How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetimes?
4 Answers
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]
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.
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
- Python 3
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)
