How to convert date columns from object type to date type in Python

Question

I have this data frame where the date columns are 'datetime' type

datetime64[ns]
2014-04-30T00:00:00 000000000

Now I want the date in this format - 2014-04-30 . So I used below code

df['StartingDate2'] = XY['StartingDate'].dt.strftime('%m/%d/%Y')

Now, this works accurately, but converts my date to object type. I read here that In Python, dates are objects. But I want to write my final data frame as a table in redshift and so I need my date columns defined as date for my python data frame. Any suggestion about how to do that , will be greatly appreciated.

Update:

sample dataframe

p1 = {'name': ['johnny', 'tommy', 'bobby', 'rocky', 'jimmy'], 'StartingDate': ['2015-07-14T00:00:00.000000000', '2013-10-30T00:00:00.000000000', '2014-04-30T00:00:00.000000000', '2014-01-27T00:00:00.000000000', '2016-01-15T00:00:00.000000000'], 'Address': ['NY', 'NJ', 'PA', 'NY', 'CA'], 'comment1': ['Very good performance', 'N/A', 'Need to work hard', 'No Comment', 'Not satisfactory'], 'comment2': ['good', 'Meets Expectation', 'N', 'N/A', 'Incompetence']}
XY = pd.DataFrame(data = p1)
XY['today'] = datetime.datetime.now()

When I am using to_datetime() solution - it doesn't work

XY['today2'] = pd.to_datetime(XY['today'],  format = '%m/%d/%Y')
XY['StartingDate2'] = pd.to_datetime(XY['today'],  format = '%m/%d/%Y')

Alternatively - this works, when strftime() and to_datetime() are used in combination.

XY['StartingDate2'] = XY['StartingDate2'].dt.strftime('%m/%d/%Y')
XY['StartingDate2'] = pd.to_datetime(XY['StartingDate2'])

But although this solution works for the sample data, is not working for me. The data I have looks like this -

array(['2015-09-29T14:34:39.000000000', '2015-10-07T14:13:03.000000000',
   '2015-10-07T19:17:50.000000000', ...,
   '2017-12-05T14:06:42.000000000', '2017-12-06T16:36:44.000000000',
   '2017-12-06T18:26:49.000000000'], dtype='datetime64[ns]'

You can floor the datetime to day frequency. I will update my post later — Quickbeam2k1
– Quickbeam2k1, Commented May 17, 2018 at 16:08

Quickbeam2k1 · Accepted Answer · 2018-05-18 06:14:14Z

3

The solution is to use to_datetime

s = pd.Series(['3/11/2000', '3/12/2000', '3/13/2000'])
s
0    3/11/2000
1    3/12/2000
2    3/13/2000
dtype: object

pd.to_datetime(s)
0   2000-03-11
1   2000-03-12
2   2000-03-13
dtype: datetime64[ns]

So in your situation, you could write

df['StartingDate2'] = pd.to_datetime(XY['StartingDate'], format='%m/%d/%Y')

and in fact you can neglect the format keyword here. But if you provide it, you will get a huge speed increase.

Benchmarks

s = pd.Series(['3/11/2000', '3/12/2000', '3/13/2000']*1000)

Without format

%%timeit
pd.to_datetime(s)
453 ms ± 3.05 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

With format

%%timeit
pd.to_datetime(s, format='%m/%d/%Y')
9.68 ms ± 44.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Note for me: Always provide the format string if possible

Answer for updated question

You can use date times floor function

df = pd.DataFrame(pd.to_datetime(['2015-09-29T14:34:39.000000000', '2015-10-07T14:13:03.000000000',
   '2015-10-07T19:17:50.000000000',
   '2017-12-05T14:06:42.000000000', '2017-12-06T16:36:44.000000000',
   '2017-12-06T18:26:49.000000000']), columns=['A'])

df['B'] = df['A'].dt.floor('d')
df.dtypes

A    datetime64[ns]
B    datetime64[ns]
dtype: object

df

                      A          B
0   2015-09-29 14:34:39 2015-09-29
1   2015-10-07 14:13:03 2015-10-07
2   2015-10-07 19:17:50 2015-10-07
3   2017-12-05 14:06:42 2017-12-05
4   2017-12-06 16:36:44 2017-12-06
5   2017-12-06 18:26:49 2017-12-06

edited May 18, 2018 at 6:14

answered May 16, 2018 at 21:42

Quickbeam2k1

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4 Comments

singularity2047 Over a year ago

I think, I have tried to_datetime() before but was not successful. let me check again. I will confirm the answer then.

singularity2047 Over a year ago

In my case it doesn't change anything. So I created a new dataframe using datetime.datetime.now() to test your code. I have updated the post at the bottom

singularity2047 Over a year ago

Thanks for your solution. One more thing - when I use this ' pd.Series(df['CreatedDate2']).dt.floor('d') ' - the output shows yy-mm-dd in datetime format, which is what I want. But when I am trying to create a new column based on that using - df['CreatedDate3'] = pd.Series(df['CreatedDate2']).dt.floor('d') - it just goes back to timestamp format. I wonder why is that happening.

Quickbeam2k1 Over a year ago

I updated again. Could you explain what timestamp format is?

Bram van Hout · Accepted Answer · 2018-05-16 21:44:06Z

0

df['StartingDate2'] = pd.to_datetime(XY['StartingDate'])

answered May 16, 2018 at 21:44

Bram van Hout

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