这是我的代码:import pandas as pdimport numpy as npdf = 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 line 上(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))。这条线应该给出TRUEwhen in bothvar1而var2不是一个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。问题: 如何纠正(not(pd.isna(df["var1"]))) & (not(pd.isna(df["var2"])))线,以防万一在这两种情况下var1都不var2是NaN条件应该给出的值TRUE?
1 回答

摇曳的蔷薇
TA贡献1793条经验 获得超6个赞
尝试:
conditions = [(~pd.isna(df["var1"]) & ~pd.isna(df["var2"])), (pd.isna(df["var1"]) & pd.isna(df["var2"]))]
添加回答
举报
0/150
提交
取消