groupby with diff function

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I have a groupby with a diff function, however I want to add an extra mean column for heart rate, how can I do this the best way?

this is the code


data= pd.DataFrame(
    [[Timestamp('2022-08-05 10:11:04'), 140, 120],
    [Timestamp('2022-08-05 10:11:05'), 160, 155],
    [Timestamp('2022-08-05 10:11:06'), 230, 156],
    [Timestamp('2022-08-05 10:11:07'), 230, 155],
    [Timestamp('2022-08-05 10:11:08'), 230, 160],
    [Timestamp('2022-08-05 10:11:09'), 140, 130],
    [Timestamp('2022-08-05 10:11:10'), 140, 131],
    [Timestamp('2022-08-05 10:11:11'), 230, 170]],
    columns=['timestamp', 'power', 'heart rate'])

m = data['power'].gt(200) #fill in power value
gb = (-data['timestamp'].diff(-1))[m].groupby([(~m).cumsum()).sum()
gb= gb.groupby((~m).cumsum()).sum()
gb

where should I add in the piece of code to calculate the average heart rate?

output will be the amount of seconds in high power zone and then i would like to add the average heart rate during this period. like this

gb = pd.DataFrame(
    [[Timestamp('00:00:04'), 210, 145],
    [Timestamp('00:00:15'), 250, 155],
    [Timestamp('00:01:00'), 230, 180],
   
    columns=['time at high intensity', ' avg power', ' avg heart rate'])
pandas mean group-by python
1 Answers

You can create helper column from by difference and then aggregate by it and another column in named aggregation in GroupBy.agg:

m = data['power'].gt(200) #fill in power value
gb = (data.assign(new=-data['timestamp'].diff(-1))[m]
          .groupby((~m).cumsum())
          .agg(time_at_high_intensity=('new','sum'),
               avg_power=('power','mean'), 
               avg_heart_rate=('heart rate','mean')))
                                               
print (gb)
      time_at_high_intensity  avg_power  avg_heart_rate
power                                                  
2            0 days 00:00:03        230             157
4            0 days 00:00:00        230             170
user_29010020
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