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

Graham Streich picture Graham Streich · Jun 18, 2017 · Viewed 12.3k times · Source

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

jezrael picture jezrael · Jun 18, 2017

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()