Calculated timedelta column in pandas

Question

I have some data in a pandas df like:

time delay 2017/06/14 10:02:10.144927 PM 15

The time is obviously a timestamp and in the delay is in seconds. df.dtypes are datetime64[ns] and int respectively. I want to add a column that has the timestamp of the original time plus the seconds passed, e.g.:

time delay end_time 2017/06/14 10:02:10.144927 PM 15 2017/06/14 10:02:25.144927 PM

I'm trying to do

df["end_time"] = df.time.add(datetime.timedelta(seconds = df.time))

However I get an error TypeError: unsupported type for timedelta seconds component: Series. How would I do this? Do I need a lambda?

df.time is a series of numbers. I think you wanted this to be element wise right? — Buckeye14Guy
– Buckeye14Guy, Commented May 28, 2019 at 12:55

jezrael · Accepted Answer · 2019-05-28 13:01:28Z

Use to_timedelta for convert integer column to timedeltas:

df["end_time"] = pd.to_datetime(df['time']).add(pd.to_timedelta(df.delay, unit='s')) print (df) time delay end_time 0 2017/06/14 10:02:10.144927 PM 15 2017-06-14 22:02:25.144927

If need same format in output add Series.dt.strftime:

df["end_time"] = (pd.to_datetime(df['time']) .add(pd.to_timedelta(df.delay, unit='s')) .dt.strftime('%Y/%m/%d %H:%M:%S.%f %p')) print (df) time delay end_time 0 2017/06/14 10:02:10.144927 PM 15 2017/06/14 22:02:25.144927 PM

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