Python 如何根据我想要的日期创建间隔为一年的时间戳数组？

我想根据我想与熊猫相处的日期，制作一个间隔一年的时间戳数组。

例如）['1994-08-17 00:00:00+09:00'，'1995-08-17 00:00:00+09:00'， “1996-08-17 00:00:00+09:00”，“1997-08-17 00:00:00+09:00”

但我不能去。请帮帮我。

下面是我制作的一些数组

1） freq='A-AUG'

日期时间索引（['1994-08-31 00:00:00+09:00'，'1995-08-31 00:00:00+09:00', '1996-08-31 00:00:00+09:00', '1997-08-31 00:00:00+09:00', '1998-08-31 00:00:00+09:00', '1999-08-31 00:00:00+09:00', '2000-08-31 00:00:00+09:00', '2001-08-31 00:00:00+09:00', '2002-08-31 00:00:00+09:00', '2003-08-31 00:00:00+09:00', '2004-08-31 00:00:00+09:00', '2005-08-31 00:00:00+09:00', '2006-08-31 00:00:00+09:00', '2007-08-31 00:00:00+09:00', '2008-08-31 00:00:00+09:00', '2009-08-31 00:00:00+09:00', '2010-08-31 00:00:00+09:00', '2011-08-31 00:00:00+09:00', '2012-08-31 00:00:00+09:00', '2013-08-31 00:00:00+09:00', '2014-08-31 00:00:00+09:00', '2015-08-31 00:00:00+09:00', '2016-08-31 00:00:00+09:00', '2017-08-31 00:00:00+09:00', '2018-08-31 00:00:00+09:00', '2019-08-31 00:00:00+09:00'], dtype='datetime64[ns，亚洲/首尔]'，freq='A-AUG'）

2） freq='A'

日期时间索引（['1994-12-31 00:00:00+09:00'，'1995-12-31 00:00:00+09:00', '1996-12-31 00:00:00+09:00', '1997-12-31 00:00:00+09:00', '1998-12-31 00:00:00+09:00', '1999-12-31 00:00:00+09:00', '2000-12-31 00:00:00+09:00', '2001-12-31 00:00:00+09:00', '2002-12-31 00:00:00+09:00', '2003-12-31 00:00:00+09:00', '2004-12-31 00:00:00+09:00', '2005-12-31 00:00:00+09:00', '2006-12-31 00:00:00+09:00', '2007-12-31 00:00:00+09:00', '2008-12-31 00:00:00+09:00', '2009-12-31 00:00:00+09:00', '2010-12-31 00:00:00+09:00', '2011-12-31 00:00:00+09:00', '2012-12-31 00:00:00+09:00', '2013-12-31 00:00:00+09:00', '2014-12-31 00:00:00+09:00', '2015-12-31 00:00:00+09:00', '2016-12-31 00:00:00+09:00', '2017-12-31 00:00:00+09:00', '2018-12-31 00:00:00+09:00', '2019-12-31 00:00:00+09:00'], dtype='datetime64[ns，亚洲/首尔]'，freq='A-DEC'）

sw_y=pd.date_范围（start='1994-08-17'，
周期=26，
freq='A-AUG'，
(亚洲/汉城)

您可以设置之后的日期：

import pandas as pd
sw_y = pd.date_range(start='1994-08-17',
                      periods=26,
                      freq='A-AUG',
                      tz='Asia/Seoul')
sw_y = sw_y + pd.DateOffset(day=17)

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如果答案解决了你的问题，你可以接受。