Python 地理编码按距离过滤

我需要过滤地理编码以了解与某个位置的接近程度。例如，我想过滤餐厅地理编码列表以识别我当前位置 10 英里范围内的那些餐厅。

谁能告诉我一个将距离转换为纬度和经度增量的函数？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:我使用的是距离的上确界范数。

请您参考如下方法:

似乎没有一个好的 Python 实现。幸运的是，SO“相关文章”边栏是我们的 friend 。 This SO article指向 excellent article这给出了数学和 Java 实现。您需要的实际功能相当短，并且嵌入在我下面的 Python 代码中。测试到显示的程度。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