您可以使用groupby.cumcount来递增数字，并将其加到数字部分：

g = df.groupby('Channel')

# ffill per group
# extract letter and number part
df2 = (g['last unique id'].ffill()
       .str.extract(r'(\D+)(\d+)')
       )

# convert number part to integer
# add cumcount, merge back as string
df['last unique id'] = (df2[0]
 .add(df2[1].astype(int)
            .add(g.cumcount())
            .astype(str)
      )
 )

print(df)

输出量：

Channel last unique id
0   MYNTRA       MN351370
1    NYKAA       NYK38219
2    NYKAA       NYK38220
3    NYKAA       NYK38221
4    NYKAA       NYK38222
5    NYKAA       NYK38223
6   MYNTRA       MN351371
7   MYNTRA       MN351372
8   MYNTRA       MN351373
9   MYNTRA       MN351374
10  MYNTRA       MN351375
11  MYNTRA       MN351376

示例

data = {'col1': {0: 'A', 1: 'B', 2: 'A', 3: 'A', 4: 'B', 5: 'B'},
 'col2': {0: 'A001', 1: 'BC020', 2: None, 3: None, 4: 'BC021', 5: None}}
df = pd.DataFrame(data)

df值

col1 col2
0   A   A001
1   B   BC020
2   A   None
3   A   None
4   B   BC021
5   B   None

代码

df[['col3', 'col4']] = df.groupby('col1')['col2'].ffill().str.extract('(\D+)(\d+)')
df['col4'] = df['col4'].astype('int') + df.groupby(['col1', 'col4']).cumcount()
df['col2'] = df['col2'].fillna(df['col3'] + df['col4'].astype('str').str.zfill(3))
df = df.drop(['col3', 'col4'], axis=1)

结果（df）：

col1 col2
0   A   A001
1   B   BC020
2   A   A002
3   A   A003
4   B   BC021
5   B   BC022

下面是如何通过填充零获得所需的输出，以使id始终保持固定长度11。
第一个