我有一个数据帧df,当我运行print(df.index)时,我得到:
DatetimeIndex(['2011-08-05 00:00:00-04:00', '2011-08-05 01:00:00-04:00',
'2011-08-05 02:00:00-04:00', '2011-08-05 03:00:00-04:00',
'2011-08-05 04:00:00-04:00', '2011-08-05 05:00:00-04:00',
'2011-08-05 06:00:00-04:00', '2011-08-05 07:00:00-04:00',
'2011-08-05 08:00:00-04:00', '2011-08-05 09:00:00-04:00',
...
'2017-07-30 14:00:00-04:00', '2017-07-30 15:00:00-04:00',
'2017-07-30 16:00:00-04:00', '2017-07-30 17:00:00-04:00',
'2017-07-30 18:00:00-04:00', '2017-07-30 19:00:00-04:00',
'2017-07-30 20:00:00-04:00', '2017-07-30 21:00:00-04:00',
'2017-07-30 22:00:00-04:00', '2017-07-30 23:00:00-04:00'],
dtype='datetime64[ns, America/New_York]', name=u'Time', length=52488, freq=None)
我正在尝试修改datetimeindex对象,以便
'2011-08-05 00:00:00-04:00'
更改为'2011-08-04 20:00:00'
和'2011-08-05 00:00:00-04:00'
更改为'2011-08-04 21:00:00'
,依此类推。 我尝试pd.to_datetime(df.index, format='%Y-%m-%d %H:%M:%S')
,但它返回与上面相同的datetimeindex
对象。
如果时间戳转换为字符串,我可以,所以我尝试了:
df.index.strftime('%Y-%m-%d %H:%M:%S')
但两行代码都没有实现我的最终目标。
答案 0 :(得分:0)
使用tz_convert
删除timezone
并添加Hour
s:
df.index.tz_convert(None) + pd.offsets.Hour(16)
或者:
df.index.tz_convert(None) + pd.Timedelta(16, unit='h')
样品:
idx = ['2011-08-05 00:00:00-04:00', '2011-08-05 01:00:00-04:00',
'2011-08-05 02:00:00-04:00', '2011-08-05 03:00:00-04:00']
idx = pd.DatetimeIndex(idx).tz_localize('UTC').tz_convert('America/New_York')
print (idx)
DatetimeIndex(['2011-08-05 00:00:00-04:00', '2011-08-05 01:00:00-04:00',
'2011-08-05 02:00:00-04:00', '2011-08-05 03:00:00-04:00'],
dtype='datetime64[ns, America/New_York]', freq=None)
idx = idx.tz_convert(None) + pd.offsets.Hour(16)
print (idx)
DatetimeIndex(['2011-08-05 20:00:00', '2011-08-05 21:00:00',
'2011-08-05 22:00:00', '2011-08-05 23:00:00'],
dtype='datetime64[ns]', freq='H')