我需要实现以下逻辑：
给定一组2d采样点（作为x-y坐标对）和一组线段（也称为x-y坐标对）。
编辑1：如何计算（矢量化）点pi与线Li的距离？
这些点大约在直线附近，我想获得从每个样本点到最近的线段的距离。可能是点，它们有点“偏离”（请参见第一张图片中的p6），这些点可以通过以下算法找到：
情况：到该线段的样本点投影为“左外”。在这种情况下，我需要从p1到x1的欧几里德距离。
情况：到该线段的采样点投影在该线段的“内部”。在这种情况下，我需要从p2到x1到x2的线的距离。
情况：到该线段的采样点投影在“右外”。在这种情况下，我需要从p3到x2的欧几里德距离。
有一个矢量化的解决方案（感谢用户Andy）使用“全局”投影，始终为每个点线段对假设情况2。但是，这将返回p1 ... p3的距离[1 1 1]，其中所需距离将为[1.4142 1 1.4142]。可以修改这些代码以满足这些需求吗？
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))
从数学上讲，可以通过查看
vec(pi-x1)在vec(x2-x1)上的投影长度来分离案例。如果此长度因数编辑1：我将添加一个伪代码来说明如何使用双for循环解决此问题，但是由于我有大约6000个样本和10000行，因此循环解决方案对我来说不是一个选择。
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：我设法向量化了该算法的第一部分。 relLength现在包含每个
pi-startOfLi在每个线段上的投影的相对长度。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
最佳答案
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
现在执行“实际”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
将结果绘制成点周围的黄色圆圈
% 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
关于vectorization - 特殊情况下计算点到线段的距离，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46079554/