Javascript或Python - 我如何判断它是白天还是白天？

Question

根据用户的时间和地点,我怎么知道它目前是晚上/白天还是日出/黎明？

我没有找到任何有用的东西,我可以在客户端或后端使用.

令它变得棘手的是小时不一定定义是否是白天和黑夜,这在很大程度上取决于年,月,日,小时和地理坐标.

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澄清......模仿这样的事情.

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对此进行近似的方法也非常有用.

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希望有人能帮忙!

Answer 1

您可以像我一样使用此公共域Sun.py模块来计算太阳相对于地球上位置的位置.它已经很老了,但多年来我一直很好用.我对它进行了一些表面修改,使其与Python 2.7更新,例如将其中的几个类改为新风格,但大多数情况下它都没有改变.

这是我创建的一个模块,名为sunriseset.py,它显示了如何使用它来计算特定位置的日出和日落时间,给出其地理坐标和时区.引用的timezone模块是模块关于对象的文档中tzinfo描述的抽象基类的实现.datetimetzinfo

# -*- 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))

Answer 2

美国海军天文台提供了计算日出和日落的算法的简要说明,可在此处获得:

http://edwilliams.org/sunrise_sunset_algorithm.htm

除了提供日期和位置之外,您还需要选择一个天顶角(太阳将被视为"上升"或"设置") - 链接的页面有多个选项.

更新

由于链接页面不再可用,我在下面引用其文字.请注意,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

Answer 3

PyEphem可用于计算下一个日出和日落的时间.基于我发现的博客文章和上升集的文档,您的问题可以解决如下.让我们假设我是你的用户,我的位置是德国奥尔登堡(Oldb).

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.")

笔记

• 由于某种原因lat,lon需要成为弦乐,但如果它们是花车,ephem不会抱怨.
• 为获得最佳效果,您可能需要获得当前的空气温度和气压.

先决条件

这应该至少适用于Python 2.7(带pip-2.7 install pyephem)和Python 3.2(带pip-3.2 install ephem).

确保在系统上运行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

确保在系统上设置了正确的时区.例如在Debian Linux上:

$ sudo dpkg-reconfigure tzdata