如何在 pandas 中放置 2 个不同的数据帧

Question

我想合并 2 个具有不同顺序的文件：

我想在file1和file2下面合并：

文件1：

col1   col2   col3
A001   B001   C001
A002   B002   C002
A003   B003   C003
A004   B004   C004
A005   B005   C005
A006   B006   C006

文件2：

col1   col2
A001   8
A002   2
A003   4
A004   1
A005   8
A006   3
B001   7
B002   4
B003   10
B004   11
B005   8
B006   3
C001   2
C002   9
C003   8
C004   1
C005   7
C006   6

获得以下信息：

col1   col2   col3  col4  col5  col6
A001   8      B001  7     C001  2
A002   2      B002  4     C002  9
A003   4      B003  10    C003  8
A004   1      B004  11    C004  1
A005   8      B005  8     C005  7
A006   3      B006  3     C006  6

我非常感谢你的帮助:)

Answer 1

我会做什么来replace

df=pd.concat([file1,file1.replace(dict(zip(file2.col1,file2.col2))).add_suffix('_1')],axis=1).\
   sort_index(axis=1)


   col1  col1_1  col2  col2_1  col3  col3_1
0  A001       8  B001       7  C001       2
1  A002       2  B002       4  C002       9
2  A003       4  B003      10  C003       8
3  A004       1  B004      11  C004       1
4  A005       8  B005       8  C005       7
5  A006       3  B006       3  C006       6

Answer 2

这是使用带有enumerate和Series.map的for loop的更具可读性的解决方案：

for idx, col in enumerate(df1.columns):
    df1[f'{col}_{idx+1}'] = df1[col].map(df2.set_index('col1')['col2'])

df1 = df1.sort_index(axis='columns')

   col1  col1_1  col2  col2_2  col3  col3_3
0  A001       8  B001       7  C001       2
1  A002       2  B002       4  C002       9
2  A003       4  B003      10  C003       8
3  A004       1  B004      11  C004       1
4  A005       8  B005       8  C005       7
5  A006       3  B006       3  C006       6

Answer 3

除了我喜欢它之外，没有人声称这更好。

def f():
    m = dict(zip(*map(file2.get, file2)))
    for i, c in enumerate(file1):
        yield file1[c].rename(f'col{i * 2 + 1}')
        yield file1[c].replace(m).rename(f'col{i * 2 + 2}')

pd.concat(f(), axis=1)

   col1 col2  col3 col4  col5  col6
0  A001    8  B001    7  C001     2
1  A002    2  B002    4  C002     9
2  A003    4  B003   10  C003     8
3  A004    1  B004   11  C004     1
4  A005    8  B005    8  C005     7
5  A006    3  B006    3  C006     6

这真的让我很困扰，没有一种超级简单的方法可以做到这一点。
这是另一个配方

m = dict(zip(*map(file2.get, file2)))

pd.concat({
    (c, i): a
    for c in file1
    for i, a in enumerate([file1[c], file1[c].replace(m)])
}, axis=1)

   col1     col2      col3   
      0  1     0   1     0  1
0  A001  8  B001   7  C001  2
1  A002  2  B002   4  C002  9
2  A003  4  B003  10  C003  8
3  A004  1  B004  11  C004  1
4  A005  8  B005   8  C005  7
5  A006  3  B006   3  C006  6