Python timedelta time difference from the given varied time of every day

Question

I have the these two following pandas dataframe :

df1 = {'Date' : ['07-10-2019',
              '07-10-2019',
              '07-10-2019',
              '08-10-2019',
              '08-10-2019',
              '08-10-2019']}

df1 = {'Time' :['07-10-2019 10:47:00',
              '07-10-2019 10:52:00',
              '07-10-2019 10:59:00',
              '08-10-2019 10:47:00',
              '08-10-2019 10:52:00',
              '08-10-2019 10:59:00',
]}

and I am trying to create d['Time Taken'] by using:
df1['Time Taken']=((pd.to_datetime(df1['Time'])-pd.to_datetime(df1['Date'])).dt.total_seconds())/60
and clearly as expected, I am getting resultant df1['Time Taken'] as:

df1 = {'Time Taken': [647.0,
                      652.0,
                      659.0,
                      647.0,
                      652.0,
                      659.0,
]}

However, I would like to see df['Time Taken] start with NaN or 0. How should I write it this line code to make it look like:

df1 = {'Time Taken': [NaN,
                      5.0,
                      7.0,
                      NaN,
                      5.0,
                      7.0,
]}

Please, help me with this! Many thanks!

Asked By: kern

||

Source

Answer 1

Use DataFrameGroupBy.diff with Series.dt.total_seconds and divide 60:

df1 = pd.DataFrame({'Date' : ['07-10-2019',
              '07-10-2019',
              '07-10-2019',
              '08-10-2019',
              '08-10-2019',
              '08-10-2019'], 
              'Time' :['07-10-2019 10:47:00',
              '07-10-2019 10:52:00',
              '07-10-2019 10:59:00',
              '08-10-2019 10:47:00',
              '08-10-2019 10:52:00',
              '08-10-2019 10:59:00',
]})

df1['Time']= pd.to_datetime(df1['Time'])


df1['Time Taken'] = df1.groupby('Date')['Time'].diff().dt.total_seconds() / 60
print(df1)
         Date                Time  Time Taken
0  07-10-2019 2019-07-10 10:47:00         NaN
1  07-10-2019 2019-07-10 10:52:00         5.0
2  07-10-2019 2019-07-10 10:59:00         7.0
3  08-10-2019 2019-08-10 10:47:00         NaN
4  08-10-2019 2019-08-10 10:52:00         5.0
5  08-10-2019 2019-08-10 10:59:00         7.0

Answered By: jezrael

Python timedelta time difference from the given varied time of every day

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