从原点和偏移创建DatetimeIndex

2024-04-24 09:09:05 发布

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我想把一组秒转换成适合数据帧的索引

我想我在寻找dt.total_seconds()操作的逆运算,就像Accessing total_seconds() in pandas data columnhttps://stackoverflow.com/a/36156620/1653571答案一样

给出:

t0="2010-10-01 00:00:00 UTC"
seconds = [0,1,2,3,5,7,11,13]

我想要这样的东西:

ii = pd.DatetimeIndex( ???(t0,seconds))

Tags: 数据答案inhttpscompandasdatadt
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1楼 · 发布于 2024-04-24 09:09:05

seconds列表转换为Timedelta,并执行一点datetime算法

# origin           + offset
pd.to_datetime(t0) + pd.to_timedelta(seconds, unit='s')

DatetimeIndex(['2010-10-01 00:00:00+00:00', '2010-10-01 00:00:01+00:00',
               '2010-10-01 00:00:02+00:00', '2010-10-01 00:00:03+00:00',
               '2010-10-01 00:00:05+00:00', '2010-10-01 00:00:07+00:00',
               '2010-10-01 00:00:11+00:00', '2010-10-01 00:00:13+00:00'],
              dtype='datetime64[ns, UTC]', freq=None)

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