python之从另一个 DataFrame 替换 pandas.DataFrame 中的值的优雅方法

我有一个数据框，我想用另一个数据框中的值替换一列中的值。

df = pd.DataFrame({'id1': [1001,1002,1001,1003,1004,1005,1002,1006], 
                   'value1': ["a","b","c","d","e","f","g","h"], 
                   'value3': ["yes","no","yes","no","no","no","yes","no"]}) 
 
dfReplace = pd.DataFrame({'id2': [1001,1002], 
                   'value2': ["rep1","rep2"]}) 

我需要使用带有公共(public)键的 groupby，当前的解决方案是使用循环。有没有更优雅(更快)的方法来使用 .map(apply) 等来执行此操作？我希望初始使用 pd.update()，但似乎不是正确的方法。

groups = dfReplace.groupby(['id2']) 
 
for key, group in groups: 
    df.loc[df['id1']==key,'value1']=group['value2'].values 

输出

df 
    id1   value1 value3 
0   1001  rep1   yes 
1   1002  rep2   no 
2   1001  rep1   yes 
3   1003  d      no 
4   1004  e      no 
5   1005  f      no 
6   1002  rep2   yes 
7   1006  h      no 

请您参考如下方法:

尝试 merge() :

merge = df.merge(dfReplace, left_on='id1', right_on='id2', how='left') 
print(merge) 
 
merge.ix[(merge.id1 == merge.id2), 'value1'] = merge.value2 
print(merge) 
 
del merge['id2'] 
del merge['value2'] 
print(merge) 

输出:

    id1 value1 value3   id2 value2 
0  1001      a    yes  1001   rep1 
1  1002      b     no  1002   rep2 
2  1001      c    yes  1001   rep1 
3  1003      d     no   NaN    NaN 
4  1004      e     no   NaN    NaN 
5  1005      f     no   NaN    NaN 
6  1002      g    yes  1002   rep2 
7  1006      h     no   NaN    NaN 
 
    id1 value1 value3   id2 value2 
0  1001   rep1    yes  1001   rep1 
1  1002   rep2     no  1002   rep2 
2  1001   rep1    yes  1001   rep1 
3  1003      d     no   NaN    NaN 
4  1004      e     no   NaN    NaN 
5  1005      f     no   NaN    NaN 
6  1002   rep2    yes  1002   rep2 
7  1006      h     no   NaN    NaN 
 
    id1 value1 value3 
0  1001   rep1    yes 
1  1002   rep2     no 
2  1001   rep1    yes 
3  1003      d     no 
4  1004      e     no 
5  1005      f     no 
6  1002   rep2    yes 
7  1006      h     no