Pandas dataframe with identical date, add seconds to differentiate the rows-CodePudding

I have a pandas dataframe as following:

Date time         LifeTime1   LifeTime2  LifeTime3 LifeTime4 LifeTime5
2020-02-11 17:30:00      6          7          NaN       NaN         3
2020-02-11 17:30:00     NaN        NaN          3       3           NaN  
2020-02-12 15:30:00      2          2          NaN       NaN         3
2020-02-16 14:30:00      4         NaN         NaN       NaN         1
2020-02-16 14:30:00     NaN         7          NaN       NaN        NaN
2020-02-16 14:30:00     NaN        NaN          8         2         NaN

The dates are identical for some rows, is it possible to add 1 second, 2 second, 3 seconds to 2, 3, and 4 identical dates? So if its just one unique date, leave as is. If there are two identical dates, leave first one as is but add 1 second to the second identical date. And if three identical date, leave first as is, second add 1 second and add 2 second to third one. Is this possible to do easily in pandas?

CodePudding user response：

You can use groupby.cumcount combined with pandas.to_datetime with unit='s' to add incremental seconds to the duplicated rows:

s = pd.to_datetime(df['Date time'])
df['Date time'] = s pd.to_timedelta(s.groupby(s).cumcount(), unit='s')

As a one liner with python 3.8 walrus operator:

df['Date time'] = ((s:=pd.to_datetime(df['Date time']))
                    pd.to_timedelta(s.groupby(s).cumcount(), unit='s')
                  )

output:

            Date time  LifeTime1  LifeTime2  LifeTime3  LifeTime4  LifeTime5
0 2020-02-11 17:30:00        6.0        7.0        NaN        NaN        3.0
1 2020-02-11 17:30:01        NaN        NaN        3.0        3.0        NaN
2 2020-02-12 15:30:00        2.0        2.0        NaN        NaN        3.0
3 2020-02-16 14:30:00        4.0        NaN        NaN        NaN        1.0
4 2020-02-16 14:30:01        NaN        7.0        NaN        NaN        NaN
5 2020-02-16 14:30:02        NaN        NaN        8.0        2.0        NaN