从两个连续行中获取数学值

这是我的数据帧

import pandas as pd
import datetime
data = [['A1','String01',45,datetime.date(2018,1,1),datetime.date(2018,3,1)],
['A1','String02',46,datetime.date(2018,3,1),datetime.date(2018,4,29)],
['A1','String03',48,datetime.date(2018,4,29),datetime.date(2018,6,30)],
['A1','String04',51,datetime.date(2018,6,30),datetime.date(2018,12,31)],
['A2','String11',32,datetime.date(2018,1,1),datetime.date(2018,6,1)],
['A2','String12',33,datetime.date(2018,6,1),datetime.date(2018,7,30)],
['A2','String13',54,datetime.date(2018,8,11),datetime.date(2018,12,31)],
['A3','String21',45,datetime.date(2018,1,1),datetime.date(2018,6,1)],
['A3','String22',47,datetime.date(2018,7,1),datetime.date(2018,12,31)],]

cols = ['ID','SomeValue','Price','StartDate','EndDate']

df = pd.DataFrame(data,columns=cols)
print(df)

如果我们打印数据帧,我们可以看到ID=A2的Price从7/31到8/11丢失了(查看StartDate和EndDate)。我们有一个类似的情况,ID=A3

我的输出应该是这样的:

 ID SomeValue  Price   StartDate     EndDate  NoOfDaysMissing
0  A1  String01     45  2018-01-01  2018-03-01              NaN
1  A1  String02     46  2018-03-01  2018-04-29              0.0
2  A1  String03     48  2018-04-29  2018-06-30              0.0
3  A1  String04     51  2018-06-30  2018-12-31              0.0
4  A2  String11     32  2018-01-01  2018-06-01              NaN
5  A2  String12     33  2018-06-01  2018-07-30              0.0
6  A2  String13     54  2018-08-11  2018-12-31             12.0
7  A3  String21     45  2018-01-01  2018-06-01              NaN
8  A3  String22     47  2018-07-01  2018-12-31             30.0