I have a DataFrame which contains a lot of intraday data, the DataFrame has several days of data, dates are not continuous.

 2012-10-08 07:12:22            0.0    0          0  2315.6    0     0.0    0
 2012-10-08 09:14:00         2306.4   20  326586240  2306.4  472  2306.8    4
 2012-10-08 09:15:00         2306.8   34  249805440  2306.8  361  2308.0   26
 2012-10-08 09:15:01         2308.0    1   53309040  2307.4   77  2308.6    9
 2012-10-08 09:15:01.500000  2308.2    1  124630140  2307.0  180  2308.4    1
 2012-10-08 09:15:02         2307.0    5   85846260  2308.2  124  2308.0    9
 2012-10-08 09:15:02.500000  2307.0    3  128073540  2307.0  185  2307.6   11
 ......
 2012-10-10 07:19:30            0.0    0          0  2276.6    0     0.0    0
 2012-10-10 09:14:00         2283.2   80   98634240  2283.2  144  2283.4    1
 2012-10-10 09:15:00         2285.2   18  126814260  2285.2  185  2285.6    3
 2012-10-10 09:15:01         2285.8    6   98719560  2286.8  144  2287.0   25
 2012-10-10 09:15:01.500000  2287.0   36  144759420  2288.8  211  2289.0    4
 2012-10-10 09:15:02         2287.4    6  109829280  2287.4  160  2288.6    5
 ......

How can I extract the unique date in the datetime format from the above DataFrame? To have result like [2012-10-08, 2012-10-10]

If you have a Series like:

In [116]: df["Date"]
Out[116]: 
0           2012-10-08 07:12:22
1           2012-10-08 09:14:00
2           2012-10-08 09:15:00
3           2012-10-08 09:15:01
4    2012-10-08 09:15:01.500000
5           2012-10-08 09:15:02
6    2012-10-08 09:15:02.500000
7           2012-10-10 07:19:30
8           2012-10-10 09:14:00
9           2012-10-10 09:15:00
10          2012-10-10 09:15:01
11   2012-10-10 09:15:01.500000
12          2012-10-10 09:15:02
Name: Date

where each object is a Timestamp:

In [117]: df["Date"][0]
Out[117]: <Timestamp: 2012-10-08 07:12:22>

you can get only the date by calling .date():

In [118]: df["Date"][0].date()
Out[118]: datetime.date(2012, 10, 8)

and Series have a .unique() method. So you can use map and a lambda:

In [126]: df["Date"].map(lambda t: t.date()).unique()
Out[126]: array([2012-10-08, 2012-10-10], dtype=object)

or use the Timestamp.date method:

In [127]: df["Date"].map(pd.Timestamp.date).unique()
Out[127]: array([2012-10-08, 2012-10-10], dtype=object)

Just to give an alternative answer to @DSM, look at this other answer from @Psidom

It would be something like:

pd.to_datetime(df['DateTime']).dt.date.unique()

It seems to me that it performs slightly better

Using regex:

(\d{4}-\d{2}-\d{2})

Run it with re.findall function to get all matches:

result = re.findall(r"(\d{4}-\d{2}-\d{2})", subject)

This is what I get on Python 3.6.8 and Pandas 1.1.5:

%timeit df['date'].map(lambda d: d.date()).unique()

2.06 ms ± 135 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df['date'].dt.date.unique()

535 µs ± 79.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df['date'].dt.normalize().unique()

1.33 ms ± 229 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

Output of normalize().unique():

array(['2021-04-08T00:00:00.000000000', '2021-04-07T00:00:00.000000000',
       '2021-04-06T00:00:00.000000000', '2021-04-05T00:00:00.000000000',
       '2021-04-04T00:00:00.000000000', '2021-04-03T00:00:00.000000000',
       '2021-04-02T00:00:00.000000000', '2021-04-01T00:00:00.000000000',
       ..., dtype="datetime64[ns]")

Versus output of the other 2:

array([datetime.date(2021, 4, 8), datetime.date(2021, 4, 7),
       datetime.date(2021, 4, 6), datetime.date(2021, 4, 5),
       datetime.date(2021, 4, 4), datetime.date(2021, 4, 3),
       datetime.date(2021, 4, 2), datetime.date(2021, 4, 1),
       datetime.date(2021, 3, 31), datetime.date(2021, 3, 30),
       ..., dtype=object)