import pandas as pd

    
df['Dates'] = pd.to_datetime(df['Dates'], errors='raise')
df['Dif'] = df['Dates'].diff().dt.days
df['Dif'] = df['Dif'].replace(0, 1)
df['Notes'] = ''

def my_func(x):
    a = df[x:][df.loc[x:, 'Dif'] != 1]
    if len(a) > 0:
        df.loc[[x - 1, a.index[0] - 1], 'Notes'] = ['Min', 'Max']
        df.loc[x - 1 : a.index[0] - 1, 'Dif'] = 1
    else:
        df.loc[[x - 1, len(df) - 1], 'Notes'] = ['Min', 'Max']
        df.loc[x - 1 : len(df) - 1, 'Dif'] = 1

[my_func(i) for i in range(1, len(df)) if df.loc[i - 1, 'Dif'] != df.loc[i, 'Dif'] and df.loc[i, 'Dif'] == 1]

df.loc[df[df['Dif'] != 1].index, 'Notes'] = 'Singledate'

print(df)

产出

Class      Dates  Dif       Notes
0      1 2022-01-12  1.0         Min
1      2 2022-01-13  1.0         Max
2      3 2022-01-15  2.0  Singledate
3      4 2022-01-20  1.0         Min
4      5 2022-01-21  1.0            
5      6 2022-01-22  1.0            
6      7 2022-01-22  1.0         Max
7      8 2022-01-30  8.0  Singledate

“日期”列将转换为日期格式。将创建一个具有天数差异的“Dif”列。0将替换为1。
我使用了列表解析，因为它比循环快很多倍。列表解析检查条件：current 'Dif' = 1，past不等于current。当条件被触发时，调用一个函数，通过loc填充所需的值。最后，填充带有'Singledate'的行。