np.select()条件下的否定

Question

这是我的代码：

import pandas as pd
import numpy as np

df = pd.DataFrame({ 'var1': ['a', 'b', 'c',np.nan, np.nan],
                   'var2': [1, 2, np.nan , 4, np.nan]
                 })



conditions = [
    (not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"]))),
    (pd.isna(df["var1"])) & (pd.isna(df["var2"]))]

choices = ["No missing", "Both missing"]

df['Result'] = np.select(conditions, choices, default=np.nan)

输出：

  File "C:\ProgramData\Anaconda3\lib\site-packages\pandas\core\generic.py", line 1478, in __nonzero__
    f"The truth value of a {type(self).__name__} is ambiguous. "

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

问题在于行(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))。当TRUE在var1和var2中都不是NaN值时，这一行应该给出NaN。这里的问题是否定，因为在没有否定的条件下，没有问题。

问题：如何纠正(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))行，以便当在var1和var2中都不是NaN值时，条件应该给TRUE

Answer 1

尝试：

conditions = [(~pd.isna(df["var1"]) & ~pd.isna(df["var2"])),
               (pd.isna(df["var1"]) &  pd.isna(df["var2"]))]