按时间分组 Delta Python Pandas

Question

I am attempting to do a group by in Python. What I have is a data frame that has two columns...Name and Time Difference ( Time Difference ) is a timedelta variable that looks like the following -1 days 14:00:0000, 0 days 00:08:0000, ect. Name has duplicates in it...it looks like Brad, Amy, Brad, Brad, Bill, Amy....what I want to do is find the Mean of Time Difference by Name. Also Time Difference does have NA values in it.

I have tried

data_frame['NewMean'] = data_frame['TimeDifference'].values.astype(np.int64)

means = data_frame.groupby(data_frame['Name']).mean()

means['NewMean'] = pd.to_timedelta(means['NewMean']) 

But I keep getting the error invalid literal for int()

I know float fixes this but I want to create a new dataframe with this information that just list out the names ( no dupes ) and the mean of each name

Answer 1

Try this:

data_frame['TimeDifference'] = data_frame['TimeDifference'].dt.days
data_frame['mean'] = data_frame.groupby('Name')['TimeDifference'].mean()

Answer 2

There is a way to get the values without casting to int and ignoring nan or nat values but involves a lambda expression, the results are a timedelta objects:

import numpy as np

time_groups = data_frame.groupby('Name').apply(
    lambda df: np.mean(df.TimeDifference)
)