Python地理编码按距离过滤(Python geocode filtering by distance)

Python地理编码按距离过滤(Python geocode filtering by distance)

我需要将近地点的地理编码过滤到某个位置。 例如，我想筛选餐馆地理编码列表以识别距我目前位置10英里内的餐馆。

有人可以指向一个函数，将距离转换成经纬度三角洲吗？ 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：我正在使用距离的上游准则。

I need to filter geocodes for near-ness to a location. For example, I want to filter a list of restaurant geocodes to identify those restaurants within 10 miles of my current location.

Can someone point me to a function that will convert a distance into latitude & longitude deltas? For example:

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

Note: I am using the supremum norm for distance.

最满意答案

似乎没有一个好的Python实现。 幸运的是，“相关文章”边栏是我们的朋友。 这篇SO文章指出了一篇出色的文章 ，它提供了数学和Java实现。 您需要的实际功能相当短，并嵌入在下面的Python代码中。 测试程度显示。 在评论中阅读警告。

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

There seems not to have been a good Python implementation. Fortunately the SO "Related articles" sidebar is our friend. This SO article points to an excellent article that gives the maths and a Java implementation. The actual function that you require is rather short and is embedded in my Python code below. Tested to extent shown. Read warnings in comments.

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