Pandas dataframe - 如何为现有日期列综合添加唯一时间戳，其中仅包含日期但不包含时间？

Question

I have a simple dataframe with a string index.
The string represents time (eg 2018-01-01), and contains duplications.
Applying pd.to_datetime() takes me in the right direction, and well converts the index from a string type into datetime type.
However it does not solves the duplications problem.
I would ideally wish to synthetically add some unique timeStamp (%h:%m:%s) to each index cell.
Can you please guide me how to achieve that?

Here is a simple example of what I'm trying to achieve:

import pandas as pd
df = pd.DataFrame(index = ['2018-01-01', '2018-01-01', '2018-01-01'], 
                  columns = ['A', 'B', 'C'] ).fillna(0)

That yields the following dataframe:

            A  B  C
2018-01-01  0  0  0
2018-01-01  0  0  0
2018-01-01  0  0  0

I would like to convert it for something like that (unique datetime index):

                     A  B  C
2018-01-01 00:00:01  0  0  0
2018-01-01 00:00:02  0  0  0
2018-01-01 00:00:03  0  0  0

Thanks ahead,
Shahar

Answer 1

If all values of datetimes are unique use to_datetime with unit and origin parameter by first value if index and add to index by DataFrame.set_index :

df = df.set_index(pd.to_datetime(np.arange(len(df)), 
                                 unit='s', 
                                 origin=df.index[0]))
print (df)
                     A  B  C
2018-01-01 00:00:00  0  0  0
2018-01-01 00:00:01  0  0  0
2018-01-01 00:00:02  0  0  0

If there are multiple unique datetime s in index add timedeltas created by GroupBy.cumcount to Datetimeindex :

import pandas as pd
df = pd.DataFrame(index = ['2018-01-01', '2018-01-01', '2018-01-01',
                           '2018-02-01', '2018-02-01'], 
                  columns = ['A', 'B', 'C'] ).fillna(0)


df = df.set_index(pd.to_datetime(df.index) + 
                  pd.to_timedelta(df.groupby(level=0).cumcount(), unit='s'))
print (df)
                     A  B  C
2018-01-01 00:00:00  0  0  0
2018-01-01 00:00:01  0  0  0
2018-01-01 00:00:02  0  0  0
2018-02-01 00:00:00  0  0  0
2018-02-01 00:00:01  0  0  0

Answer 2

You can use pd.to_datetime in combination with pd.to_timedelta to get the desired results.

Use:

df.index = (
    pd.to_datetime(df.index) + 
    pd.to_timedelta(range(1, len(df) + 1), unit='s'))

print(df)

This prints the resulting dataframe as:

                     A  B  C
2018-01-01 00:00:01  0  0  0
2018-01-01 00:00:02  0  0  0
2018-01-01 00:00:03  0  0  0

Answer 3

To express your task more generally (for multiple dates):

• you have a DataFrame with a string index, formatted like dates,
• you want to convert the index to datetime ,
• but within each date set the time part to consecitive seconds.

To do it you can run:

df.index = pd.Series(pd.Timedelta('1S'), index=pd.to_datetime(df.index)).groupby(level=0)\
    .transform(lambda grp: grp.cumsum() + grp.index)

Steps:

• pd.Series(pd.Timedelta('1S'), index=pd.to_datetime(df.index)) - create a Series filled with one second values and the index from df converted to datetime , for now still with no time part.
• groupby(...) - group it by dates.
• transform(...) - transform it with the lambda function given.
• grp.cumsum() - the time part alone - consecutive seconds.
• + grp.index - add the date part.
• df.index - set the index in df to this result.

The result, for 2 dates, even when dates are "intermixed", is still OK:

            A  B  C
2018-01-01  0  0  0
2018-01-01  0  0  0
2018-01-01  0  0  0
2018-01-02  0  0  0
2018-01-02  0  0  0
2018-01-02  0  0  0
2018-01-01  0  0  0

If you have a DataFrame with a single date, you can still use this code (you will have a single group only).