如何找到 python 上经纬度点的最近邻点？

Question

输入：

point = (lat, long)
places = [(lat1, long1), (lat2, long2), ..., (latN, longN)]
count = L

输出： neighbors = places 的子集接近 point。 (len(neighbors)=L)

问题： 我可以使用 kd-tree 快速最近邻 s 查找具有纬度和经度的点吗？（例如，scipy中的实现）

是否需要将坐标x,y中的点的地理坐标（经纬度）进行变换？

这是解决这个问题的最佳方法吗？

Answer 1

老实说，我不知道使用 kd-tree 是否能正常工作，但我的直觉说它不准确。

我认为你需要使用更大的圆距离之类的东西来获得准确的距离。


from math import radians, cos, sin, asin, sqrt, degrees, atan2

def validate_point(p):
    lat, lon = p
    assert -90 <= lat <= 90, "bad latitude"
    assert -180 <= lon <= 180, "bad longitude"

# original formula from  http://www.movable-type.co.uk/scripts/latlong.html
def distance_haversine(p1, p2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    Haversine
    formula: 
        a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
                        _   ____
        c = 2 ⋅ atan2( √a, √(1−a) )
        d = R ⋅ c

    where   φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
            note that angles need to be in radians to pass to trig functions!
    """
    lat1, lon1 = p1
    lat2, lon2 = p2
    for p in [p1, p2]:
        validate_point(p)

    R = 6371 # km - earths's radius

    # convert decimal degrees to radians 
    lat1, lon1, lat2, lon2 = map(radians, [lat1, lon1, lat2, lon2])

    # haversine formula 
    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) # 2 * atan2(sqrt(a), sqrt(1-a))
    d = R * c
    return d

Answer 2

我认为你正在尝试解决k Nearest Neighbor问题。

既然你的数据集是二维的，那么kd-tree就可以了，总的来说，我不知道辣

但是，如果你的点开始生活在更高的维度，那么。

Answer 3

scikit-learn 提供 BallTree class that supports the Haversine metric. See also this SO question.

如何找到 python 上经纬度点的最近邻点？

How to find the nearest neighbors for latitude and longitude point on python?

python

geolocation

kdtree

nearest-neighbor

scipy