0
diff=table['DelayInMinute']=(pd.to_datetime(table['joinTime']) - 
     pd.to_datetime(table['ScheduleDateTime']))

Output: "0 days 00:00:00"

diff_minute=pd.Timedelta(diff).total_seconds()/60.0;

But getting

ValueError: Value must be Timedelta, string, integer, float, timedelta or convertible

RAVI VERMA
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1 Answers1

3

You can use Series.dt.total_seconds if multiple values of Series:

diff_minute=diff.dt.total_seconds()/60.0

If want output scalar by first value use iat for select first value with Timedelta.total_seconds, converting to Timedelta is not necessary:

diff_minute=diff.iat[0].total_seconds()/60.0
jezrael
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