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我有一个数据框 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

  1. 系列中的第一个时间戳从 更改'2011-08-05 00:00:00-04:00''2011-08-04 20:00:00'
  2. 系列中的第二个邮票将从 更改'2011-08-05 00:00:00-04:00''2011-08-04 21:00:00',以此类推。

我试过pd.to_datetime(df.index, format='%Y-%m-%d %H:%M:%S')了,但它返回与datetimeindex上面相同的对象。

如果将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')

但是这两行代码都没有达到我的最终目标。

4

1 回答 1

2

用于tz_convert删除timezone和添加Hour

df.index.tz_convert(None) + pd.offsets.Hour(16)

或者:

df.index.tz_convert(None) + pd.Timedelta(16, unit='h')

样本:

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')
于 2017-12-03T17:16:11.040 回答