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年05月23日 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年05月23日
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['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.

• fantabolous: fyi when timezone is specified in the string it ignores it

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)

• Asclepius: A customized approach can be used without resorting to .apply which has no fast cache, and will struggle when converting a billion values. An alternative, but not a great one, is col = pd.concat([pd.to_datetime(col, errors='coerce', format=f) for f in formats], axis='columns').bfill(axis='columns').iloc[:, 0]
• Asclepius: If you have a mixture of formats, you should not use infer_datetime_format=True as this assumes a single format. Just skip this argument. To understand why, try pd.to_datetime(pd.Series(['1/5/2015 8:08:00 AM', '1/4/2015 11:24:00 PM']), infer_datetime_format=True) with and without errors='coerce'. See this issue.

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.

---

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.

timings

---

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

• wjandrea: "Change '2022–12–31 00:00:00' to '2022–12–31 00:00:01'" - What does that have to do with the question?

Answer 7

print(df1.shape)
(638765, 95)
%timeit df1['Datetime'] = pd.to_datetime(df1['Date']+" "+df1['HOUR'])
473 ms ± 8.33 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df1['Datetime'] = pd.to_datetime(df1['Date']+" "+df1['HOUR'], format='mixed')
688 ms ± 3.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df1['Datetime'] = pd.to_datetime(df1['Date']+" "+df1['HOUR'], format='%Y-%m-%d %H:%M:%S')
470 ms ± 7.31 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)