Finding the mean and standard deviation of a timedelta object in pandas df
You need to convert timedelta to some numeric value, e.g. int64 by values what is most accurate, because convert to ns is what is the numeric representation of timedelta:
dropped['new'] = dropped['diff'].values.astype(np.int64)means = dropped.groupby('bank').mean()means['new'] = pd.to_timedelta(means['new'])std = dropped.groupby('bank').std()std['new'] = pd.to_timedelta(std['new'])Another solution is to convert values to seconds by total_seconds, but that is less accurate:
dropped['new'] = dropped['diff'].dt.total_seconds()means = dropped.groupby('bank').mean()
No need to convert timedelta back and forth. Numpy and pandas can seamlessly do it for you with a faster run time. Using your dropped DataFrame:
import numpy as npgrouped = dropped.groupby('bank')['diff']mean = grouped.apply(lambda x: np.mean(x))std = grouped.apply(lambda x: np.std(x))