Python groupby datetime列的运行时间（基于另一列的条件）

我有以下数据集

lst=[
 ['a','2019-03-02 23:20:28',0],
 ['a','2019-03-02 23:21:29',0],
 ['a','2019-03-02 23:22:30',1],
 ['a','2019-03-02 23:30:31',0.5],
 ['a','2019-03-02 23:38:32',0.5],
 ['a','2019-03-02 23:50:32',0.5],  
 ['a','2019-03-02 23:50:32',0], 
 ['b','2019-03-02 23:10:32',0],  
 ['b','2019-03-02 23:12:32',0],  
 ['b','2019-03-02 23:20:32',1],  
 ['b','2019-03-02 23:30:32',0.5],  
 ['b','2019-03-02 23:50:32',1],  
 ['b','2019-03-02 23:55:32',1],  
 ['b','2019-03-02 23:56:32',0],
 ['a','2019-03-02 22:20:28',0],
 ['a','2019-03-02 22:21:29',0],
 ['a','2019-03-02 22:22:30',1],
 ['a','2019-03-02 22:30:31',0.5],  
 ['a','2019-03-02 22:30:32',0], 
  ]

df = pd.DataFrame(lst,columns=['ID','ts','signal'])
df['ts']=pd.to_datetime(df['ts'])

我希望获得每个ID的所有行，即信号列中0到0之间的总运行时间大于15分钟

i、 e.仅适用于以下情况：

仅针对b：

使用：

#filter out rows with 0
df1 = df[df['signal'].ne(0)]
#create Series from original column for unique consecutive groups for non 0 rows
a = df['signal'].eq(0).cumsum()

thr = pd.Timedelta(15, unit='min')

#get difference between first and last value per group and filtering by thresh
df2 = df1[df1['ts'].groupby(a).transform(lambda x: x.iat[-1] - x.iat[0]) > thr]
print (df2)
   ID                  ts  signal
2   a 2019-03-02 23:22:30     1.0
3   a 2019-03-02 23:30:31     0.5
4   a 2019-03-02 23:38:32     0.5
5   a 2019-03-02 23:50:32     0.5
9   b 2019-03-02 23:20:32     1.0
10  b 2019-03-02 23:30:32     0.5
11  b 2019-03-02 23:50:32     1.0
12  b 2019-03-02 23:55:32     1.0

---

抱歉，那是打字错误。现在应该很清楚了，对于我来说，只有

pd.Timedelta（15，unit='min'）

只能与

pd.Timedelta（15）

一起使用。