知道我是如何根据用户的时间和位置确定当前的夜间/白天还是日出/黎明的吗?
我还没有发现在客户机或后端中可以使用的任何有用的东西。
让它变得棘手的是,时间不一定定义它是否是日夜,这在很大程度上取决于年份,月,日,小时和地理坐标。
为了澄清..。去模仿这样的事情。
--一种近似于此的方法也非常有用.
希望有人能帮忙!
发布于 2013-02-23 19:25:17
Python3用户注意:--我刚刚使用Python3.8.6尝试了下面的代码,它也在那个版本中工作。我必须将print
语句转换为print()
函数调用,但仅此而已。
我注意到,tzinfo
类的示例实现与Python2文档中的实现不同,但使用最新文件中引用的example.py文件中的示例tzinfo
类可以正常工作(但较早的2.x版本也是如此)。
您可以从sunriseset.py
下载下面所示的Python3版本的这里文件。
您可以像我一样使用这个公共域Sun.py模块来计算太阳相对于地球上位置的位置。(警告:包含制表符,并假定制表符为每8个字符。)它很旧,但多年来一直对我很有帮助。我对它做了一些表面上的修改,使其更接近Python 2.7的最新版本,比如使它中的少数类具有新的风格,但在很大程度上它并没有改变。
下面是我创建的一个模块,名为sunriseset.py
,它展示了如何使用它来计算特定位置的日出和日落时间,给定它的地理坐标和时区。引用的timezone
模块是在datetime
模块关于对象的文档中描述的tzinfo
抽象基类的实现。
# -*- coding: iso-8859-1 -*-
import datetime
import timezone # concrete tzinfo subclass based on the Python docs
import math
from Sun import Sun
__all__ = ['getsuninfo', 'Place']
class Place(object):
def __init__(self, name, coords, tz=timezone.Pacific):
self.name = name # string
self.coords = coords # tuple (E/W long, N/S lat)
self.tz = tz # tzinfo constant
def _hoursmins(hours):
"""Convert floating point decimal time in hours to integer hrs,mins"""
frac,h = math.modf(hours)
m = round(frac*60, 0)
if m == 60: # rounded up to next hour
h += 1; m = 0
return int(h),int(m)
def _ymd(date):
"""Return y,m,d from datetime object as tuple"""
return date.timetuple()[:3]
def getsuninfo(location, date=None):
"""Return local datetime of sunrise, sunset, and length of day in hrs,mins)"""
if date == None:
querydate = datetime.date.today()
else: # date given should be datetime instance
querydate = date
args = _ymd(querydate) + location.coords
utcrise, utcset = Sun().sunRiseSet(*args)
daylength = Sun().dayLength(*args)
hrs,mins = _hoursmins(daylength)
risehour, risemin = _hoursmins(utcrise)
sethour, setmin = _hoursmins(utcset)
# convert times to timedelta values (ie from midnight utc of the date)
midnight = datetime.datetime(tzinfo=timezone.utc, *_ymd(querydate))
deltarise = datetime.timedelta(hours=risehour, minutes=risemin)
utcdatetimerise = midnight+deltarise
deltaset = datetime.timedelta(hours=sethour, minutes=setmin)
utcdatetimeset = midnight+deltaset
# convert results from UTC time to local time of location
localrise = utcdatetimerise.astimezone(location.tz)
localset = utcdatetimeset.astimezone(location.tz)
return localrise, localset, hrs, mins
if __name__ == "__main__":
import datetime, timezone
def unittest(location, testdate):
risetime, settime, hrs, mins = getsuninfo(location, testdate)
print "Location:", location.name
print "Date:", testdate.strftime("%a %x")
print risetime.strftime("Sunrise %I:%M %p"), settime.strftime("- Sunset %I:%M %p (%Z)")
print "daylight: %d:%02d" % (hrs,mins)
print
place = Place("My House", (-121.990278, 47.204444), timezone.Pacific)
# test dates just before and after DST transitions
print "pre 2007"
print "========="
unittest(place, datetime.date(2006, 4, 1))
unittest(place, datetime.date(2006, 4, 2))
unittest(place, datetime.date(2006, 10, 28))
unittest(place, datetime.date(2006, 10, 29))
print "2007"
print "========="
unittest(place, datetime.date(2007, 3, 10))
unittest(place, datetime.date(2007, 3, 11))
unittest(place, datetime.date(2007, 11, 3))
unittest(place, datetime.date(2007, 11, 4))
发布于 2013-02-23 19:31:47
美国海军天文台提供了计算日出和日落算法的简明说明,可在此查阅:
algorithm.htm
除了提供日期和位置,您还需要选择一个天顶角(太阳将被视为“上升”或“设置”)--链接的页面有几个选项。
更新
由于链接的页面不再可用,我在下面引用它的文本。请注意,所包含的公式是伪代码形式,而不是JavaScript。
Source:
Almanac for Computers, 1990
published by Nautical Almanac Office
United States Naval Observatory
Washington, DC 20392
Inputs:
day, month, year: date of sunrise/sunset
latitude, longitude: location for sunrise/sunset
zenith: Sun's zenith for sunrise/sunset
offical = 90 degrees 50'
civil = 96 degrees
nautical = 102 degrees
astronomical = 108 degrees
NOTE: longitude is positive for East and negative for West
NOTE: the algorithm assumes the use of a calculator with the
trig functions in "degree" (rather than "radian") mode. Most
programming languages assume radian arguments, requiring back
and forth convertions. The factor is 180/pi. So, for instance,
the equation RA = atan(0.91764 * tan(L)) would be coded as RA
= (180/pi)*atan(0.91764 * tan((pi/180)*L)) to give a degree
answer with a degree input for L.
1. first calculate the day of the year
N1 = floor(275 * month / 9)
N2 = floor((month + 9) / 12)
N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))
N = N1 - (N2 * N3) + day - 30
2. convert the longitude to hour value and calculate an approximate time
lngHour = longitude / 15
if rising time is desired:
t = N + ((6 - lngHour) / 24)
if setting time is desired:
t = N + ((18 - lngHour) / 24)
3. calculate the Sun's mean anomaly
M = (0.9856 * t) - 3.289
4. calculate the Sun's true longitude
L = M + (1.916 * sin(M)) + (0.020 * sin(2 * M)) + 282.634
NOTE: L potentially needs to be adjusted into the range [0,360) by adding/subtracting 360
5a. calculate the Sun's right ascension
RA = atan(0.91764 * tan(L))
NOTE: RA potentially needs to be adjusted into the range [0,360) by adding/subtracting 360
5b. right ascension value needs to be in the same quadrant as L
Lquadrant = (floor( L/90)) * 90
RAquadrant = (floor(RA/90)) * 90
RA = RA + (Lquadrant - RAquadrant)
5c. right ascension value needs to be converted into hours
RA = RA / 15
6. calculate the Sun's declination
sinDec = 0.39782 * sin(L)
cosDec = cos(asin(sinDec))
7a. calculate the Sun's local hour angle
cosH = (cos(zenith) - (sinDec * sin(latitude))) / (cosDec * cos(latitude))
if (cosH > 1)
the sun never rises on this location (on the specified date)
if (cosH < -1)
the sun never sets on this location (on the specified date)
7b. finish calculating H and convert into hours
if if rising time is desired:
H = 360 - acos(cosH)
if setting time is desired:
H = acos(cosH)
H = H / 15
8. calculate local mean time of rising/setting
T = H + RA - (0.06571 * t) - 6.622
9. adjust back to UTC
UT = T - lngHour
NOTE: UT potentially needs to be adjusted into the range [0,24) by adding/subtracting 24
10. convert UT value to local time zone of latitude/longitude
localT = UT + localOffset
发布于 2013-05-06 18:56:24
PyEphem可以用来计算下一个日出和日落的时间。在我发现一篇博文和上升集文献的基础上,您的问题可以按照以下方式解决。假设我是您的用户,我的位置是Oldenburg (Oldb),德国。
import ephem
user = ephem.Observer()
user.lat = '53.143889' # See wikipedia.org/Oldenburg
user.lon = '8.213889' # See wikipedia.org/Oldenburg
user.elevation = 4 # See wikipedia.org/Oldenburg
user.temp = 20 # current air temperature gathered manually
user.pressure = 1019.5 # current air pressure gathered manually
next_sunrise_datetime = user.next_rising(ephem.Sun()).datetime()
next_sunset_datetime = user.next_setting(ephem.Sun()).datetime()
# If it is daytime, we will see a sunset sooner than a sunrise.
it_is_day = next_sunset_datetime < next_sunrise_datetime
print("It's day." if it_is_day else "It's night.")
# If it is nighttime, we will see a sunrise sooner than a sunset.
it_is_night = next_sunrise_datetime < next_sunset_datetime
print("It's night." if it_is_night else "It's day.")
备注
lat
和lon
需要是字符串,但是如果它们是浮动的话,ephem不会抱怨。先决条件
这至少适用于Python2.7( pip-2.7 install pyephem
)和Python3.2( pip-3.2 install ephem
)。
确保系统上运行了网络时间协议客户端。例如,在Debian Linux上:
$ sudo apt-get install ntp
$ sudo /etc/init.d/ntp start
确保系统上设置了正确的时区。例如,在Debian Linux上:
$ sudo dpkg-reconfigure tzdata
https://stackoverflow.com/questions/15044521
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