使用python将Julian日期转换为正常日期和时间

Question

我想将儒略日期转换为正常日期和时间。我的起始年份是2010.1.1 00:00:00，我处理日期的代码部分如下：

import astropy.time
import dateutil.parser
import datetime
from datetime import datetime
dt = dateutil.parser.parse('2010.01.01')
time1 = astropy.time.Time(dt)
jd=time1.jd
print(jd)

datetime_start=266065955.675 #seconds from 2010.1.1 00:00:00
datetime_startjd=(datetime_start/24./60./60.)+jd
print(datetime_startjd)

from __future__ import print_function, division
from PyAstronomy import pyasl

print("The decimal year %10.5f correspond to " % datetime_startjd+ \
  pyasl.decimalYearGregorianDate(datetime_startjd, "yyyy-mm-dd hh:mm:ss"))
print(" ... or equivalently (y, m, d, h, m, s, ms): ", \
  pyasl.decimalYearGregorianDate(datetime_startjd, "tuple"))

我得到一个错误：

ValueError: year 2458276 is out of range

有人能给我推荐一个准确的代码或方法将这个儒略日期转换成正常的日期和时间吗

Answer 1

我使用以下代码找到了我需要的东西，以防有人想要使用它：

import numpy as np
from astropy.time import Time

times = [ '2010-01-01T00:00:00']
t = Time(times, format='isot', scale='utc')
juliant=t.jd
print(juliant)

datetime_start=265633590.42
datetime_startjd=(datetime_start/24./60./60.)+juliant
print(datetime_startjd)

time2=Time('2458271.96285208',format='jd')
time3=time2.iso
print(time3)

datetime_length=1.0800000429153442
cover_time=((datetime_start+datetime_length)/24./60./60.)+juliant
print(cover_time)
cover_time=Time(cover_time,format='jd')
cover_time=cover_time.iso
print(cover_time)