computing the Cumulative Sum, where Sum can be reset by a condition

Question

Pandas DataFrame, computing the Time Difference between one row and other row which satisfies a condition

Similar to this Question, Given,

I would like to find the 'Cumulative Sum Of Occurrence Master Event' , and 'Cumulative Sum Of Occurrence Event A ' and Event B in between the Master Event. In other words, the instance when Master Event Occour, Cummulative Sum of Event A is reset to Zero.

Sample output is as follows,

Sample Input code by @Jon Strutz

 import pandas as pd
df = pd.DataFrame({'year': [2019] * 10,
                       'month': [8] * 10,
                       'day': [16] * 10,
                       'hour': [12, 12, 12, 12, 13, 13, 13, 13, 13, 13],
                       'minute': [50, 52, 53, 57, 0, 3,4,5,13,21]})

df = pd.DataFrame(pd.to_datetime(df), columns=['Time_Stamp'])
df['Event_Master'] = [0, 0, 1, 0, 0 ,0, 0, 0, 1,0]
df['Event_B']      = [0, 0, 0, 1, 0 ,0, 1, 0, 0,1]

And expected output could be like

df['Event_Master_Out'] = [0, 0, 1, 1, 1 ,1, 1, 1, 2,2]
df['Event_B_Out'] =      [0, 0, 0, 1, 1 ,1, 2, 2, 0,1]

Answer 1

Use Series.cumsum and output is used for GroupBy.cumsum :

df['Event_Master_Out'] = df['Event_Master'].cumsum()
df['Event_B_Out'] = df.groupby('Event_Master_Out')['Event_B'].cumsum()
print (df)
           Time_Stamp  Event_Master  Event_B  Event_Master_Out  Event_B_Out
0 2019-08-16 12:50:00             0        0                 0            0
1 2019-08-16 12:52:00             0        0                 0            0
2 2019-08-16 12:53:00             1        0                 1            0
3 2019-08-16 12:57:00             0        1                 1            1
4 2019-08-16 13:00:00             0        0                 1            1
5 2019-08-16 13:03:00             0        0                 1            1
6 2019-08-16 13:04:00             0        1                 1            2
7 2019-08-16 13:05:00             0        0                 1            2
8 2019-08-16 13:13:00             1        0                 2            0
9 2019-08-16 13:21:00             0        1                 2            1