您可以使用merge/combine_first：

lcols, rcols = ["c2", "d2"], ["c1", "d1"] #mappings
​
df2["e2"] = (df2.merge(df1, left_on=lcols, right_on=rcols, how="left")
                 ["e1"].combine_first(df2["e2"])
            )

输出：

print(df2)

   idx  a2  b2  c2  d2   e2
0    1  x2  x1   2   2  5.0 # <- updated
1    2  x8  x2   6   9  6.0 # <- updated
2    3  x6  x4   7   9  2.0 # <- updated
3    4  x4  x7   6   8  9.0

在Series.fillna中使用两个DataFrame的左连接，并将不匹配的值替换为原始列：

df2['e2'] = (df2.reset_index()
                .merge(df1, left_on=['c2','d2'], right_on=['c1','d1'], how='left')
                .set_index('index')['e1'].fillna(df2['e2']))

如果df2.index中的默认索引：

df2['e2'] = (df2.merge(df1, left_on=['c2','d2'], right_on=['c1','d1'], how='left')['e1']
                .fillna(df2['e2']))

或者使用Index.map by MulitIndex：

d = df1.set_index(['c1','d1'])['e1'].to_dict()

df2['e2'] = (pd.Series(df2.set_index(['c2','d2']).index.map(d).to_numpy(), index=df2.index)
               .fillna(df2['e2']))
print (df2)
   idx  a2  b2  c2  d2   e2
0    1  x2  x1   2   2  5.0
1    2  x8  x2   6   9  6.0
2    3  x6  x4   7   9  2.0
3    4  x4  x7   6   8  9.0