你应该加一个列来表示重名，这样就容易多了，否则在计算时会出现追溯的麻烦

df['dup'] = df.groupby(['type, invoice_number'])['invoice_number'].cumcount()

如果dup值为而不是0，则可以轻松识别任何invoice是否重复

您可以使用groupby.transform()进行一些算术运算和整数到字符的转换：

import pandas as pd
import numpy as np

df = pd.DataFrame(
    {
        "type": ["H", "D", "H", "D", "H", "D"],
        "invoice": ["ABC", "ABC", "DEF", "DEF", "ABC", "ABC"],
        "amount": ["10", "10", "20", "20", "10", "10"],
    }
)

df["suffix"] = df.groupby("invoice")["type"].transform(
    lambda x: ["_" + chr(65 + int(np.floor(i / 2))) for i in range(len(x))]
)
df["invoice"] = df["invoice"] + df["suffix"]

结果：

type invoice amount suffix
0    H   ABC_A     10     _A
1    D   ABC_A     10     _A
2    H   DEF_A     20     _A
3    D   DEF_A     20     _A
4    H   ABC_B     10     _B
5    D   ABC_B     10     _B

注意，这种方法假设每个事务有2个条目。

更新：

df['invoice number'] = (df['invoice number'] + 
                        (df[df.duplicated(['type', 'invoice number'], keep=False)].groupby('ID').ngroup())
                              .map(dict(enumerate(ascii_uppercase)))).fillna(df['invoice number'])

输出：

ID type invoice number  amount
0   1    H   BOB12192022A    10.0
1   1    D   BOB12192022A    10.0
2   2    H    JOE12182022    30.0
3   2    D    JOE12182022    30.0
4   3    H   BOB12192022B    15.0
5   3    D   BOB12192022B    15.0
6   4    H   BOB12192022C    75.0
7   4    D   BOB12192022C    25.0
8   4    D   BOB12192022C    25.0
9   4    D   BOB12192022C    25.0

试试这个：

from string import ascii_uppercase

df['invoice number'] += np.where(df.groupby(['invoice number'])['invoice number'].transform('count') > 2,
                                 df.groupby(['type', 'invoice number']).cumcount().map(dict(enumerate(ascii_uppercase))), '')

输出：

type invoice number  amount
0    H   BOB12192022A    10.0
1    D   BOB12192022A    10.0
2    H    JOE12182022    30.0
3    D    JOE12182022    30.0
4    H   BOB12192022B    15.0
5    D   BOB12192022B    15.0
6    H   BOB12192022C    50.0
7    D   BOB12192022C    50.0