Question

我有一个用起点和终点定义的线段：

A: 
x1 = 10.7196405787775
y1 = 59.9050401935882

B: 
x2 = 10.7109989561813
y2 = 59.9018650448204

其中x定义经度，y定义纬度。

我也有一点意见：

P:
x0 = 10.6542116666667
y0 = 59.429105

如何计算线段与点之间的最短距离？我知道如何在笛卡尔坐标中执行此操作，但不能在长/纬度坐标中执行此操作。

Answer 1

以下是关于Wikipedia的公式的实现：

def distance(p0, p1, p2): # p3 is the point
    x0, y0 = p0
    x1, y1 = p1
    x2, y2 = p2
    nom = abs((y2 - y1) * x0 - (x2 - x1) * y0 + x2 * y1 - y2 * x1)
    denom = ((y2 - y1)**2 + (x2 - x1) ** 2) ** 0.5
    result = nom / denom
    return result

print distance((0, 0), (3, 4), (5, 6))

# should probably test less obvious cases
assert 1 == distance((0, 0), (1, 1), (2, 1))
# change 0.001 to whatever accuracy you demand on testing.
# Notice that it's never perfect...
assert 0.5 * (2 ** 0.5) - distance((0, 0), (1, 0), (0, 1)) < 0.001

Answer 2

使用有用的Python地理编码库geopy以及来自Chris Veness's geodesy formulae的大圆中点的公式，我们可以找到大圆弧与给定点之间的距离：

from math import sin, cos, atan2, sqrt, degrees, radians, pi
from geopy.distance import great_circle as distance
from geopy.point import Point


def midpoint(a, b):
    a_lat, a_lon = radians(a.latitude), radians(a.longitude)
    b_lat, b_lon = radians(b.latitude), radians(b.longitude)
    delta_lon = b_lon - a_lon
    B_x = cos(b_lat) * cos(delta_lon)
    B_y = cos(b_lat) * sin(delta_lon)
    mid_lat = atan2(
        sin(a_lat) + sin(b_lat),
        sqrt(((cos(a_lat) + B_x)**2 + B_y**2))
    )
    mid_lon = a_lon + atan2(B_y, cos(a_lat) + B_x)
    # Normalise
    mid_lon = (mid_lon + 3*pi) % (2*pi) - pi
    return Point(latitude=degrees(mid_lat), longitude=degrees(mid_lon))

在此示例中给出了：

# Example:
a = Point(latitude=59.9050401935882, longitude=10.7196405787775)
b = Point(latitude=59.9018650448204, longitude=10.7109989561813)
p = Point(latitude=59.429105, longitude=10.6542116666667)

d = distance(midpoint(a, b), p)
print d.km
# 52.8714586903

计算纬度和经度中点和线段之间的距离

2 个答案: