我在pandas数据框中有一个列,如下所示:
test1.Received
Out[9]:
0 01/01/2015 17:25
1 02/01/2015 11:43
2 04/01/2015 18:21
3 07/01/2015 16:17
4 12/01/2015 20:12
5 14/01/2015 11:09
6 15/01/2015 16:05
7 16/01/2015 21:02
8 26/01/2015 03:00
9 27/01/2015 08:32
10 30/01/2015 11:52
这表示时间戳为日月年小时分钟。我想将日期重新安排为年月日小时分钟。所以它看起来像:
test1.Received
Out[9]:
0 2015/01/01 17:25
1 2015/01/02 11:43
...
答案 0 :(得分:1)
只需使用pd.to_datetime
:
In [33]:
import pandas as pd
pd.to_datetime(df['date'])
Out[33]:
index
0 2015-01-01 17:25:00
1 2015-02-01 11:43:00
2 2015-04-01 18:21:00
3 2015-07-01 16:17:00
4 2015-12-01 20:12:00
5 2015-01-14 11:09:00
6 2015-01-15 16:05:00
7 2015-01-16 21:02:00
8 2015-01-26 03:00:00
9 2015-01-27 08:32:00
10 2015-01-30 11:52:00
Name: date, dtype: datetime64[ns]
在你的情况下:
pd.to_datetime(test1['Received'])
应该正常工作
如果要更改显示格式,则需要解析为日期时间,然后apply
`datetime.strftime:
In [35]:
import datetime as dt
pd.to_datetime(df['date']).apply(lambda x: dt.datetime.strftime(x, '%m/%d/%y %H:%M:%S'))
Out[35]:
index
0 01/01/15 17:25:00
1 02/01/15 11:43:00
2 04/01/15 18:21:00
3 07/01/15 16:17:00
4 12/01/15 20:12:00
5 01/14/15 11:09:00
6 01/15/15 16:05:00
7 01/16/15 21:02:00
8 01/26/15 03:00:00
9 01/27/15 08:32:00
10 01/30/15 11:52:00
Name: date, dtype: object
所以上面现在显示月/日/年,在你的情况下,以下应该有效:
pd.to_datetime(test1['Received']).apply(lambda x: dt.datetime.strftime(x, '%y/%m/%d %H:%M:%S'))
修改强>
您似乎需要将参数dayfirst=True
传递给to_datetime
:
In [45]:
pd.to_datetime(df['date'], format('%d/%m/%y %H:%M:%S'), dayfirst=True).apply(lambda x: dt.datetime.strftime(x, '%m/%d/%y %H:%M:%S'))
Out[45]:
index
0 01/01/15 17:25:00
1 01/02/15 11:43:00
2 01/04/15 18:21:00
3 01/07/15 16:17:00
4 01/12/15 20:12:00
5 01/14/15 11:09:00
6 01/15/15 16:05:00
7 01/16/15 21:02:00
8 01/26/15 03:00:00
9 01/27/15 08:32:00
10 01/30/15 11:52:00
Name: date, dtype: object
答案 1 :(得分:1)
Pandas内置此功能,您可以指定日期时间格式
http://pandas.pydata.org/pandas-docs/stable/generated/pandas.to_datetime.html。
使用infer_datetime_format
>>> import pandas as pd
>>> i = pd.date_range('20000101',periods=100)
>>> df = pd.DataFrame(dict(year = i.year, month = i.month, day = i.day))
>>> pd.to_datetime(df.year*10000 + df.month*100 + df.day, format='%Y%m%d')
0 2000-01-01
1 2000-01-02
...
98 2000-04-08
99 2000-04-09
Length: 100, dtype: datetime64[ns]
答案 2 :(得分:1)
您可以使用datetime
函数从字符串转换为字符串。
# converts to date
datetime.strptime(date_string, 'DD/MM/YYYY HH:MM')
和
# converts to your requested string format
datetime.strftime(date_string, "YYYY/MM/DD HH:MM:SS")