<p>您可以使用类似于<a href="https://stackoverflow.com/a/57886573/2901002">this</a>的解决方案,仅使用新助手<code>DataFrame</code>过滤每个组的连续日期时间,添加所有缺少的日期时间,最后<code>merge</code>用于添加新列:</p>
<pre><code>df['datetimes'] = pd.to_datetime(df['date'].astype(str) + ' ' + df['time'].astype(str))
df1 = (df.set_index('datetimes')
.groupby(['code','tank', 'date'])['no_operation_flag']
.resample('30T')
.first()
.reset_index())
shifted1 = df1.groupby(['code','tank', 'date'])['no_operation_flag'].shift()
g1 = df1['no_operation_flag'].ne(shifted1).cumsum()
mask1 = g1.map(g1.value_counts()).gt(5) & df1['no_operation_flag'].eq(1)
df1['final_flag'] = mask1.astype(int)
#print (df1.head(40))
df = df.merge(df1[['code','tank','datetimes','final_flag']]).drop('datetimes', axis=1)
</code></pre>
<hr/>
<pre><code>print (df)
code tank date time no_operation_flag final_flag
0 123 1 01-01-2019 00:00:00 1 0
1 123 1 01-01-2019 00:30:00 1 0
2 123 1 01-01-2019 01:00:00 0 0
3 123 1 01-01-2019 01:30:00 1 1
4 123 1 01-01-2019 02:00:00 1 1
5 123 1 01-01-2019 02:30:00 1 1
6 123 1 01-01-2019 03:00:00 1 1
7 123 1 01-01-2019 03:30:00 1 1
8 123 1 01-01-2019 04:00:00 1 1
9 123 1 01-01-2019 05:00:00 1 0
10 123 1 01-01-2019 14:00:00 1 1
11 123 1 01-01-2019 14:30:00 1 1
12 123 1 01-01-2019 15:00:00 1 1
13 123 1 01-01-2019 15:30:00 1 1
14 123 1 01-01-2019 16:00:00 1 1
15 123 1 01-01-2019 16:30:00 1 1
16 123 2 02-01-2019 00:00:00 1 0
17 123 2 02-01-2019 00:30:00 0 0
18 123 2 02-01-2019 01:00:00 0 0
19 123 2 02-01-2019 01:30:00 0 0
20 123 2 02-01-2019 02:00:00 1 0
21 123 2 02-01-2019 02:30:00 1 0
22 123 2 02-01-2019 03:00:00 1 0
23 123 2 03-01-2019 03:30:00 1 0
24 123 2 03-01-2019 04:00:00 1 0
25 123 1 03-01-2019 14:00:00 1 0
26 123 2 03-01-2019 15:00:00 1 0
27 123 2 03-01-2019 00:30:00 1 0
28 123 2 04-01-2019 11:00:00 1 0
29 123 2 04-01-2019 11:30:00 0 0
30 123 2 04-01-2019 12:00:00 1 0
31 123 2 04-01-2019 13:30:00 1 0
32 123 2 05-01-2019 03:00:00 1 0
33 123 2 05-01-2019 03:30:00 1 0
34 123 2 05-01-2019 04:00:00 1 0
</code></pre>