我从车辆中集成的GPS接收到此数据：
INS_Time::INS_Time_Millisec[ms] # example of the value: 295584830.0
INS_Time::INS_Time_Week[Week] # example of the value: 2077.0
INS_Time::Leap_seconds[s] # example of the value: 18.0
我需要的是UTC时间戳，所以我假设必须将GPS中所有不同的时间组合起来才能获得UTC时间戳，但是我不知道该怎么做。如果有人可以指导我完成这项工作，我将不胜感激。
example of the result I want to have: 1572430625230
我正在使用python 3.7，如果有一个用于此的库，那将非常有帮助，否则我也在搜索算法来做到这一点
我的猜测：
根据https://en.wikipedia.org/wiki/Epoch_(computing)#Notable_epoch_dates_in_computing，GPS时代是1980年1月6日，GPS counts weeks (a week is defined to start on Sunday) and 6 January is the first Sunday of 1980
根据http://leapsecond.com/java/gpsclock.htm，GPS time was zero at 0h 6-Jan-1980 and since it is not perturbed by leap seconds GPS is now ahead of UTC by 18 seconds.
因此，我们必须定义一个gps_epoch并减去给定的leap秒才能获得utc日期时间
from datetime import datetime,timedelta
import pytz
def gps_datetime(time_week,time_ms,leap_seconds):
gps_epoch = datetime(1980,1,6,tzinfo=pytz.utc)
# gps_time - utc_time = leap_seconds
return gps_epoch + timedelta(weeks=time_week,milliseconds=time_ms,seconds=-leap_seconds)
以您的示例
>>>gps_datetime(2077,295584830.0,18.0)
datetime.datetime(2019,10,30,830000,tzinfo=<UTC>)
>>>gps_datetime(2077,18.0).timestamp()
1572429966.83
但是，我离您的预期结果还很遥远（1572430625230即使以毫秒表示）
别忘了pip install pytz