如何从 python 中的当前日期的 gps 纪元获取闰秒

Question

据我所知，目前在 python 中没有从当前日期的 gps 纪元获取闰秒的功能。 investigation/development下的相应补丁：leap seconds path in datetime

您能否建议获得闰秒的最佳方法？我找到了解决方案：

def getLeapSec(Tgps):
  daysFromGPSEpoch = timedelta(seconds=Tgps).days
  tableLeapSec = ([
    [0,  542,  907, 1272, 2003, 2917, 3648, 4013, 4560, 4925, 5290, 5839, 6386, 6935, 9492, 10588, 11865, 12960],
    [0,    1,    2,    3,    4,    5,    6,    7,    8,    9,   10,   11,   12,   13,   14,    15,    16,    17]
                 ])

  for i in range(1, len(tableLeapSec[0])):
    if daysFromGPSEpoch < tableLeapSec[0][i]:
      return tableLeapSec[1][i-1]

示例：

In [25]: Tgps = 1092121243.0

In [26]: getLeapSec(Tgps)
Out[26]: 16

Answer 1

您可以在您的系统上使用 leapseconds module if you have access to the up-to-date tzfile：

>>> from datetime import datetime, timedelta
>>> import leapseconds
>>> Tgps = 1092121243.0
>>> gps_time = datetime(1980, 1, 6) + timedelta(seconds=Tgps)
>>> leapseconds.dTAI_UTC_from_tai(leapseconds.gps_to_tai(gps_time))
datetime.timedelta(0, 35)

主要优点是 tzdata 包每年由系统自动更新几次（比引入闰秒更频繁）。您的代码将使用新数据而无需更改它。

否则，作为回退，您可以对 leapseconds.leapseconds() 调用的结果进行硬编码：

>>> import leapseconds
>>> leapseconds.leapseconds()
[LeapSecond(utc=datetime.datetime(1972, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 10)),
 LeapSecond(utc=datetime.datetime(1972, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 11)),
 LeapSecond(utc=datetime.datetime(1973, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 12)),
 LeapSecond(utc=datetime.datetime(1974, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 13)),
 LeapSecond(utc=datetime.datetime(1975, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 14)),
 LeapSecond(utc=datetime.datetime(1976, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 15)),
 LeapSecond(utc=datetime.datetime(1977, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 16)),
 LeapSecond(utc=datetime.datetime(1978, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 17)),
 LeapSecond(utc=datetime.datetime(1979, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 18)),
 LeapSecond(utc=datetime.datetime(1980, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 19)),
 LeapSecond(utc=datetime.datetime(1981, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 20)),
 LeapSecond(utc=datetime.datetime(1982, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 21)),
 LeapSecond(utc=datetime.datetime(1983, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 22)),
 LeapSecond(utc=datetime.datetime(1985, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 23)),
 LeapSecond(utc=datetime.datetime(1988, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 24)),
 LeapSecond(utc=datetime.datetime(1990, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 25)),
 LeapSecond(utc=datetime.datetime(1991, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 26)),
 LeapSecond(utc=datetime.datetime(1992, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 27)),
 LeapSecond(utc=datetime.datetime(1993, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 28)),
 LeapSecond(utc=datetime.datetime(1994, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 29)),
 LeapSecond(utc=datetime.datetime(1996, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 30)),
 LeapSecond(utc=datetime.datetime(1997, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 31)),
 LeapSecond(utc=datetime.datetime(1999, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 32)),
 LeapSecond(utc=datetime.datetime(2006, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 33)),
 LeapSecond(utc=datetime.datetime(2009, 1, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 34)),
 LeapSecond(utc=datetime.datetime(2012, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 35)),
 LeapSecond(utc=datetime.datetime(2015, 7, 1, 0, 0), dTAI_UTC=datetime.timedelta(0, 36))]

在这种情况下，当引入下一个闰秒时，您必须手动更新列表。

闰秒更新的权威来源是Bulletin C issued by IERS that also provides TAI-UTC data since 1961。

您可以每半年（例如，一月、七月）对 IESR 执行一次 ping 操作，以使用 leap_second_client.py 以编程方式找出下一个闰秒的计划时间。

如何从 python 中的当前日期的 gps 纪元获取闰秒

How to get leap seconds from gps epoch for current date in python

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