首先,不要害怕这个问题的外观;)
我正在尝试在matlab中实现一个名为Circular Blurred Shape Model的形状描述符,其中一部分是获取每个径向段的最近邻居列表,如图1d所示。
我在MATLAB中进行了一个简单而简单的实现,但是我坚持算法的第5步和第6步,主要是因为我无法理解这个定义:
Xb{c,s} = {b1, ..., b{c*s}} as the sorted set of the elements in B*
so that d(b*{c,s}, bi*) <= d(b*{c,s}, bj*), i<j
对我而言,这听起来像一个级联排序,首先按升序距离排序,然后按升序索引排序,但我找到的最近邻居不符合论文。
作为一个例子,我向你展示了我为段b {4,1}获得的最近邻居,这是图1d中标记为“EX”的那个)
我得到b {4,1}的最近邻居列表:b{3,2}, b{3,1}, b{3,8}, b{2,1}, b{2,8}
根据论文的正确性是:b{4,2}, b{4,8}, b{3,2}, b{3,1}, b{3,8}
然而,我的分数实际上是按欧几里德距离测量的最接近所选分段的设置!距离b{4,1} <=> b{2,1}
小于b{4,1} <=> b{4,2}
或b{4,1} <=> b{4,8}
...
这是我的(丑陋,但直截了当)MATLAB代码:
width = 734;
height = 734;
assert(width == height, 'Image must be square in size!');
% Radius of the correlogram
R = width;
% Number of circles in correlogram
C = 4;
% Number of sections in correlogram
S = 8;
% "width" of ring segments
d = R/C;
% angle of one segment in degrees
g = 360/S;
% set of bins for the circular description of I
B = zeros(C, S);
% centroid coordinates for bins
B_star = zeros(C,S,2);
% calculate centroids of bins
for c=1:C
for s=1:S
alpha = deg2rad(max(s-1, 0)*g + g/2);
r = d*max((c-1),0) + d/2;
B_star(c,s,1) = r*cos(alpha);
B_star(c,s,2) = r*sin(alpha);
end
end
% create sorted list of bin numbers which fullfill
% d(b{c,s}*, bi*) <= d(b{c,s}, bj*) where i<j
% B_star_dists is a simple square distance matrix for getting
% the distance between two centroids c_i,s_i and c_j,s_j
B_star_dists = zeros(C*S, C*S);
for i=1:C*S
[c_i, s_i] = ind2sub([C,S], i);
% x,y centroid coordinates for point i
b_star_i = [B_star(c_i, s_i, 1), B_star(c_i, s_i, 2)];
for j=1:C*S
[c_j, s_j] = ind2sub([C,S], j);
% x,y centroid coordinates for point j
b_star_j = [B_star(c_j, s_j, 1), B_star(c_j, s_j, 2)];
% store the euclidean distance between these two centroids
% in the distance matrix.
B_star_dists(i,j) = norm(b_star_i - b_star_j);
end
end
% calculate nearest neighbour "centroids" for each centroid
% B_NN is a cell array, B{idx} gives an array of indexes to the
% nearest neighbour centroids.
B_NN = cell(C*S, 1);
for i=1:C*S
[c_i, s_i] = ind2sub([C,S], i);
% get a (C*S)x2 matrix of all distances, the first column are the array
% indexes and the second column are the distances e.g
% 1 d1
% 2 d2
% .. ..
% CS d{c,s}
dists = [transpose(1:C*S), B_star_dists(:, i)];
% sort ascending by the distances first (e.g second column) then
% sort ascending by the array index (e.g first column)
dists = sortrows(dists, [2,1]);
% middle section has nine neighbours, set as default
neighbour_count = 9;
if c_i == 1
% inner region has S+3 neighbours
neighbour_count = S+3;
elseif c_i == C
% outer most ring has 6 neighbours
neighbour_count = 6;
end
B_NN{i} = dists(1:neighbour_count,1);
end
% FROM HERE ON JUST VISUALIZATION CODE
figure(1);
hold on;
for c=1:C
% plot circles
r = c*d;
plot(r*cos(0:pi/50:2*pi), r*sin(0:pi/50:2*pi), 'k:');
end
for s=1:S
% plot lines
line_len = C*d;
alpha = deg2rad(s*g);
start_pt = [0, 0];
end_pt = start_pt + line_len.*[cos(alpha), sin(alpha)];
plot([start_pt(1), end_pt(1)], [start_pt(2), end_pt(2)], 'k-');
end
for c=1:C
% plot centroids of segments
for s=1:S
segment_centroid = B_star(c,s, :);
plot(segment_centroid(1), segment_centroid(2), '.k');
end
end
% plot some nearest neighbours
% list of [C;S]
plot_nn = [4;1];
for i = 1:size(plot_nn,2)
start_c = plot_nn(1,i);
start_s = plot_nn(2,i);
start_pt = [B_star(start_c, start_s,1), B_star(start_c, start_s,2)];
start_idx = sub2ind([C, S], start_c, start_s);
plot(start_pt(1), start_pt(2), 'xb');
nn_idx_list = B_NN{start_idx};
for j = 1:length(nn_idx_list)
nn_idx = nn_idx_list(j);
[nn_c, nn_s] = ind2sub([C, S], nn_idx);
nn_pt = [B_star(nn_c, nn_s,1), B_star(nn_c, nn_s,2)];
plot(nn_pt(1), nn_pt(2), 'xr');
end
end
可以找到完整的论文here
答案 0 :(得分:2)
论文谈到“地区邻居”;这些是欧几里德距离意义上的“最近邻居”的解释是不正确的。它们只是某个区域的邻居区域,找到它们的方法很简单:
区域有2个坐标:(c,s)其中c表示它们所属的同心圆,从中心的1到边缘的C,s表示它们所属的扇区,从1开始从角度0°开始到S以角度360°结束。
c和s坐标与区域坐标最多相差1的每个区域都是相邻区域(区段编号从S到1环绕)。根据区域的位置,有3种情况: (如图1d所示)
该区域是中间区域(标记为MI),例如区域b(2,4)
有2个相邻的圈子和2个相邻的扇区,总共9个区域:
圈1,2或3中的每个区域和扇区3,4或5:
b(1,3), b(2,3), b(3,3), b(1,4), b(2,4), b(3,4), b(1,5), b(2,5), b(3,5)
该区域是内部区域(标记为IN),例如区域b(1,8)
只有一个相邻的圆圈和2个相邻的扇区,但所有的内部区域都是邻居,所以共有S + 3个区域:
第2圈和第7,8或1区的每个区域:
b(2,7), b(2,8), b(2,1)
以及内圈的每个区域:
b(1,1), b(1,2), b(1,3), b(1,4), b(1,5), b(1,6), b(1,7), b(1,8)
该区域是外部区域(标记为EX),例如区域b(3,1)
只有一个相邻的圈子和2个相邻的扇区,总共有6个区域:
圈2或3中的每个区域和扇区8,1或2:
b(2,8), b(2,1), b(2,2), b(3,8), b(3,1), b(3,2)
答案 1 :(得分:2)
要将一些matlab添加到@ m69的great answer,您可以自动化邻居的索引,如下所示:
%assume C and S defined according to the answer of @m69
iif=@(varargin) varargin{2*find([varargin{1:2:end}], 1, 'first')}();
ncfun=@(c) iif(c==1,c+1,c==C,c-1,true,[c-1, c+1]);
nsfun=@(s)mod([s-1, s+1]-1,S)+1;
neighbs=@(c,s) [b(c,nsfun(s)), b(ncfun(c),s)', reshape(b(ncfun(c),nsfun(s)),1,[])];
第一个定义an inline if用于匿名函数,以便在单独的文件中定义函数(c
情况需要这样做)。这具有语法iif(condition1,result1,condition2,result2,...)
,其中每个条件在彼此之后进行测试,并且返回与产生true
的第一个条件相对应的结果。
第二个使用if:return [c-1,c + 1]定义径向邻居索引,除非其中任何一个未定义(这将导致数组绑定违规)
第三个定义角度扇区的周期性索引,例如S=4
nsfun(2)==[1, 3]
和nsfun(4)==[3, 1]
。
我刚刚添加了一个示例,其中给定的有效c,s
neighbs(c,s)
对将返回b(1:C,1:S)
的邻居的子阵列:首先是上下/左 - 正确的邻居,然后是(最多)四个角落。
答案 2 :(得分:0)
在花了太多时间后,我发现了,计算元素b {c,s}的邻近区域的技巧是:
获取c的相邻环的所有段,包括c本身,即c-1, c, c+1
,如果它是最内圈,则只有c,c+1
,如果它是最外层的c-1, c
}
计算从b {c,s}到上一步所有选定质心的欧氏距离,包括点b {c,s}本身
按升序对距离进行排序,并取前N个段。
描述符效果很好,但是它的旋转不变性取决于具有最高密度的轴,在某些情况下,这对噪声/遮挡非常敏感,即描述符可能是+/- 1旋转关闭。
这是MATLAB中完整的CBSM描述符实现,没有以任何方式进行优化(我听说MATLAB讨厌for循环):
<强> csbm.m 强>
function [descriptor] = csbm(I, R, C, S)
% Input:
% ------
% I = The image, binary, must be square in sice
% R = The radius of the correlogram, could be half the image width
% C = Number of circles
% S = Number of segments per circle
% Output:
% -------
% A C*S-by-1 row vector, the descriptor
[width, height] = size(I);
assert(width == height, 'Image must be square in size!');
% "width" of ring segments
d = R/C;
% angle of one segment in degrees
g = 2*pi/S;
% centroid coordinates for bins
B_star = zeros(C*S,2);
% initialize the descriptor
descriptor = zeros(C*S,1);
% calculate centroids of bins
for c=1:C
for s=1:S
alpha = max(s-1, 0)*g + g/2;
r = d*max((c-1),0) + d/2;
ind = (c-1)*S+s; %sub2ind(cs_size, c,s);
B_star(ind, :) = [r*cos(alpha), r*sin(alpha)];
end
end
% calculate nearest neighbor regions
B_NN = cell(C*S,1);
for c=1:C
min_circle = max(1,c-1);
max_circle = min(C,c+1);
start_sel = (min_circle-1)*S+1;
end_sel = (max_circle)*S;
centroids = B_star(start_sel:end_sel, :);
for s=1:S
current_ind = (c-1)*S+s;
centroid = B_star(current_ind, :);
num_centroids = length(centroids);
distances = sqrt(sum((centroids-repmat(centroid, num_centroids,1)).^2,2));
distances = horzcat(distances, transpose((start_sel:end_sel)));
distances = sortrows(distances, [1,2]);
neighbour_count = 9;
if c == 1
% inner region has S+3 neighbours
neighbour_count = S+3;
elseif c == C
% outer most ring has 6 neighbours
neighbour_count = 6;
end
B_NN{current_ind} = distances(1:neighbour_count, 2);
end
end
for x=1:width
x_centered = x-width/2;
for y=1:width
if I(x,y) == 0
continue;
end
y_centered = y-width/2;
% bin the image point
r = sqrt(x_centered^2 + y_centered^2);
a = wrapTo2Pi(atan2(y_centered, x_centered));
% get bin
c_pixel = max(1, floor(r/d));
s_pixel = max(1, floor(a/g));
if c_pixel > C
continue;
end
ind_pixel = (c_pixel-1)*S+s_pixel;
pt_pixel = [x_centered, y_centered];
% get neighbours of this bin
neighbours = B_NN{ind_pixel};
% calculate distance to neighbours
nn_dists = sqrt(sum((B_star(neighbours, :) - repmat(pt_pixel, length(neighbours), 1)).^2,2));
D = sum(1./nn_dists);
% update the probabilities vector
descriptor(neighbours) = descriptor(neighbours) + 1./(nn_dists.*D);
end
end
% normalize the vector v
descriptor = descriptor./sum(descriptor);
% make it rotationally invariant
G = zeros(S/2, 2*C);
for s=1:S/2
for c=1:C
G(s,c) = descriptor((c-1)*S+s);
G(s,c+C) = descriptor((c-1)*S+s+S/2);
end
end
[~, max_G_idx] = max(sum(G,2));
L_G = 0;
R_G = 0;
for j=1:C
for k=1:S
if k > (max_G_idx) && k < (max_G_idx+S/2)
L_G = L_G + descriptor((j-1)*S+k);
elseif k ~= max_G_idx && k ~= (max_G_idx+S/2)
R_G = R_G + descriptor((j-1)*S+k);
end
end
end
if L_G > R_G
% B is rotated k=i+S/2-1 positions to the left:
fprintf('rotate %d to the left\n', max_G_idx+S/2-1);
rotate_by = -(max_G_idx+S/2-1);
else
% B is rotated k=i-1 positions to the right:
fprintf('rotate %d to the right\n', max_G_idx-1);
rotate_by = -(max_G_idx-1);
end
% segments are grouped by circle
% so for every circle we get all segments and circular shift them
for c=1:C
range_sel = ((c-1)*S+1):(c*S);
segments = descriptor(range_sel);
descriptor(range_sel) = circshift(segments, [rotate_by,0]);
end
end
<强> plot_descriptor.m 强>
function plot_descriptor(R,C,S, descriptor)
% Input:
% ------
% R Radius for the correlogram in pixels, can be arbitrary
% C Number of circles
% S Number of segments per circle
% descriptor The C*S-by-1 descriptor vector
% "width" of ring segments
d = R/C;
% angle of one segment in degrees
g = 2*pi/S;
% full image
[x,y] = meshgrid(-R:R);
[theta, rho] = cart2pol(x,y);
theta = wrapTo2Pi(theta);
brightness = zeros(size(rho));
min_val = min(descriptor);
max_val = max(descriptor);
scale_fact = 1/(max_val-min_val);
for c=1:C
rInner = (c-1)*d;
rOuter = c*d;
for s=1:S
idx = (c-1)*S+s;
minTheta = (s-1)*g;
maxTheta = (s*g);
matching_theta = find(theta > minTheta & theta < maxTheta);
matching_rho = find(rho > rInner & rho < rOuter);
matching_idx = intersect(matching_theta, matching_rho);
intensity = descriptor(idx)*scale_fact;
brightness(matching_idx) = intensity;
end
end
figure; imshow(mat2gray(brightness));
如何运行
I = imread('bat-18.gif');
I = transpose(im2bw(I, graythresh(I)));
I = edge(I);
[width, height] = size(I);
R = width/2;
C = 24;
S = 24;
descriptor = csbm(I, R, C, S);
plot_descriptor(R,C,S,descriptor);
输出:
只是为了踢,在编码时出现了一些可怕的图片: