在MATLAB中给定约束条件的情况下随机选择子矩阵

Question

这里是编程的新手，我真的不知道该如何处理。

我的问题：我有一堆用矩形边界框注释的图像，想要随机选择其他与我已有的边界框不重叠的矩形边界框。 因此，基本上，我有一个矩阵M和一个M子矩阵的预定义子集X，我想生成具有不与X重叠但可以相互重叠的随机位置的新子矩阵。 生成的子矩阵的大小应与X中的子矩阵大小相同，并包含在矩阵M中。

http://i.imgur.com/74AEI2x.png

在上面的示例中，该图像中的方框表示X，即足球的正序； 我想生成相等数量的相同大小的框，它们不包含足球，以表示足球的负面示例。

任何方向都值得赞赏，我正在MATLAB中进行此操作。

Answer 1

让我们仔细阅读代码和注释，并尝试了解如何实现问题中设定的目标。

%// Bounding box array, where the first and second columns denote the X-Y 
%// location of the uppper-left corner pixel. The third and fourth columns
%// denote the extent of the repctive boxes along X and Y directions 
%// respectively. Some random values are used here for demo purposes.
bb = [
    12 10 10 5
    15 20 14 12
    135 60 11 4
    20 30 10 7
    20 30 13 13
    20 30 13 14]

%// Tolerance in terms of the size difference betwen similar boxes that
%// is admissible as a less than or equal to value
tol = 2

%// Get X and Y direction limits for each box
xlims = [bb(:,1) bb(:,1) + bb(:,3)]
ylims = [bb(:,2) bb(:,2) + bb(:,4)];

%// Create a binary matrix that decides whether each box is in or out with
%// respect to all other boxes along both X and Y directions. Ones mean "in"
%// and zeros denote "out".
x1 = bsxfun(@ge,xlims(:,1),xlims(:,1)') & bsxfun(@le,xlims(:,1),xlims(:,2)')
x2 = bsxfun(@ge,xlims(:,2),xlims(:,1)') & bsxfun(@le,xlims(:,2),xlims(:,2)')
x12 = x1 | x2;

y1 = bsxfun(@ge,ylims(:,1),ylims(:,1)') & bsxfun(@le,ylims(:,1),ylims(:,2)')
y2 = bsxfun(@ge,ylims(:,2),ylims(:,1)') & bsxfun(@le,ylims(:,2),ylims(:,2)')
y12 = y1 | y2;

d1 = x12 & y12

%// Create another binary matrix based on sizes to decide for each box
%// what other boxes are "similar"
szmat = bb(:,[3 4])
v1 = abs(bsxfun(@minus,szmat,permute(szmat,[3 2 1])));
szmat_d = squeeze(all(v1<=tol,2));

%// Get a binary matrix based on combined decisions from X-Y incompatibility
%// and sizes. Please note for incompatibility, negation of d1 is needed.
out1 = ~d1 & szmat_d
out1(1:size(out1,1)+1:end)=0
out2 = mat2cell(out1,ones(1,size(out1,1)),size(out1,2))
out3 = cellfun(@find,out2,'uni',0)

如何使用代码-

out3是容纳每个框的最终决策的最终输出，还有哪些其他框相似且不重叠。 为了进行验证，让我们通过执行out3{1}查看其他哪些框符合框1的这些条件。 它打印出3和4 ，这意味着框3和4就是框1的框。这可以通过查看边界框bb中的值来手动验证。

Answer 2

感谢您的帮助Divakar。

我还提供了自己的解决方案，以供将来参考：

% Loop through each image in the dataset
for i=1:numel(anno.image_names)

% Skip images with no positive examples
inds = anno.bboxes(:,1) == i;
if sum(inds) == 0
    continue
end

% Read image
img = imread(fullfile(folder,anno.image_names{i}));

% Query image size
w_img = size(img,2);
h_img = size(img,1);

% Define the size of new negative bboxes to be the mean of the size of
% all positive bboxes in the image
w_new = floor(mean(anno.bboxes(inds,5)));
h_new = floor(mean(anno.bboxes(inds,6)));

% Define top-left and bottom-right corner of each existing bbox
tmp_bboxes = anno.bboxes(inds,:);
x = floor(tmp_bboxes(:,3));
y = floor(tmp_bboxes(:,4));
x_max = ceil(tmp_bboxes(:,3)+tmp_bboxes(:,5));
y_max = ceil(tmp_bboxes(:,4)+tmp_bboxes(:,6));

% Choose a random origin (a,b) that represents the upper-left
% corner pixel location and is properly constrained by the image size
a = randi([1,w_img - w_new],[sum(inds), 1]);
b = randi([1,h_img - h_new],[sum(inds),1]);

% Check if for the origin (a,b), the bbox created by the origin and
% w_new, h_new overlaps with the positive bboxes
inds2 = (a<x & (a+w_new)<x) | (a>x_max & (a+w_new)>x_max) | ...
    (b<y & (b+h_new)<y) | (b>y_max & (b+h_new)>y_max);
while sum(inds2) < sum(inds)
    a(~inds2) = randi([1,w_img - w_new],[sum(~inds2), 1]);
    b(~inds2) = randi([1,h_img - h_new],[sum(~inds2),1]);
    inds2 = (a<x & (a+w_new)<x) | (a>x_max & (a+w_new)>x_max) | ...
    (b<y & (b+h_new)<y) | (b>y_max & (b+h_new)>y_max);
end

end