使用特殊情况

Question

我需要实现以下逻辑： 给定一组2d样本点（以x-y坐标对给出）和一组线段（也是x-y-坐标对）。 编辑1：如何计算（向量化）点 pi 到 Li 的距离？

这些点大致靠近直线，我想得到每个采样点到最近的线段的距离。可能是点，有点＆＃34;关＆＃34; （参见第一张图中的第6页），这些可以通过以下算法找到：

1. 案例：投影到线段的样本点是＆＃34;离开＆＃34;。在这种情况下，我需要从p1到x1的欧氏距离。
2. 案例：投影到线段的样本点是＆＃34;内部＆＃34;线段。在这种情况下，我需要从p2到x1到x2之间的距离。
3. 案例：投影到线段的样本点是＆＃34;在右边＆＃34;。在这种情况下，我需要从p3到x2的欧几里德距离。



有一个矢量化解决方案（感谢用户Andy），它使用投影＆＃34;全局＆＃34;，总是假设每个点 - 中文对的情况2。但是，这会返回p1 ... p3的距离[1 1 1]，其中所需距离为[1.4142 1 1.4142]。可以根据这些需求修改此代码吗？

ptsx = [1 3 5];
ptsy= [1 1 1];

linesx = [2 4];
linesy = [0 0];

points = [ptsx;ptsy];
lines = [linesx;linesy];

% start of lines
l = [linesx(1:end-1);linesy(1:end-1)];
% vector perpendicular on line
v = [diff(linesy);-diff(linesx)];
% make unit vector
v = v ./ hypot (v(1,:),v(2,:));
v = repmat (v, 1, 1, size (points, 2));
% vector from points (in third dimension) to start of lines (second dimension)
r = bsxfun (@minus, permute (points, [1 3 2]), l);
d = abs (dot (v, r));
dist = squeeze (min (d, [], 2))

从数学角度讲，通过查看vec(pi-x1)到vec(x2-x1)的投影长度，可以分离案例。如果该长度因子是＆lt; 0，那么该点是＆＃34;左＆＃34;如果线段是0到1之间，则可以进行垂直投影，如果是> 0。 1，那么关键是＆＃34;对＆＃34;线段。

编辑1：我将添加一个伪代码来解释如何通过双重for循环解决这个问题，但由于我有大约6000个样本和10000行，所以循环解决方案对我来说不是一个选项。

for each sample point pi
  for each linesegment Li
    a = vector from start of Li to end of Li
    b = vector from pi to start of Li
    relLength = dot(a,b)/norm(a)^2
    if relLength < 0: distance = euclidean distance from start of Li to pi
    if relLength > 1: distance = euclidean distance from end of Li to pi
    else: distance = perpendicular distance from pi to Li
  endfor 
endfor

编辑2 / 2017-09-07：我设法对该算法的第一部分进行了矢量化。 relLength 现在包含每个pi-startOfLi投影到每个线段的相对长度。

ptsx = [0.5 2 3 5.5 8 11];
ptsy= [1  2 -1.5 0.5 4 5];

linesx = [0 2 4 6 10 10 0 0];
linesy = [0 0 0 0  0  4 4 0];

points = [ptsx;ptsy];
lines = [linesx;linesy];

% Start of all lines
L1 = [linesx(1:end-1); linesy(1:end-1)];
% Vector of each line segment
a = [diff(linesx); diff(linesy)];
b = bsxfun(@minus, permute(points, [1 3 2]), L1);
aRep = repmat(a, 1, 1, length(b(1,1,:)));
relLength = dot(aRep,b)./norm(a, 'cols').^2

Answer 1

在GNU Octave中：

points = [1 4.3 3.7 2.9;0.8 0.8 2.1 -0.5];
lines = [0 2 4 3.6;0 -1 1 1.75];

% plot them
hold off
plot (points(1,:), points(2,:), 'or')
hold on
plot (lines(1,:), lines(2,:), '-xb')
text (points(1,:), points(2,:),...
      arrayfun (@(x) sprintf('  p%i',x),...
                1:columns(points),'UniformOutput', false))
axis ('equal')
grid on
zoom (0.9);

% some intermediate vars
s_lines = lines (:,1:end-1);    % start of lines
d_lines = diff(lines, 1, 2);    % vectors between line points
l_lines = hypot (d_lines(1,:),
                 d_lines(2,:)); % length of lines

现在做＆＃34;真实＆＃34;工作：

% vectors perpendicular on lines
v = [0 1;-1 0] * d_lines;
vn = v ./ norm (v, 2, 'cols');   %make unit vector

% extend to number of points
vn = repmat (vn, 1, 1, columns (points));
points_3 = permute (points, [1 3 2]);

% vectors from points (in third dimension) to start of lines (second dimension)
d =  dot (vn, points_3 - s_lines);

% check if the projection is on line
tmp = dot (repmat (d_lines, 1, 1, columns (points)),...
           points_3 - s_lines)./l_lines.^2;
point_hits_line = tmp > 0 & tmp < 1;

% set othogonal distance to Inf if there is no hit
d(~ point_hits_line) = Inf;

dist = squeeze (min (abs(d), [], 2));

% calculate the euclidean distance from points to line start/stop
% if the projection doesn't hit the line
nh = isinf (dist);
tmp = points_3(:,:,nh) - lines;
tmp = hypot(tmp(1,:,:),tmp(2,:,:));
tmp = min (tmp, [], 2);
% store the result back
dist (nh) = tmp

将结果绘制为点周围的黄色圆圈

% use for loops because this hasn't to be fast
t = linspace (0, 2*pi, 40);
for k=1:numel(dist)
  plot (points (1, k) + cos (t) * dist(k),
        points (2, k) + sin (t) * dist(k),
        '-y')
end

Answer 2

使用八度音阶几何包

Octaves geometry包中包含解决所请求问题的所有必要工具。实现您的问题的解决方案有两个功能：

函数distancePointPolyline和distancePointPolygon都应该能够计算请求的距离。多边形是封闭的折线。

实施例

以下脚本演示了这些函数的用法。有关图形结果，请参阅figure。

% Load octave geometry package (package is also available for matlab)
pkg load geometry
% Define points
points = [1,4.3,3.7,2.9; 0.8, 0.8, 2.1, -0.5]';
% Define polyline
lines  = [0, 2, 4, 3.6;   0, -1, 1, 1.75]';

% Calculate distance
d = distancePointPolyline (points,lines);

% Produce figure
figure('name','Distance from points to polyline');
hold all
drawPoint(points);
drawPolyline(lines);
drawCircle(points, d);
axis equal

结果

matlab octave