Setting values given two column

Question

I have 2-time series data frames. Both contain values [0,1] only. The first one is called init_signal and the second is called end_signal. The idea is to create a new data frame when init_signal has a 1, it will find the NEXT 1 in end_signal.

The example below merges both the init_signal and end_signal as one data frame. the first column is init_signal and the second column is end_signal.

For example:

2016-06-13  1  0
2016-06-14  0  0
2016-06-15  0  1
2016-06-16  0  0

Will become:

2016-06-13  1  
2016-06-14  1  
2016-06-15  1  
2016-06-16  0  

For example 2:

2016-06-13  1  1
2016-06-14  0  0
2016-06-15  0  1
2016-06-16  0  0

Will become:

2016-06-13  1  
2016-06-14  1  
2016-06-15  1  
2016-06-16  0  

In the 2nd example, if there is also a 1 in the same row as the init_signal, it will be ignored.

I've been using a complex loop. Is there a non-looping way?

Answer 1

I hope this helps and covers all the boundary conditions.

import pandas as pd
import datetime as dt

# Dummy data
df = pd.DataFrame()
df['init_signal'] = [0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0]
df['end_signal']  = [0,1,0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,1,0]
df.index = [dt.date.today()+dt.timedelta(-i) for i in reversed(range(len(df)))]

# Cumsum for calculation
df['init_c'] = df['init_signal'].cumsum()
df['end_c']  = df['end_signal'].cumsum()

# result calculation
func = lambda x: (x==x.min()).shift().fillna(True)
df['result'] = df.groupby('init_c')['end_c'].transform(func)
df['result'] = df['result']&df['init_c'].astype(bool)
df['result'] = df['result'].astype(int)

# dropping temp columns
df.drop(['init_c','end_c'], axis=1, inplace=True)

print(df)

Output:

            init_signal  end_signal  result
2018-12-12            0           0       0
2018-12-13            0           1       0
2018-12-14            0           0       0
2018-12-15            1           0       1
2018-12-16            0           0       1
2018-12-17            0           0       1
2018-12-18            0           1       1
2018-12-19            0           0       0
2018-12-20            1           1       1
2018-12-21            0           0       1
2018-12-22            0           0       1
2018-12-23            0           1       1
2018-12-24            0           0       0
2018-12-25            0           1       0
2018-12-26            0           0       0
2018-12-27            0           0       0
2018-12-28            1           0       1
2018-12-29            0           1       1
2018-12-30            0           0       0