如何在python中从GPS未分段时间中获取当前日期和时间

我的gps未分段时间如下：

Tgps = 1092121243.0

我想知道那是什么日期和时间。GPS时间开始于1980年1月6日。Python函数

datetime.utcfromtimestamp 

可以从1970年1月1日起给出秒数

我发现：

from datetime import datetime
GPSfromUTC = (datetime(1980,1,6) - datetime(1970,1,1)).total_seconds()
curDate = datetime.utcfromtimestamp(Tgps + GPSfromUTC) 

Out[83]: datetime.datetime(2014, 8, 15, 7, 0, 43)

def gps2utc(week, secs):
    """
    :param week: GPS week number, i.e. 1866
    :param secs: number of seconds since the beginning of `week`
    :return: datetime instance with UTC time
    """
    secs_in_week = 604800
    gps_epoch = datetime(1980, 1, 6, 0, 0, 0)
    date_before_leaps = gps_epoch + timedelta(seconds=week * secs_in_week + secs)
    return date_before_leaps - timedelta(seconds=leap(date_before_leaps))

我不确定它们是否包含在函数datetime中，或者我应该计算它们并从结果中减去？

---

可能还有更好的解决方案吗？

GPS时间与UTC同步开始：

1980-01-06（UTC）==1980-01-06（GPS）

。两者都以秒为单位。GPS时间和UTC时间之间的差异随着每闰秒的增加而增加

要找到正确的UTC时间，您需要知道给定GPS时间之前发生的闰秒数：

#!/usr/bin/env python
from datetime import datetime, timedelta

# utc = 1980-01-06UTC + (gps - (leap_count(2014) - leap_count(1980)))
utc = datetime(1980, 1, 6) + timedelta(seconds=1092121243.0 - (35 - 19))
print(utc)

输出 其中

leap\u count（date）

是给定日期之前引入的闰秒数。发件人（注：该网站是leap seconds的权威来源。它发布）：

因此：

(leap_count(2014) - leap_count(1980)) == (35 - 19)

---

如果您在Unix上，则可以使用

“right”

时区从TAI时间获取UTC时间 （而且很容易从GPS时间获取TAI时间：）：

输出

---

如果从相应的列表中提取闰秒列表，则可以避免更改时区。它是前两种方法的组合，其中第一种方法的跳跃计数计算是自动的，而第二种方法的自动更新

tzdata

（系统包for）是使用的：

>>> from datetime import datetime, timedelta
>>> import leapseconds
>>> leapseconds.gps_to_utc(datetime(1980,1,6) + timedelta(seconds=1092121243.0))
datetime.datetime(2014, 8, 15, 7, 0, 27)

其中可以从

/usr/share/zoneinfo/right/UTC

文件（属于

tzdata

软件包）中提取闰秒

---

这三种方法产生相同的结果。

我使用以下计算闰秒的函数：

import bisect
from datetime import datetime, timedelta

_LEAP_DATES = ((1981, 6, 30), (1982, 6, 30), (1983, 6, 30),
               (1985, 6, 30), (1987, 12, 31), (1989, 12, 31),
               (1990, 12, 31), (1992, 6, 30), (1993, 6, 30),
               (1994, 6, 30), (1995, 12, 31), (1997, 6, 30),
               (1998, 12, 31), (2005, 12, 31), (2008, 12, 31),
               (2012, 6, 30), (2015, 6, 30), (2016, 12, 31))

LEAP_DATES = tuple(datetime(i[0], i[1], i[2], 23, 59, 59) for i in _LEAP_DATES)

def leap(date):
    """
    Return the number of leap seconds since 1980-01-01

    :param date: datetime instance
    :return: leap seconds for the date (int)
    """
    # bisect.bisect returns the index `date` would have to be
    # inserted to keep `LEAP_DATES` sorted, so is the number of
    # values in `LEAP_DATES` that are less than `date`, or the
    # number of leap seconds.
    return bisect.bisect(LEAP_DATES, date)

当然，您需要偶尔更新

\u LEAP\u DATES

，但这些更新非常罕见

通常，GPS时间由两个数字组成：GPS周和自当前GPS周开始的秒数。因此，您可以使用以下选项：

from datetime import datetime
GPSfromUTC = (datetime(1980,1,6) - datetime(1970,1,1)).total_seconds()
curDate = datetime.utcfromtimestamp(Tgps + GPSfromUTC) 

Out[83]: datetime.datetime(2014, 8, 15, 7, 0, 43)

def gps2utc(week, secs):
    """
    :param week: GPS week number, i.e. 1866
    :param secs: number of seconds since the beginning of `week`
    :return: datetime instance with UTC time
    """
    secs_in_week = 604800
    gps_epoch = datetime(1980, 1, 6, 0, 0, 0)
    date_before_leaps = gps_epoch + timedelta(seconds=week * secs_in_week + secs)
    return date_before_leaps - timedelta(seconds=leap(date_before_leaps))

在您的情况下，

week=0

，因此：

In [1]: gps2utc(0, 1092121243.0)
Out[1]: datetime.datetime(2014, 8, 15, 7, 0, 27)

您可以使用软件包执行以下操作：

到泰山的GPS时间 返回

2014-08-15 07:01:02.000

全球定位系统到UTC的时间

---

它输出

2014-08-15 07:00:27.000

为了确保您的代码是最新的，您必须每年咨询两次（理论上每年咨询4次）。您可以依赖自动更新的

tzdata

系统包。

def gps2utc(week, secs):
    """
    :param week: GPS week number, i.e. 1866
    :param secs: number of seconds since the beginning of `week`
    :return: datetime instance with UTC time
    """
    secs_in_week = 604800
    gps_epoch = datetime(1980, 1, 6, 0, 0, 0)
    date_before_leaps = gps_epoch + timedelta(seconds=week * secs_in_week + secs)
    return date_before_leaps - timedelta(seconds=leap(date_before_leaps))

In [1]: gps2utc(0, 1092121243.0)
Out[1]: datetime.datetime(2014, 8, 15, 7, 0, 27)

from astropy.time import Time
mytime = 1092121243.0
t = Time(mytime, format='gps')
t = Time(t, format='iso') # same as scale='tai'
print(t)

from astropy.time import Time
sec = 1092121243.0
t_in = Time(sec, format='gps')
t_out = Time(t_in, format='iso', scale='utc')
print(t_out)