Convert Pandas Column to DateTime

You can use the DataFrame method .apply() to operate on the values in Mycol:

>>> df = pd.DataFrame(['05SEP2014:00:00:00.000'],columns=['Mycol'])
>>> df
                    Mycol
0  05SEP2014:00:00:00.000
>>> import datetime as dt
>>> df['Mycol'] = df['Mycol'].apply(lambda x: 
                                    dt.datetime.strptime(x,'%d%b%Y:%H:%M:%S.%f'))
>>> df
       Mycol
0 2014-09-05

chrisb‘s answer works:

raw_data['Mycol'] =  pd.to_datetime(raw_data['Mycol'], format='%d%b%Y:%H:%M:%S.%f') 

however it results in a Python warning of
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

I would guess this is due to some chaining indexing.

If you have more than one column to be converted you can do the following:

df[["col1", "col2", "col3"]] = df[["col1", "col2", "col3"]].apply(pd.to_datetime)

Use the pandas to_datetime function to parse the column as DateTime. Also, by using infer_datetime_format=True, it will automatically detect the format and convert the mentioned column to DateTime.

import pandas as pd
raw_data['Mycol'] =  pd.to_datetime(raw_data['Mycol'], infer_datetime_format=True)

Time Saver:

raw_data['Mycol'] =  pd.to_datetime(raw_data['Mycol'])

Just like we convert object data type to float or int. Use astype()

raw_data['Mycol']=raw_data['Mycol'].astype('datetime64[ns]')

To silence SettingWithCopyWarning

If you got this warning, then that means your dataframe was probably created by filtering another dataframe. Make a copy of your dataframe before any assignment and you’re good to go.

df = df.copy()
df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f')

errors='coerce' is useful

If some rows are not in the correct format or not datetime at all, errors= parameter is very useful, so that you can convert the valid rows and handle the rows that contained invalid values later.

df['date'] = pd.to_datetime(df['date'], format='%d%b%Y:%H:%M:%S.%f', errors='coerce')

# for multiple columns
df[['start', 'end']] = df[['start', 'end']].apply(pd.to_datetime, format='%d%b%Y:%H:%M:%S.%f', errors='coerce')

Setting the correct format= is much faster than letting pandas find out¹

Long story short, passing the correct format= from the beginning as in chrisb’s post is much faster than letting pandas figure out the format, especially if the format contains time component. The runtime difference for dataframes greater than 10k rows is huge (~25 times faster, so we’re talking like a couple minutes vs a few seconds). All valid format options can be found at https://strftime.org/.

¹ Code used to produce the timeit test plot.

import perfplot
from random import choices
from datetime import datetime

mdYHMSf = range(1,13), range(1,29), range(2000,2024), range(24), *[range(60)]*2, range(1000)
perfplot.show(
    kernels=[lambda x: pd.to_datetime(x), 
             lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M:%S.%f'), 
             lambda x: pd.to_datetime(x, infer_datetime_format=True),
             lambda s: s.apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))],
    labels=["pd.to_datetime(df['date'])", 
            "pd.to_datetime(df['date'], format='%m/%d/%Y %H:%M:%S.%f')", 
            "pd.to_datetime(df['date'], infer_datetime_format=True)", 
            "df['date'].apply(lambda x: datetime.strptime(x, '%m/%d/%Y %H:%M:%S.%f'))"],
    n_range=[2**k for k in range(20)],
    setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}:{S}.{f}" 
                               for m,d,Y,H,M,S,f in zip(*[choices(e, k=n) for e in mdYHMSf])]),
    equality_check=pd.Series.equals,
    xlabel='len(df)'
)