Finding the mean and standard deviation of a timedelta object in pandas df

Question

I would like to calculate the mean and standard deviation of a timedelta by bank from a dataframe with two columns shown below. When I run the code (also shown below) I get the below error:

pandas.core.base.DataError: No numeric types to aggregate 

My dataframe:

   bank                          diff    Bank of Japan                 0 days 00:00:57.416000    Reserve Bank of Australia     0 days 00:00:21.452000    Reserve Bank of New Zealand  55 days 12:39:32.269000    U.S. Federal Reserve          8 days 13:27:11.387000 

My code:

means = dropped.groupby('bank').mean() std = dropped.groupby('bank').std() 

Answer 1

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() 

Answer 2

Pandas mean() and other aggregation methods support numeric_only=False parameter.

dropped.groupby('bank').mean(numeric_only=False) 

Found here: Aggregations for Timedelta values in the Python DataFrame