How to convert a column consisting of datetime64 objects to a strings that would read 01-11-2013 for today's date of November 1.
I have tried
df['DateStr'] = df['DateObj'].strftime('%d%m%Y')
but I get this error
AttributeError: 'Series' object has no attribute 'strftime'
As of version 17.0, you can format with the dt accessor:
df['DateStr'] = df['DateObj'].dt.strftime('%d%m%Y')
assign, such as: (df .query('my query criteria here) .assign(month = lambda x: x['date'].dt.strftime('%Y-%m') )
df['A'].apply(lambda x: x.strftime('%d%m%Y')) . For the 230k col, the cost time is 1.3 seconds, and that of the latter one is about 1.9 seconds.
dt accessor should be significantly faster as apply is more or less a for loop whereas dt uses C-based optimizations iirc.
A value is trying to be set on a copy of a slice from a DataFrame.
df['DateObj'].astype('datetime64[ns]').dt.strftime('%Y')In [6]: df = DataFrame(dict(A = date_range('20130101',periods=10)))
In [7]: df
Out[7]:
A
0 2013-01-01 00:00:00
1 2013-01-02 00:00:00
2 2013-01-03 00:00:00
3 2013-01-04 00:00:00
4 2013-01-05 00:00:00
5 2013-01-06 00:00:00
6 2013-01-07 00:00:00
7 2013-01-08 00:00:00
8 2013-01-09 00:00:00
9 2013-01-10 00:00:00
In [8]: df['A'].apply(lambda x: x.strftime('%d%m%Y'))
Out[8]:
0 01012013
1 02012013
2 03012013
3 04012013
4 05012013
5 06012013
6 07012013
7 08012013
8 09012013
9 10012013
Name: A, dtype: object
It works directly if you first set as index. Then essentially you pass a 'DatetimeIndex' object and not a 'Series'
df = df.set_index('DateObj').copy()
df['DateStr'] = df.index.strftime('%d%m%Y')
