import pandas as pd

df1 = pd.DataFrame([['update me', 'leave', 'take the other nan']], index=[0], columns=['A', 'B', 'C'])

df2 = pd.DataFrame([['update with me', pd.np.nan, 'stay out']], index=[0], columns=['A', 'C', 'D'])

# want something like: df1.update_using_nans_please(df2) # to return:

# pd.DataFrame([['update with me', 'leave', pd.np.nan]], columns=['A', 'B', 'C'])

df1.update(df2.fillna('nan'))

df1.replace('nan', pd.np.nan) # Any way to do it without this hack?

解决办法：

作为替代方案，我们可以使用一点索引魔法pd.concat。然而，由于上述原因，这不执行就地修改。

a, b = df1.columns, df2.columns

pd.concat([df1[a.difference(b)], df2[a.intersection(b)]], axis=1)

B A C

0 leave update with me NaN

要保留原始订单，

pd.concat([df1[a.difference(b)], df2[a.intersection(b)]], axis=1).reindex_like(df1)

A B C

0 update with me leave NaN