如何连接两个ID不匹配的数据帧,并创建新列来表示来自哪个数据帧ID?

弗斯特 concat 数据帧在一起:

df = (pd.concat([df1, df2],  keys=('df1','df2'))
        .rename_axis(('df_name','idx'))
        .reset_index(level=1, drop=True)
        .reset_index())

print (df)
  df_name  id  column1  column2
0     df1   1       30       90
1     df1   2        1        2
2     df2   1       30       90
3     df2   3        1        2

那就都一样了 id 以下内容:

a = df1.merge(df2, on='id')['id']

最后一个筛选依据 isin :

df = df[~df['id'].isin(a)]
print (df)
  df_name  id  column1  column2
1     df1   2        1        2
3     df2   3        1        2

编辑:

类似的解决方案,如@w-b,只添加了参数 身份证件 和 suffixes :

df = (df1.merge(df2,indicator=True,how='outer', on='id', suffixes=('_df1','_df2'))
         .query("_merge != 'both'"))
df['_merge'] = df['_merge'].map({'left_only':'df1','right_only':'df2'})

print (df)
   id  column1_df1  column2_df1  column1_df2  column2_df2 _merge
1   2          1.0          2.0          NaN          NaN    df1
2   3          NaN          NaN          1.0          2.0    df2

如果要所有行,也要相同的行 身份证件 :

df['_merge'] = df['_merge'].map({'left_only':'df1','right_only':'df2', 'both':'df1+df2'})

print (df)
   id  column1_df1  column2_df1  column1_df2  column2_df2   _merge
0   1         30.0         90.0         30.0         90.0  df1+df2
1   2          1.0          2.0          NaN          NaN      df1
2   3          NaN          NaN          1.0          2.0      df2

让我们一起做 merge

df=df1.merge(df2,indicator = True,how='outer').loc[lambda x : x['_merge'].ne('both')]
df['df_name']=df['_merge'].map({'left_only':'df1','right_only':'df2'})
df
Out[328]: 
   id  column1  column2      _merge df_name
1   2        1        2   left_only     df1
2   3        1        2  right_only     df2