使用asfreq by month添加缺失的日期时间，将其过滤为新的Series，如有必要，使用创建的月份列表按年分组：

s = s.asfreq('m')
s1 = pd.Series(s[s.isnull()].index)
print (s1)
0   1964-10-31
1   1965-01-31
2   1965-02-28
3   1965-07-31
Name: 0, dtype: datetime64[ns]

out = s1.dt.month.groupby(s1.dt.year).apply(list)
print (out)
0
1964         [10]
1965    [1, 2, 7]
Name: 0, dtype: object

• • 设置**：

s = pd.Series({pd.Timestamp('1964-07-31 00:00:00'): 100.0, 
               pd.Timestamp('1964-08-31 00:00:00'): 98.81, 
               pd.Timestamp('1964-09-30 00:00:00'): 101.21, 
               pd.Timestamp('1964-11-30 00:00:00'): 101.42, 
               pd.Timestamp('1964-12-31 00:00:00'): 101.45,
               pd.Timestamp('1965-03-31 00:00:00'): 91.49, 
               pd.Timestamp('1965-04-30 00:00:00'): 90.33, 
               pd.Timestamp('1965-05-31 00:00:00'): 85.23, 
               pd.Timestamp('1965-06-30 00:00:00'): 86.1, 
               pd.Timestamp('1965-08-31 00:00:00'): 84.26})

print (s)
1964-07-31    100.00
1964-08-31     98.81
1964-09-30    101.21
1964-11-30    101.42
1964-12-31    101.45
1965-03-31     91.49
1965-04-30     90.33
1965-05-31     85.23
1965-06-30     86.10
1965-08-31     84.26
dtype: float64

编辑：
如果日期时间不总是月份的最后一天：

s = pd.Series({pd.Timestamp('1964-07-31 00:00:00'): 100.0, 
               pd.Timestamp('1964-08-31 00:00:00'): 98.81, 
               pd.Timestamp('1964-09-01 00:00:00'): 101.21, 
               pd.Timestamp('1964-11-02 00:00:00'): 101.42, 
               pd.Timestamp('1964-12-05 00:00:00'): 101.45,
               pd.Timestamp('1965-03-31 00:00:00'): 91.49, 
               pd.Timestamp('1965-04-30 00:00:00'): 90.33, 
               pd.Timestamp('1965-05-31 00:00:00'): 85.23, 
               pd.Timestamp('1965-06-30 00:00:00'): 86.1, 
               pd.Timestamp('1965-08-31 00:00:00'): 84.26})
print (s)
1964-07-31    100.00
1964-08-31     98.81
1964-09-01    101.21
1964-11-02    101.42
1964-12-05    101.45
1965-03-31     91.49
1965-04-30     90.33
1965-05-31     85.23
1965-06-30     86.10
1965-08-31     84.26
dtype: float64

#convert all months to first day
s.index = s.index.to_period('m').to_timestamp()
#MS is start month frequency
s = s.asfreq('MS')
s1 = pd.Series(s[s.isnull()].index)
print (s1)
0   1964-10-01
1   1965-01-01
2   1965-02-01
3   1965-07-01
dtype: datetime64[ns]

我经常通过计算每个指数值之间差距来实现这一点。

times_gaps = df.index - df.index.shift(1)

然后您可以绘制这些：

times_gaps.plot()

如果有缺口，你会很快看到缺口在哪里。如果没有缺口，你会看到一条笔直的水平线。
您还可以选择间隙时间，执行以下操作：

times_gaps[times_gaps> threshold]

假设 Dataframe 与您的输入相同（第一列是日期），您可以执行以下操作：

all = pd.Series(data=pd.date_range(start=df[0].min(), end=df[0].max(), freq='M'))
mask = all.isin(df[0].values)
print(all[~mask])

• • 产出**

3    1964-10-31
6    1965-01-31
7    1965-02-28
12   1965-07-31
dtype: datetime64[ns]

其思想是创建一个从第一个日期到最后一个日期的按月频率的日期范围，然后根据第一列检查这些值。

import pandas as pd

# Create a sample time-series data
dates = pd.date_range('2022-01-01', periods=12, freq='M')
data = range(12)
df = pd.DataFrame({'date': dates, 'value': data})

# Check if the time-series is continuous for every month
df_monthly = df.set_index('date').resample('M').mean()
if df_monthly.isnull().sum().sum() == 0:
    print("The time-series is continuous for every hour.")
else:
    print("The time-series is NOT continuous for every hour.")