到凸包的距离

我一直在寻找一种方法来计算到凸包/多边形的距离，如果该点在包内，则距离为正，如果在包外，则距离为负。例如，给定一个船体和一组点，可以计算正/负距离吗？

from scipy.spatial import ConvexHull
import matplotlib.pyplot as plt
import numpy as np

# Original points, hull and test points
points = np.random.rand(30, 2)   # 30 random points in 2-D
hull = ConvexHull(points)
newpoints = np.random.rand(30, 2)   # 30 random points in 2-D

# Plot original points, hull and new points
plt.plot(points[:,0], points[:,1], 'ro')
plt.plot(points[hull.vertices,0], points[hull.vertices,1], 'r--', lw=2)
plt.plot(newpoints[:,0], newpoints[:,1], 'go')

所以在上面我想计算每个绿点的有符号距离。非常感谢你花时间陪伴！！

更新，使用来自 (http://www.fundza.com/vectors/point2line/index.html http://www.fundza.com/vectors/point2line/index.html) 可以计算无符号距离：

from scipy.spatial import ConvexHull
import matplotlib.pyplot as plt
import numpy as np
from vectors import *

# Original points, hull and test points
points = np.random.rand(30, 2)   # 30 random points in 2-D
hull = ConvexHull(points)
newpoints = np.random.rand(30, 2)   # 30 random points in 2-D



def pnt2line(pnt, start, end):
    line_vec = vector(start, end)
    pnt_vec = vector(start, pnt)
    line_len = length(line_vec)
    line_unitvec = unit(line_vec)
    pnt_vec_scaled = scale(pnt_vec, 1.0/line_len)
    t = dot(line_unitvec, pnt_vec_scaled)    
    if t < 0.0:
        t = 0.0
    elif t > 1.0:
        t = 1.0
    nearest = scale(line_vec, t)
    dist = distance(nearest, pnt_vec)
    nearest = add(nearest, start)
    return (dist, nearest)



pt_dist = []
for p_idx in range(30):
    pt = newpoints[p_idx,:]
    dist_list = []
    for v_idx in range(len(hull.vertices)):
        v1 = hull.vertices[v_idx - 1]
        v2 = hull.vertices[v_idx]
        start = points[v1]
        end = points[v2]
        temp = pnt2line(pt, start, end)
        dist_list.append(temp[0])
    pt_dist.append(min(dist_list))


# Plot original points, hull and new points
plt.plot(points[:,0], points[:,1], 'ro')
plt.plot(points[hull.vertices,0], points[hull.vertices,1], 'r--', lw=2)
plt.plot(newpoints[:,0], newpoints[:,1], 'go')
for p_idx in range(30):
    pt = newpoints[p_idx,:]
    dist = pt_dist[p_idx]
    distLabel = "%.2f" % dist
    plt.annotate(distLabel,xy=pt)

（注意将 vector.py 代码修改为 2d）：

import math

def dot(v,w):
    x,y = v
    X,Y = w
    return x*X + y*Y

def length(v):
    x,y = v
    return math.sqrt(x*x + y*y)

def vector(b,e):
    x,y = b
    X,Y = e
    return (X-x, Y-y)

def unit(v):
    x,y = v
    mag = length(v)
    return (x/mag, y/mag)

def distance(p0,p1):
    return length(vector(p0,p1))

def scale(v,sc):
    x,y = v
    return (x * sc, y * sc)

def add(v,w):
    x,y = v
    X,Y = w
    return (x+X, y+Y)

您始终可以将此正距离乘以一个真/假结果多边形中的点 http://en.wikipedia.org/wiki/Point_in_polygon计算。我通常喜欢使用绕数算法（只是因为我从复杂变量的角度理解它） - 但计算交叉点也很好。