Python地理编码按距离过滤

Question

我需要过滤地理编码以了解与某个位置的接近程度。例如，我想要过滤餐厅地理编码列表，以识别距我当前位置 10 英里以内的餐厅。

有人可以指点我一个将距离转换为纬度和纬度的函数吗？经度增量？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：我使用距离的最高范数。

Answer 1

似乎还没有一个好的Python 实现。幸运的是，“相关文章”侧边栏是我们的朋友。 这篇文章< /a> 指向一篇优秀文章，其中提供了数学知识和 Java 实现。您需要的实际函数相当短，并且嵌入在下面我的 Python 代码中。已测试至所示程度。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

这就是使用半正弦公式计算纬度/经度对之间的距离的方法：

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;

现在可以根据阈值测试“d”（也以公里为单位）。如果您想要除公里以外的其他值，请调整半径。

很抱歉，我无法为您提供即插即用的解决方案，但我不理解您的代码骨架（请参阅评论）。

另请注意，现在您可能想使用余弦球面定律而不是半正弦定律。数值稳定性的优势不再值得，而且它非常容易理解、编码和使用。

Answer 3

如果您将数据存储在 MongoDB 中，它会为您很好地进行索引地理位置搜索，并且优于上面的纯 Python 解决方案，因为它会为您处理优化。

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

Answer 4

约翰·梅钦的回答对我帮助很大。只是有一个小错误：在 boundigbox 中交换了纬度和经度：

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

这解决了问题。原因是经度不会改变每度的距离，但纬度会改变。它们的距离取决于经度。