Question

我需要将近地点的地理编码过滤到某个位置。例如，我想过滤餐馆地理编码列表，以识别距离我当前位置10英里范围内的餐馆。

有人可以指出我将一个距离转换为纬度的功能吗？经度三角洲？例如：

class GeoCode(object):
   """Simple class to store geocode as lat, lng attributes."""
   def __init__(self, lat=0, lng=0, tag=None):
      self.lat = lat
      self.lng = lng
      self.tag = None

def distance_to_deltas(geocode, max_distance):
   """Given a geocode and a distance, provides dlat, dlng
      such that

         |geocode.lat - dlat| <= max_distance
         |geocode.lng - dlng| <= max_distance
   """
   # implementation
   # uses inverse Haversine, or other function?
   return dlat, dlng

注意：我使用的是supremum规范的距离。

Answer 1

似乎没有一个好的Python实现。幸运的是，SO“相关文章”侧边栏是我们的朋友。 This SO article指向excellent article，它提供数学和Java实现。您需要的实际功能相当短，并嵌入在我下面的Python代码中。测试范围显示。阅读评论中的警告。

from math import sin, cos, asin, sqrt, degrees, radians

Earth_radius_km = 6371.0
RADIUS = Earth_radius_km

def haversine(angle_radians):
    return sin(angle_radians / 2.0) ** 2

def inverse_haversine(h):
    return 2 * asin(sqrt(h)) # radians

def distance_between_points(lat1, lon1, lat2, lon2):
    # all args are in degrees
    # WARNING: loss of absolute precision when points are near-antipodal
    lat1 = radians(lat1)
    lat2 = radians(lat2)
    dlat = lat2 - lat1
    dlon = radians(lon2 - lon1)
    h = haversine(dlat) + cos(lat1) * cos(lat2) * haversine(dlon)
    return RADIUS * inverse_haversine(h)

def bounding_box(lat, lon, distance):
    # Input and output lats/longs are in degrees.
    # Distance arg must be in same units as RADIUS.
    # Returns (dlat, dlon) such that
    # no points outside lat +/- dlat or outside lon +/- dlon
    # are <= "distance" from the (lat, lon) point.
    # Derived from: http://janmatuschek.de/LatitudeLongitudeBoundingCoordinates
    # WARNING: problems if North/South Pole is in circle of interest
    # WARNING: problems if longitude meridian +/-180 degrees intersects circle of interest
    # See quoted article for how to detect and overcome the above problems.
    # Note: the result is independent of the longitude of the central point, so the
    # "lon" arg is not used.
    dlat = distance / RADIUS
    dlon = asin(sin(dlat) / cos(radians(lat)))
    return degrees(dlat), degrees(dlon)

if __name__ == "__main__":

    # Examples from Jan Matuschek's article

    def test(lat, lon, dist):
        print "test bounding box", lat, lon, dist
        dlat, dlon = bounding_box(lat, lon, dist)
        print "dlat, dlon degrees", dlat, dlon
        print "lat min/max rads", map(radians, (lat - dlat, lat + dlat))
        print "lon min/max rads", map(radians, (lon - dlon, lon + dlon))

    print "liberty to eiffel"
    print distance_between_points(40.6892, -74.0444, 48.8583, 2.2945) # about 5837 km
    print
    print "calc min/max lat/lon"
    degs = map(degrees, (1.3963, -0.6981))
    test(*degs, dist=1000)
    print
    degs = map(degrees, (1.3963, -0.6981, 1.4618, -1.6021))
    print degs, "distance", distance_between_points(*degs) # 872 km

Answer 2

这是使用半正弦公式计算纬度/经度对之间距离的方法：

import math 

R = 6371 # km
dLat = (lat2-lat1) # Make sure it's in radians, not degrees
dLon = (lon2-lon1) # Idem 
a = math.sin(dLat/2) * math.sin(dLat/2) +
    math.cos(lat1) * math.cos(lat2) * 
    math.sin(dLon/2) * math.sin(dLon/2) 
c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a)) 
d = R * c;

现在，根据您的阈值测试“d”（也以km为单位）是微不足道的。如果您想要除km之外的其他内容，请调整半径。

对不起，我不能给你一个简单的解决方案，但我不理解你的代码框架（见评论）。

还要注意，现在你可能想要使用余弦的球面定律而不是使用Haversine。数值稳定性的优势不再值得，它的理解，编码和使用都很简单。

Answer 3

如果您在MongoDB中存储数据，它会为您进行精确索引的地理定位搜索，并且优于上面的纯Python解决方案，因为它将为您处理优化。

http://www.mongodb.org/display/DOCS/Geospatial+Indexing

Answer 4

John Machin的回答对我很有帮助。只有一个小错误：纬度和经度在boundigbox中交换：

dlon = distance / RADIUS
dlat = asin(sin(dlon) / cos(radians(lon)))
return degrees(dlat), degrees(dlon)

这解决了这个问题。原因是经度不会改变每度的距离 - 但纬度确实如此。它们的距离取决于经度。

Python地理编码按距离过滤

4 个答案: