您可以尝试以下操作：

from string import ascii_uppercase as letters

df["Date_Time_2"] = pd.to_datetime(df["Date_Time_2"])
for n, (_, gdf) in enumerate(df.sort_values("Date_Time_2").groupby("Group")):
    nths = gdf.groupby("Date_Time_2", as_index=False).ngroup()
    df.loc[gdf[nths == n].index, "New"] = str(n + 1)
for i, c in zip(gdf[nths > n].index, letters):
    df.at[i, "New"] = f"{n + 1}_{c}"

• 首先确保列Date_Time_2包含日期时间。
• 然后，在沿Date_Time_2排序后，按Group对df分组。
• 然后在每个组中识别属于n的第Date_Time_2子组（从0开始）的索引，并在相应的New列行上设置n + 1。
• 然后取最后一组，将字母值添加到New列。

也许你得把最后一部分换成

for k, c in zip(range(n + 1, nths.max() + 1), letters):
    df.loc[gdf[nths == k].index, "New"] = f"{n + 1}_{c}"

字母值是否也应该分组。
问题中样本的结果：

ID Group       Date_Time_1         Date_Time_2  Difference New_Column  New
0   123     A  14-10-2021 15:19 2021-10-14 15:32:00          13      First    1
1   123     A  14-10-2021 15:19 2021-10-14 15:36:00          17        NaN  NaN
2   123     A  14-10-2021 15:19 2021-10-14 15:37:00          18        NaN  NaN
3   123     A  14-10-2021 15:19 2021-10-14 16:29:00          70        NaN  NaN
4   123     A  14-10-2021 15:19 2021-10-14 17:04:00         105        NaN  NaN
5   123     B  14-10-2021 15:21 2021-10-14 15:32:00          11        NaN  NaN
6   123     B  14-10-2021 15:21 2021-10-14 15:36:00          15     Second    2
7   123     B  14-10-2021 15:21 2021-10-14 15:37:00          16        NaN  NaN
8   123     B  14-10-2021 15:21 2021-10-14 16:29:00          68        NaN  NaN
9   123     B  14-10-2021 15:21 2021-10-14 17:04:00         103        NaN  NaN
10  123     C  14-10-2021 15:22 2021-10-14 15:32:00          10        NaN  NaN
11  123     C  14-10-2021 15:22 2021-10-14 15:36:00          14        NaN  NaN
12  123     C  14-10-2021 15:22 2021-10-14 15:37:00          15      Third    3
13  123     C  14-10-2021 15:23 2021-10-14 16:29:00          67    Third_A  3_A
14  123     C  14-10-2021 15:48 2021-10-14 17:04:00         102    Third_B  3_B
15  789     A  14-10-2021 15:19 2021-10-14 15:32:00          13      First    1
16  789     A  14-10-2021 15:19 2021-10-14 15:36:00          17        NaN  NaN
17  789     B  14-10-2021 15:21 2021-10-14 15:32:00          11        NaN  NaN
18  789     B  14-10-2021 15:21 2021-10-14 15:36:00          15     Second    2
19  789     C  14-10-2021 15:22 2021-10-14 15:32:00          10        NaN  NaN

如果必须为每个ID组完成整个过程，则可以尝试

...
for _, df_id in df.sort_values("Date_Time_2").groupby("ID"):
    for n, (_, gdf) in enumerate(df_id.groupby("Group")):
        nths = gdf.groupby("Date_Time_2", as_index=False).ngroup()
        df.loc[gdf[nths == n].index, "New"] = str(n + 1)
    for i, c in zip(gdf[nths > n].index, letters):
        df.at[i, "New"] = f"{n + 1}_{c}"

而不是。

首先，确保你正确地读取csv值。这意味着日期时间值应该被正确地解释，例如。

date_parse = lambda x : pd.to_datetime(x, format="%d-%m-%Y %H:%M")
df = pd.read_csv('filename.csv', parse_dates=['Date_Time_1','Date_Time_2'], date_parser= date_parse)

如果您已经有了 Dataframe ，您可以使用以下代码来解析 Dataframe 中日期时间对象，

df['Date_Time_2'] = pd.to_datetime(df['Date_Time_2'], format="%d-%m-%Y %H:%M")
df['Date_Time_2'] = pd.to_datetime(df['Date_Time_2'], format="%d-%m-%Y %H:%M")

现在，只需在不同的组中迭代，并在排序列表中过滤出date_time_2列，最后取出适当的索引，例如，对于组“A”，取出“0”索引，对于组“B”，取出“1”索引......，选择适当的 Dataframe 并更新新列中的值

df['New_Column'] = 'NA'
for index, group in enumerate(df['Group'].unique()):
    unqiue_time = df[df['Group'] == group]['Date_Time_2'].unique()[index]
    df.loc[(df['Group'] == group) & (df['Date_Time_2'] == unqiue_time), 'New_Column'] = index
print(df)

注意：添加数字要容易得多，例如“first”、“second”，如果需要，可以创建一个新列表，并从索引中赋值，如下所示

df['New_Column'] = 'NA'
number_as_string = ['first', 'second', 'third']
for index, group in enumerate(df['Group'].unique()):
    unqiue_time = df[df['Group'] == group]['Date_Time_2'].unique()[index]
    df.loc[(df['Group'] == group) & (df['Date_Time_2'] == unqiue_time), 'New_Column'] = number_as_string[index]
print(df)