如何获取当前的日期和时间，从python的GPS unsegment时间

问题描述

我有gps未分段的时间，如下所示：

  Tgps = 1092121243.0 
  / pre> 
 
 
我想了解这个日期和时间。 GPS时间的开始是1980年1月6日。Python功能
  datetime.utcfromtimestamp 
  
可以从1970年1月1日起给出秒。
 
 
 
我发现以下内容：
从$ datetime import datetime 
 GPSfromUTC =（datetime（1980,1,6） -  datetime（1970,1,1））。 total_seconds（）
 curDate = datetime.utcfromtimestamp（Tgps + GPSfromUTC）
 
输出[83]：datetime.datetime（2014，8，15，7，0，43）
  
我不确定它们是否包含在函数datetime中，或者我应该从结果中计算和减少
也可能存在这个问题的更好的解决方案？
解决方案
 GPS时间与UTC同步开始： 1980-01-06（UTC）== 1980-01-06（GPS）。两者都以SI秒为单位。 GPS时间和UTC时间之间的差异随着每个（闰秒）闰秒而增加。
 
 
 
要查找正确的UTC时间，您需要知道之前发生的闰秒数给定的GPS时间：
 ＃！/ 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
 
 
 
  2014-08-15 07:00:27＃（UTC）
  
其中 leap_count（date）是给定日期之前引入的闰秒数。从（注意：该网站是闰秒的权威来源，它发布）：
  1980 ..：19s 
 2012 ..：35s 
  
因此：
 （leap_count（2014） -  leap_count（1980））==（35  -  19）
  
如果您使用的是Unix，那么您可以使用right时区从TAI时间获取UTC时间
 （并且很容易从GPS时间获取TAI时间：）：
 ＃！/ usr / bin / env python 
 import os 
进口时间
 
 os.environ ['TZ'] ='right / UTC'＃TAI scale with 1970-01-01 00:00:10（TAI）epoch 
 time.tzset（）＃Unix 
 
 from datetime import datetime，timedelta 
 
 gps_timestamp = 1092121243.0 ＃input 
 gps_epoch_as_gps = datetime（1980,1,6）
＃定义
 gps_time_as_gps = gps_epoch_as_gps + timedelta（seconds = gps_timestamp）
 gps_time_as_tai = gps_time_as_gps + timedelta（seconds = 19 ）＃constant offset 
 tai_epoch_as_tai = datetime（1970，1，1，0，0，10）
＃定义
 tai_timestamp =（gps_time_as_tai  -  tai_epoch_as_tai）.total_seconds（）
 print（datetime.utcfromtimestamp（tai_timestamp））＃righttimezone有效！ 
  
 
 
 
输出
 
 
 
  2014-08-15 07:00:27＃（UTC）
  
 
 
 
 
 
 
如果从相应的。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化的，并且自动更新 tzdata （系统包为）：
 >>>从datetime import datetime，timedelta 
>>>导入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 文件中提取闰秒（ tzdata 包）。
 
 
 
所有三种方法都产生相同的结果。
 
I have gps unsegmented time like this:
Tgps = 1092121243.0
And I'd like to understand what date and time is that. The begining of GPS time is 6 January 1980. Python function 
datetime.utcfromtimestamp
could give seconds from 1 January 1970 year.
I found following:
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)
I'm not sure about leapseconds are they included in function datetime or I should calculate them and substract from the result?May be also exists better solution of this problem?
 解决方案 
GPS time started in sync with UTC: 1980-01-06 (UTC) == 1980-01-06 (GPS). Both tick in SI seconds. The difference between GPS time and UTC time increases with each (intercalary) leap second.
To find the correct UTC time, you need to know the number of leap seconds occurred before the given GPS time:
#!/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)
Output
2014-08-15 07:00:27 # (UTC)
where leap_count(date) is the number of leap seconds introduced before the given date. From TAI-UTC table (note: the site is the authoritative source on leap seconds. It publishes Bulletin C announcing new leap seconds):
1980..: 19s
2012..: 35s
and therefore: 
(leap_count(2014) - leap_count(1980)) == (35 - 19)
If you are on Unix then you could use "right" time zone to get UTC time from TAI time(and it is easy to get TAI time from GPS time: TAI = GPS + 19 seconds (constant offset)):
#!/usr/bin/env python
import os
import time

os.environ['TZ'] = 'right/UTC' # TAI scale with 1970-01-01 00:00:10 (TAI) epoch
time.tzset() # Unix

from datetime import datetime, timedelta

gps_timestamp = 1092121243.0 # input
gps_epoch_as_gps = datetime(1980, 1, 6)
# by definition
gps_time_as_gps = gps_epoch_as_gps + timedelta(seconds=gps_timestamp)
gps_time_as_tai = gps_time_as_gps + timedelta(seconds=19) # constant offset
tai_epoch_as_tai = datetime(1970, 1, 1, 0, 0, 10)
# by definition
tai_timestamp = (gps_time_as_tai - tai_epoch_as_tai).total_seconds()
print(datetime.utcfromtimestamp(tai_timestamp)) # "right" timezone is in effect!
Output
2014-08-15 07:00:27 # (UTC)
You could avoid changing the timezone if you extract the leap seconds list from the corresponding tzfile(5). It is a combination of the first two methods where the leap count computation from the first method is automated and the autoupdating tzdata (system package for the tz database) from the second method is used:
>>> 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)
where leapseconds.py can extract leap seconds from /usr/share/zoneinfo/right/UTC file (part of tzdata package).
All three methods produce the same result.
                        
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