Question

我有一组无序的2D点，它们代表建筑物的各个角落。我需要连接它们以获得建筑物的轮廓。

这些点是通过组合不同个人收集的不同多边形而获得的。我的想法是使用这些多边形按顺序获取点（例如，获取最大和最小多边形之间的区域并连接点，使其进入该区域）。

我尝试使用最小距离标准，并且还基于角度连接点。但不幸的是，它不起作用。我有用的一件事是点顺序正确的许多多边形的原始数据。那么有可能与这些多边形进行比较以连接这些点吗？如上所述，我的教授提出了采用最大和最小多边形并将中间的区域用作缓冲区的想法。所有点都将落入此缓冲区。但是我不确定如何实现。

预期结果是一个闭合多边形，代表建筑物的平面图。我有15个构建示例，并且代码需要适用于所有人。有些建筑物在拐角之间没有保留正确的角度标准。我正在附上我拥有的数据。我拥有的点是通过整合多边形获得的。那么有没有办法使用这些多边形（点按顺序排列）actual data before integration

Answer 1

编辑

因此，我可以使用下面提到的想法找到解决方案。

备注：我手动添加了一个缺失点。而且，我删除了底部的两个小角。要么这些必须总共是四个角，否则可以将它们视为根本没有角。

我明确指出，由于我的想法包含了这种假设，因此拐角通常具有90度角。

常规方法

通过以下方法查找点的顺序。
对于所有点，在找到的顺序的限制内确定潜在的“邻居”。
对于每两个邻居，确定邻居＃1-当前点-邻居＃2之间的角度。理想情况下，该角度应为90度。
对于所有候选组合，找到总距离最小的组合，即距离（邻居＃1-当前点）+距离（当前点-邻居＃2）。

我意识到通过在所有点上使用for循环，导致绘制所有线条两次。同样，许多计算可能会被矢量化并从循环中移出。优化不是我现在的意图。 ;-）

% Point data of building corners; modified!
X = [285.400 372.267 397.067 408.133 382.471 379.533 199.412 195.267 184.385 168.643 157.533 174.500 108.533 99.333 150.733 184.800 138.105 179.474 218.278 232.133 267.714 306.929 312.143 357.733 421.333 431.000 371.867 364.533];
Y = [130.150 233.360 228.627 286.693 314.541 292.960 348.671 326.693 269.308 330.857 274.493 226.786 239.200 193.467 182.760 101.893 111.000 80.442 74.356 140.360 64.643 56.857 77.786 69.493 133.293 180.427 142.160 192.027];

% Place approximative center of building at (0, 0)
X = X - mean(X);
Y = Y - mean(Y);
C = [mean(X), mean(Y)];

% Sort points by angle with respect to center
[~, idx] = sort(atan2(X, Y));

% Rearrange points with respect to sorted angles
X = X(idx);
Y = Y(idx);

% Number of data points
n = numel(X);

% Calculate direction vectors for X and Y coordinates
dvX = repmat(X.', 1, n);
dvX = dvX - dvX.';
dvY = repmat(Y.', 1, n);
dvY = dvY - dvY.';

% Calculate distances
dst = sqrt(dvX.^2 + dvY.^2);

% Number of "neighbouring" points to be considered with respect to the order
nn = 8;


figure(1);
hold on;

% Center
plot(C(1), C(2), 'kx', 'MarkerSize', 15);

% Plain points
plot(X, Y, '.', 'MarkerSize', 15);

for k = 1:n

  % Index  
  text(X(k) + 0.05, Y(k) + 0.05, num2str(k), 'FontSize', 12);

  % Set up neighbourhood  
  nbh = mod([k-nn/2:k-1 k+1:k+nn/2], n);
  nbh(nbh == 0) = n;

  % Calculate angles and total distance arrays
  ang = Inf(nn);
  len = Inf(nn);
  for ii = 1:nn
    l = nbh(ii);
    d1 = [dvX(k, l) dvY(k, l)];
    for jj = ii+1:nn
      m = nbh(jj);
      d2 = [dvX(k, m) dvY(k, m)];
      len(ii, jj) = dst(k, l) + dst(k, m);
      ang(ii, jj) = abs(pi/2 - acos(dot(d1, d2) / (norm(d1) * norm(d2))));
    end
  end

  % Find candidates with angle difference < 10 degree
  cand = find(ang < pi/18);

  % For these candidates, find the one with the shortest total distance
  [~, I] = min(len(cand));

  % Get corresponding indices
  [I, J] = ind2sub([nn nn], cand(I));
  cand = nbh([I J]);

  % Lines 
  plot([X(k) X(cand(1))], [Y(k) Y(cand(1))], 'b', 'LineWidth', 1);
  plot([X(k) X(cand(2))], [Y(k) Y(cand(2))], 'b', 'LineWidth', 1);

end

hold off;

输出图像：

一种近似（！）解决方案是确定由找到的点描述的轮廓的中心，并相对于中心使用atan2来按角度对点进行排序。请参见以下代码段以进行可视化：

% Points
X = 2 * rand(1, 15) - 1;
Y = 2 * rand(1, 15) - 1;

% Center
C = [0, 0];

% Determine indices
[~, idx] = sort(atan2(X, Y));

figure(1);
hold on;

% Center
plot(C(1), C(2), 'kx', 'MarkerSize', 15);

% Plain points
plot(X, Y, '.', 'MarkerSize', 15);

% Indices and lines
for k = 1:numel(X)
  text(X(idx(k)) + 0.05, Y(idx(k)) + 0.05, num2str(k), 'FontSize', 12);
  if (k == numel(X))
    plot([X(idx(k)) X(idx(1))], [Y(idx(k)) Y(idx(1))], 'b');
  else
    plot([X(idx(k)) X(idx(k+1))], [Y(idx(k)) Y(idx(k+1))], 'b');
  end
end

hold off;

给出以下输出：

尽管我敢肯定，一定数量的凹面将得到正确处理，但恐怕对于给定的示例（尤其是上半部分），它会失败。这是因为图像不是完美的顶视图，因此角度有点“失真”。

尽管如此，也许订购可以提高您的最小距离进近率。

Answer 2

Here's a solution which that good for shapes that have outlines made from perpendicular^* lines (as the one in your example). The idea is as follows:

We rotate the points to align^* them to the XY grid.
We group points into families that have either the same^* X or Y coordinates.
For each point we compute two points: the closest horizontally, and the closest vertically, from within the allowed families.
Build a connectivity matrix and transform back.

Just like in HansHirse's answer, I must change the dataset: add a missing corner (pt. 30), remove two non-corners (pts. 7-8), remove the duplicate last point.

^* - approximately.

function A = q55511236
%% Initialization:
% Define points:
X = [364.533 372.267 397.067 408.133 382.471 379.533 329.250 257.200 199.412 195.267 184.385 ...
     168.643 157.533 174.500 108.533 99.333 150.733 184.800 138.105 179.474 218.278 232.133 ...
     267.714 306.929 312.143 357.733 421.333 431.000 371.867];
Y = [192.027 233.360 228.627 286.693 314.541 292.960 327.450 340.500 348.671 326.693 269.308 ...
     330.857 274.493 226.786 239.200 193.467 182.760 101.893 111.000 80.442 74.356 140.360 ...
     64.643 56.857 77.786 69.493 133.293 180.427 142.160];

%% Preprocessing:
% Centering:
XY = [X;Y] - [mean(X); mean(Y)];
% Rotation:
[U,~,~] = svd(XY,'econ');
rXY = (U.' * XY).';

% Fixing problems w/ some points:
rXY = vertcat(rXY, [-21.8, 66]); % add missing point
rXY(7:8, :) = NaN; % remove non-corners
% figure(); scatter(rXY(:,1),rXY(:,2));

%% Processing:
% Group points according to same-X and same-Y
CLOSE_ENOUGH_DISTANCE = 10; % found using trial and error
[~,~,sameXpts] = uniquetol(rXY(:,1), CLOSE_ENOUGH_DISTANCE, 'DataScale', 1);
[~,~,sameYpts] = uniquetol(rXY(:,2), CLOSE_ENOUGH_DISTANCE, 'DataScale', 1);

% Create masks for distance evaluations:
nP = size(rXY,1);
[maskX,maskY] = deal(zeros(nP));
maskX(sameXpts == sameXpts.') = Inf;
maskY(sameYpts == sameYpts.') = Inf;

% Compute X and Y distances separately (we can do this in the rotated space)
dX = abs(rXY(:,1) - rXY(:,1).') + maskX + 1./maskY;
dY = abs(rXY(:,2) - rXY(:,2).') + maskY + 1./maskX;
[~,nX] = min(dX);
[~,nY] = min(dY);

% Construct connectivity matrix:
A = false(nP);
idxTrue = sub2ind(size(A), repmat(1:nP, [1,2]), [nX(:).', nY(:).']);
A(idxTrue) = true;

%% Plot result:
% Rotated coordinates:
figure(); gplot(A, rXY, '-o'); text(rXY(:,1), rXY(:,2), string(1:nP));
uXY = (U*rXY.').';
% Original coordinates:
figure(); gplot(A, uXY, '-o'); text(uXY(:,1), uXY(:,2), string(1:nP)); axis ij;

Resulting in:

Answer 3

答案的概念是“旅行商问题”。将在这些点周围创建一个缓冲区，并将该缓冲区作为附加条件。

a=[141 188 178 217 229 282 267 307 313 357 372 422 434 365 372 398 411 382 382 233 229 191 185 166 156 183 173 114 97 149 139 139];
b=[109 103 79 76 140 132 64 56 78 72 141 133 180 192 234 228 287 293 315 348 343 348 329 332 270 268 225 240 194 184 108 108];
X=[364.5333 232.1333 397.0667 157.5333 431 421.3333 306.9286 184.3846 357.7333 199.4118 168.6429 179.4737 408.1333 382.4706 150.7333 372.2667 184.8 138.1053 312.1429 108.5333 174.5 195.2667 257.2 99.33333 379.5333 371.8667 329.25 280.7059 267.7143 218.2778];
Y=[192.0267 140.36 228.6267 274.4933 180.4267 133.2933 56.85714 269.3077 69.49333 348.6706 330.8571 80.44211 286.6933 314.5412 182.76 233.36 101.8933 111 77.78571 239.2 226.7857 326.6933 340.5 193.4667 292.96 142.16 327.45 130.5529 64.64286 74.35556];
R = [a' b'];
d = 12;
polyout = polybuffer(R,'lines',d)
figure
 %imshow(I2);
hold on
%plot(R(:,1),R(:,2),'r.','MarkerSize',10)
plot(X,Y,'r.', 'MarkerSize', 15)
plot(polyout)
axis equal
hold off
[s,t] = boundary(polyout);  %%this is the boundary polygon of the buffer 
numPoints = length(clustersCentroids);
x = X; %these are the points to be connected
y = Y;
x([1 2],:)=x([2 1],:);
y([1 2],:)=y([2 1],:);
figure
plot(x, y, 'bo', 'LineWidth', 2, 'MarkerSize', 17);
grid on;
 imshow(I2);
xlabel('X', 'FontSize', 10);
ylabel('Y', 'FontSize', 10);
% Make a list of which points have been visited
beenVisited = false(1, numPoints);
% Make an array to store the order in which we visit the points.
visitationOrder = ones(1, numPoints);
% Define a filasafe
maxIterations = numPoints + 1;
iterationCount = 1;
% Define a current index.  currentIndex will be 1 to start and then will vary.
currentIndex = 1;
while sum(beenVisited) < numPoints
    visitationOrder(iterationCount) = currentIndex; 
  beenVisited(currentIndex) = true; 
  % Get the x and y of the current point.
  thisX = x(currentIndex);
  thisY = y(currentIndex);
  %text(thisX + 0.01, thisY, num2str(currentIndex), 'FontSize', 35, 'Color', 'r');
  % Compute distances to all other points
  distances = sqrt((thisX - x) .^ 2 + (thisY - y) .^ 2);
  distances(beenVisited)=inf;
   distances(currentIndex) = inf;
  % Don't consider visited points by setting their distance to infinity.
  [out,idx] = sort(distances);
  xx=[x(currentIndex) x(idx(1))]
   yy=[y(currentIndex) y(idx(1))]
  if isempty(polyxpoly(xx,yy,s,t))
   iterationCount = iterationCount + 1;
   currentIndex =idx(1);
  else 
     xx=[x(currentIndex) x(idx(2))]
   yy=[y(currentIndex) y(idx(2))]  
  if isempty(polyxpoly(xx,yy,s,t))
   iterationCount = iterationCount + 1;
   currentIndex =idx(2);
   else 
     xx=[x(currentIndex) x(idx(3))]
   yy=[y(currentIndex) y(idx(3))]  
  if isempty(polyxpoly(xx,yy,s,t))
   iterationCount = iterationCount + 1;
   currentIndex =idx(3);
   else 
     xx=[x(currentIndex) x(idx(4))]
   yy=[y(currentIndex) y(idx(4))]  
  if isempty(polyxpoly(xx,yy,s,t))
   iterationCount = iterationCount + 1;
   currentIndex =idx(4);
  end
  end
  end
  end
end

% Plot lines in that order.
hold on;
orderedX = [x(visitationOrder); x(1)];
orderedY = [y(visitationOrder) ;y(1)];
plot(orderedX,orderedY, 'm-', 'LineWidth', 2);
title('Result', 'FontSize', 10);

连接一组点以获得非自相交非凸多边形

3 个答案:

编辑