Python：最近指向一行

Question

我有以下問題。我有一個裝滿座標和三點的盒子，它們組成了一條線。現在我想計算所有框座標到該線的最短距離。我有三種方法來做到這一點，vtk和numpy版本總是有相同的結果，但沒有勻稱的距離方法。但我需要這個造型的版本，因爲我想測量從一個點到hwole線的最近距離，而不是單獨的線段。以下是一個示例代碼。所以，問題是「pdist」：

from shapely.geometry import LineString, Point 
import vtk, numpy as np 
import itertools 

import math 

from numpy.linalg import norm 

x1=np.arange(4,21) 
y1=np.arange(4,21) 
z1=np.arange(-7,6) 

linepoints = np.array([[1,10,0],[10,10,0],[15,15,0]]) 


for i in itertools.product(x1,y1,z1): 

    for m in range(len(linepoints)-1): 

     line3 = LineString([linepoints[m],linepoints[m+1]]) 

     p = Point(i) 

     d = norm(np.cross(linepoints[m]-linepoints[m+1], linepoints[m]-i))/norm(linepoints[m+1]-linepoints[m]) 

     dist=math.sqrt(vtk.vtkLine().DistanceToLine(i,linepoints[m],linepoints[m+1])) 

     pdist = p.distance(line3) 

     print(d,dist,pdist) 

Answer 1

的問題是，與跨產品您正在計算通過由點linepoints[m]和linepoints[m+1]確定的段橫跨的線正交距離。另一方面，Shapely計算到該段的距離，即，如果正交投影落在該段的「外側」，則該距離返回到正交投影的距離或返回到邊界點中的一個。

爲了得到一致的結果，你可以自己計算正交投影，然後調用勻稱的距離方法：

import numpy as np 
from shapely.geometry import Point, LineString 


A = np.array([1,0]) 
B = np.array([3,0]) 
C = np.array([0,1]) 


l = LineString([A, B]) 
p = Point(C) 


d = np.linalg.norm(np.cross(B - A, C - A))/np.linalg.norm(B - A) 

n = B - A 
v = C - A 

z = A + n*(np.dot(v, n)/np.dot(n, n)) 

print(l.distance(p), d, Point(z).distance(p)) 
#1.4142135623730951 1.0 1.0 

但是請注意，身材勻稱有效地忽略z座標。因此，舉例來說：

import numpy as np 
from shapely.geometry import Point, LineString 

A = np.array([1,0,1]) 
B = np.array([0,0,0]) 

print(Point([1,0,1]).distance(Point([0,0,0]))) 

回報距離僅1

編輯： 根據您的意見，在這裏將它計算出距離（任意維數）版本段：

from shapely.geometry import LineString, Point 
import numpy as np 
import itertools 

import math 

from numpy.linalg import norm 

x1=np.arange(4,21) 
y1=np.arange(4,21) 
z1=np.arange(-7,6) 

linepoints = np.array([[1,10,0],[10,10,0],[15,15,0]]) 

def dist(A, B, C): 
    """Calculate the distance of point C to line segment spanned by points A, B. 

    """ 

    a = np.asarray(A) 
    b = np.asarray(B) 
    c = np.asarray(C) 

    #project c onto line spanned by a,b but consider the end points 
    #should the projection fall "outside" of the segment  
    n, v = b - a, c - a 

    #the projection q of c onto the infinite line defined by points a,b 
    #can be parametrized as q = a + t*(b - a). In terms of dot-products, 
    #the coefficient t is (c - a).(b - a)/((b-a).(b-a)). If we want 
    #to restrict the "projected" point to belong to the finite segment 
    #connecting points a and b, it's sufficient to "clip" it into 
    #interval [0,1] - 0 corresponds to a, 1 corresponds to b. 

    t = max(0, min(np.dot(v, n)/np.dot(n, n), 1)) 
    return np.linalg.norm(c - (a + t*n)) #or np.linalg.norm(v - t*n) 


for coords in itertools.product(x1,y1,z1): 
    for m in range(len(linepoints)-1): 

     line3 = LineString([linepoints[m],linepoints[m+1]]) 
     d = dist(linepoints[m], linepoints[m+1], coords) 
     print(coords, d)