np.select()条件中的否定 [英] Negation in np.select() condition

问题描述

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().

问题出在行(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))上.当同时在var1和var2中使用时，此行应使用TRUE而不是NaN值.这里的问题在于否定，因为没有否定的条件就没有问题.

Problem is with line (not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"]))). This line should give TRUE when in both var1 and var2 in not a NaN value. Problem here is with negation, because with conditions without negation there is no problem.

问题:如何纠正(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))行，以便在var1和var2两者中都没有NaN值的情况下应该给出TRUE? >

Question: How can to correct (not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"]))) line so in case when in both var1 and var2 in not a NaN value the condition should give TRUE?

推荐答案

尝试:

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

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