Pandas: convert two times to datetime, then take the difference and display back as only times-CodePudding

I have the following toy Pandas dataframe named df:

df = pd.DataFrame({'begin' : ['08:00', '10:00', '14:00'], 
                   'end'   : ['14:00', '17:00', '22:00']})

begin     end      
08:00     14:00    
10:00     17:00    
14:00     22:00

I would like to calculate the difference between the begin and end times (in hours). Here is my attempt:

df['begin'] = pd.to_datetime(df['begin'])
df['end'] = pd.to_datetime(df['end'])

df['diff_hours'] = (df['end'] - df['begin']) / pd.Timedelta(hours=1)

This yields:

    begin                  end                     diff_hours
0   2021-12-28 08:00:00    2021-12-28 14:00:00     6
1   2021-12-28 10:00:00    2021-12-28 17:00:00     7
2   2021-12-28 14:00:00    2021-12-28 22:00:00     8

This approach takes the current date. I would like to strip the current date from the output above, such that the df becomes:

    begin          end          diff_hours
0   08:00:00       14:00:00     6
1   10:00:00       17:00:00     7
2   14:00:00       22:00:00     8

Thanks in advance for any assistance you can give!

CodePudding user response：

Coerce the time to datetime, substract and convert outcome to hours

   df['diff_hours']=(pd.to_datetime(df['end'], format="%H:%M")-pd.to_datetime(df['begin'], format="%H:%M")).astype('timedelta64[m]')/60



 begin    end  diff_hours
0  08:00  14:00         6.0
1  10:00  17:00         7.0
2  14:00  22:00         8.0

CodePudding user response：

You can convert date time to time like this:

df['begin'] = df['begin'].dt.time
df['end'] = df['end'].dt.time

      begin       end  diff_hours
0  08:00:00  14:00:00         6.0
1  10:00:00  17:00:00         7.0
2  14:00:00  22:00:00         8.0