python/pandas from start date + time, create datetime index

Question

Currently I have a dataframe that looks like the following:

df = 
     Open    High     Low   Close  TotalVolume
0  113.40  113.54  113.40  113.54         7237
1  113.54  113.58  113.52  113.57        10099
2  113.59  113.81  113.52  113.78        13827
3  113.76  113.94  113.75  113.92        16129
4  113.91  114.01  113.88  113.97        27052
5  113.97  114.11  113.92  114.01        24925
6  114.00  114.15  113.99  114.04        13461
7  114.06  114.14  113.94  113.94        10702
8  113.92  113.99  113.86  113.99         5538
9  113.96  113.96  113.85  113.86        14000

It does not necessarily have to be a datetime index but I felt like it would be the easiest. From this I have a variable startDate that follows this format startDate = "03-20-2018t14:00"

From this, this is minute data and for it to run another program, the format has to follow this, but this is the end result I am hoping for:

updated_df =
Date          Time     Open    High     Low   Close  TotalVolume
03/20/2018   14:00   113.40  113.54  113.40  113.54         7237
03/20/2018   14:01   113.54  113.58  113.52  113.57        10099
03/20/2018   14:02   113.59  113.81  113.52  113.78        13827
03/20/2018   14:03   113.76  113.94  113.75  113.92        16129
03/20/2018   14:04   113.91  114.01  113.88  113.97        27052
03/20/2018   14:05   113.97  114.11  113.92  114.01        24925
03/20/2018   14:06   114.00  114.15  113.99  114.04        13461
03/20/2018   14:07   114.06  114.14  113.94  113.94        10702
03/20/2018   14:08   113.92  113.99  113.86  113.99         5538
03/20/2018   14:09   113.96  113.96  113.85  113.86        14000

harpan · Accepted Answer · 2018-07-20 14:15:34Z

3

You need to use pandas.date_range() with start, periods and freq parameters.

df['datetime'] = pd.date_range(start='03-20-2018t14:00', periods=len(df), freq="1min")

Or if you want them separate you can extract date and time from the DatetimeIndex as below:

datetime_col = pd.date_range(start='03-20-2018t14:00', periods=len(df), freq="1min")
df['Date'] = datetime_col.date
df['Time'] = datetime_col.time

Refer to docs for detailed information.

Output:

         Date      Time    Open    High     Low   Close  TotalVolume
0  2018-03-20  14:00:00  113.40  113.54  113.40  113.54         7237
1  2018-03-20  14:01:00  113.54  113.58  113.52  113.57        10099
2  2018-03-20  14:02:00  113.59  113.81  113.52  113.78        13827
3  2018-03-20  14:03:00  113.76  113.94  113.75  113.92        16129
4  2018-03-20  14:04:00  113.91  114.01  113.88  113.97        27052
5  2018-03-20  14:05:00  113.97  114.11  113.92  114.01        24925
6  2018-03-20  14:06:00  114.00  114.15  113.99  114.04        13461
7  2018-03-20  14:07:00  114.06  114.14  113.94  113.94        10702
8  2018-03-20  14:08:00  113.92  113.99  113.86  113.99         5538
9  2018-03-20  14:09:00  113.96  113.96  113.85  113.86        14000

edited Jul 20, 2018 at 14:15

answered Jul 20, 2018 at 13:47

harpan

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6 Comments

BENY Over a year ago

You did not cover the datetime format conversion in your answer.1/1/2018 14:00 and please do not using your own data .

lee.edward01 Over a year ago

@Wen what would be the easiest way to convert the time which is in UTC right now, to EST?

BENY Over a year ago

@lee.edward01 stackoverflow.com/questions/35398944/…

harpan Over a year ago

@Wen, edited the answer. However, As the docs have shown, I can pass string as a date, not necessarily the datetime object as startdate. Correct me if I am wrong. Thanks.

BENY Over a year ago

@HarvIpan I think this is input '03-20-2018t14:00' not '03-20-2018 14:00'

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