in i] for i in l_of_l]
# [[1, 2, 3], [0, 0, 0], [3, 4, 5]]

如果您想避免导入 numpy ，尽管数据表明您应该使用它，则实际上可以执行以下操作:

If you want to avoid importing numpy, though the data suggests that you should be using it, you could actually do the same with:

[[0 if j!=j else j for j in i] for i in l_of_l]
# [[1, 2, 3], [0, 0, 0], [3, 4, 5]]

根据定义，NaN 永远不等于自己

This works as by definition NaNs are never equal to themselves

或直接构建一个 numpy 数组并使用 nan_to_num :

Or directly build a numpy array and use nan_to_num:

np.nan_to_num(np.array(l_of_l))

array([[1., 2., 3.],
       [0., 0., 0.],
       [3., 4., 5.]])