Convert DataFrame column type from string to datetime

Question

How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetime dtype?

Answer 1

The easiest way is to use to_datetime:

df['col'] = pd.to_datetime(df['col'])

It also offers a dayfirst argument for European times (but beware this isn't strict).

Here it is in action:

In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0   2005-05-23 00:00:00
dtype: datetime64[ns]

You can pass a specific format:

In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0   2005-05-23
dtype: datetime64[ns]

Answer 2

If your date column is a string of the format '2017-01-01' you can use pandas astype to convert it to datetime.

df['date'] = df['date'].astype('datetime64[ns]')

or use datetime64[D] if you want Day precision and not nanoseconds

print(type(df_launath['date'].iloc[0])) 

yields

<class 'pandas._libs.tslib.Timestamp'>

the same as when you use pandas.to_datetime

You can try it with other formats then '%Y-%m-%d' but at least this works.

Answer 3

You can use the following if you want to specify tricky formats:

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

More details on format here:

• Python 2 https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
• Python 3 https://docs.python.org/3.7/library/datetime.html#strftime-strptime-behavior

Answer 4

If you have a mixture of formats in your date, don't forget to set infer_datetime_format=True to make life easier.

df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

Source: pd.to_datetime

or if you want a customized approach:

def autoconvert_datetime(value):
    formats = ['%m/%d/%Y', '%m-%d-%y']  # formats to try
    result_format = '%d-%m-%Y'  # output format
    for dt_format in formats:
        try:
            dt_obj = datetime.strptime(value, dt_format)
            return dt_obj.strftime(result_format)
        except Exception as e:  # throws exception when format doesn't match
            pass
    return value  # let it be if it doesn't match

df['date'] = df['date'].apply(autoconvert_datetime)

Answer 5

Multiple datetime columns

If you want to convert multiple string columns to datetime, then using apply() would be useful.

df[['date1', 'date2']] = df[['date1', 'date2']].apply(pd.to_datetime)

You can pass parameters to to_datetime as kwargs.

df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(pd.to_datetime, format="%m/%d/%Y")

Passing to apply, without specifying axis, still converts values vectorially for each column. apply is needed here because pd.to_datetime can only be called on a single column. If it has to be called on multiple columns, the options are either use an explicit for-loop, or pass it to apply. On the other hand, if you call pd.to_datetime using apply on a column (e.g. df['date'].apply(pd.to_datetime)), that would not be vectorized, and should be avoided.

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Use format= to speed up

If the column contains a time component and you know the format of the datetime/time, then passing the format explicitly would significantly speed up the conversion. There's barely any difference if the column is only date, though. In my project, for a column with 5 millions rows, the difference was huge: ~2.5 min vs 6s.

It turns out explicitly specifying the format is about 25x faster. The following runtime plot shows that there's a huge gap in performance depending on whether you passed format or not.

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The code used to produce the plot:

import perfplot
import random

mdYHM = range(1, 13), range(1, 29), range(2000, 2024), range(24), range(60)
perfplot.show(
    kernels=[lambda x: pd.to_datetime(x), lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M')],
    labels=['pd.to_datetime(x)', "pd.to_datetime(x, format='%m/%d/%Y %H:%M')"],
    n_range=[2**k for k in range(19)],
    setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}" 
                               for m,d,Y,H,M in zip(*[random.choices(e, k=n) for e in mdYHM])]),
    equality_check=pd.Series.equals,
    xlabel='len(df)'
)

Answer 6

Try this solution:

• Change '2022–12–31 00:00:00' to '2022–12–31 00:00:01'
• Then run this code: pandas.to_datetime(pandas.Series(['2022–12–31 00:00:01']))
• Output: 2022–12–31 00:00:01