我正在努力理解在此处找到的 LBP 算法的 Matlab 实现。我试图找出它如何计算每个像素的二进制文件？它只是计算相邻像素大于实际中心像素大小的位置。我想计算每个像素的二进制文件，以便使用局部直方图来计算图像的特征。
[ysize, xsize] = size(image);
miny=min(spoints(:,1));
maxy=max(spoints(:,1));
minx=min(spoints(:,2));
maxx=max(spoints(:,2));
% Block size, each LBP code is computed within a block of size bsizey*bsizex
bsizey=ceil(max(maxy,0))-floor(min(miny,0))+1;
bsizex=ceil(max(maxx,0))-floor(min(minx,0))+1;
% Coordinates of origin (0,0) in the block
origy=1-floor(min(miny,0));
origx=1-floor(min(minx,0));
% Minimum allowed size for the input image depends
% on the radius of the used LBP operator.
if(xsize < bsizex || ysize < bsizey)
error('Too small input image. Should be at least (2*radius+1) x (2*radius+1)');
end
% Calculate dx and dy;
dx = xsize - bsizex;
dy = ysize - bsizey;
% Fill the center pixel matrix C.
C = image(origy:origy+dy,origx:origx+dx);
d_C = double(C);
bins = 2^neighbors;
% Initialize the result matrix with zeros.
result=zeros(dy+1,dx+1);
%Compute the LBP code image
% the whole process here
for i = 1:neighbors
y = spoints(i,1)+origy;
x = spoints(i,2)+origx;
% Calculate floors, ceils and rounds for the x and y.
fy = floor(y); cy = ceil(y); ry = round(y);
fx = floor(x); cx = ceil(x); rx = round(x);
% Check if interpolation is needed.
if (abs(x - rx) < 1e-6) && (abs(y - ry) < 1e-6)
% Interpolation is not needed, use original datatypes
N = image(ry:ry+dy,rx:rx+dx);
D = N >= C;
else
% Interpolation needed, use double type images
ty = y - fy;
tx = x - fx;
% Calculate the interpolation weights.
w1 = roundn((1 - tx) * (1 - ty),-6);
w2 = roundn(tx * (1 - ty),-6);
w3 = roundn((1 - tx) * ty,-6) ;
% w4 = roundn(tx * ty,-6) ;
w4 = roundn(1 - w1 - w2 - w3, -6);
% Compute interpolated pixel values
N = w1*d_image(fy:fy+dy,fx:fx+dx) + w2*d_image(fy:fy+dy,cx:cx+dx) + ...
w3*d_image(cy:cy+dy,fx:fx+dx) + w4*d_image(cy:cy+dy,cx:cx+dx);
N = roundn(N,-4);
D = N >= d_C;
end
% Update the result matrix.
v = 2^(i-1);
result = result + v*D;
end
%Apply mapping if it is defined
if isstruct(mapping)
bins = mapping.num;
for i = 1:size(result,1)
for j = 1:size(result,2)
result(i,j) = mapping.table(result(i,j)+1);
end
end
end
if (strcmp(mode,'h') || strcmp(mode,'hist') || strcmp(mode,'nh'))
% Return with LBP histogram if mode equals 'hist'.
result=hist(result(:),0:(bins-1));
if (strcmp(mode,'nh'))
result=result/sum(result);
end
else
%Otherwise return a matrix of unsigned integers
if ((bins-1)<=intmax('uint8'))
result=uint8(result);
elseif ((bins-1)<=intmax('uint16'))
result=uint16(result);
else
result=uint32(result);
end
end
size(result)
end
迭代地，它为每个像素的所有 8 个邻居在结果中添加一些值。但它如何与 LBP 二进制文件相关联？它与以下 c++ LBP 方法的以下代码有何关联：
uchar lbp(const Mat_<uchar> & img, int x, int y)
{
// this is pretty much the same what you already got..
uchar v = 0;
uchar c = img(y,x);
v += (img(y-1,x ) > c) << 0;
v += (img(y-1,x+1) > c) << 1;
v += (img(y ,x+1) > c) << 2;
v += (img(y+1,x+1) > c) << 3;
v += (img(y+1,x ) > c) << 4;
v += (img(y+1,x-1) > c) << 5;
v += (img(y ,x-1) > c) << 6;
v += (img(y-1,x-1) > c) << 7;
return v;
}