[英]How to get current date and time from GPS unsegment time in python
我有這樣的 gps 未分段時間:
Tgps = 1092121243.0
我想知道那是什么日期和時間。 GPS的開始時間是1980年1月6日。 Python function
datetime.utcfromtimestamp
可以給出從 1 January 1970 年開始的秒數。
我發現以下內容:
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)
我不確定閏秒是否包含在 function 日期時間中,或者我應該計算它們並從結果中減去? 可能還存在更好的解決這個問題的方法嗎?
GPS 時間開始與 UTC 同步: 1980-01-06 (UTC) == 1980-01-06 (GPS)
。 兩者都在 SI 秒內打勾。 GPS 時間和 UTC 時間之間的差異隨着每個(閏秒)閏秒而增加。
要找到正確的 UTC 時間,您需要知道在給定 GPS 時間之前發生的閏秒數:
#!/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)
2014-08-15 07:00:27 # (UTC)
其中leap_count(date)
是在給定日期之前引入的閏秒數。 來自TAI-UTC 表(注意:該站點是閏秒的權威來源。它發布公告 C 宣布新的閏秒):
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"
時區從 TAI 時間獲取 UTC 時間(並且很容易從 GPS 時間獲取 TAI 時間: TAI = GPS + 19 秒(恆定偏移量) ):
#!/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!
2014-08-15 07:00:27 # (UTC)
如果從相應的tzfile(5)
提取閏秒列表,則可以避免更改時區。 它是前兩種方法的組合,其中第一種方法的跳躍計數計算是自動的,並且使用第二種方法的自動更新tzdata
( tz 數據庫的系統包):
>>> 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leapseconds.py
可以從/usr/share/zoneinfo/right/UTC
文件( tzdata
包的一部分)中提取閏秒。
所有三種方法都會產生相同的結果。
你可以使用astropy.time包來做到這一點:
from astropy.time import Time
mytime = 1092121243.0
t = Time(mytime, format='gps')
t = Time(t, format='iso') # same as scale='tai'
print(t)
返回2014-08-15 07:01:02.000
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)
輸出2014-08-15 07:00:27.000
我使用以下計算閏秒的函數:
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)
當然,您偶爾需要更新_LEAP_DATES
,但這些更新非常罕見。
一般來說,GPS 時間由兩個數字組成: GPS 周和自當前 GPS 周開始以來的秒數。 因此,您可以使用以下方法:
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))
在你的情況下week = 0
,所以:
In [1]: gps2utc(0, 1092121243.0)
Out[1]: datetime.datetime(2014, 8, 15, 7, 0, 27)
Ru887321和Yury Kirienko之間的線性組合答案。
from astropy.time import Time
def gps2utc(gpsweek, gpsseconds):
""" GPS time to UTC.
Parameters
----------
gpsweek : int
GPS week number, i.e. 1866.
gpsseconds : int
Number of seconds since the beginning of week.
Returns
-------
datetime
datetime instance with UTC time.
"""
secs_in_week = 604800
secs = gpsweek * secs_in_week + gpsseconds
t_gps = Time(secs, format="gps")
t_utc = Time(t_gps, format="iso", scale="utc")
return t_utc.datetime
在你的情況下week = 0
,所以:
In [1]: gps2utc(0, 1092121243.0)
Out[1]: datetime.datetime(2014, 8, 15, 7, 0, 27)
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