说明：一共有三个m文件，一个是lbp.m, 存放主要的lbp算法，一个是getmapping,用以做算法的辅助函数，一个是lbptest.m,存放着测试代码。这三个文件需要放到同一个文件夹，并在文件夹中添加相应的图片，具体的图片名字见lbptest.m的代码，运行lbptest.m可以查看结果。代码最后给出效果图
lbp.m文件
%LBP returns the local binary pattern image or LBP histogram of an image.% J = LBP(I,R,N,MAPPING,MODE) returns either a local binary pattern
% coded image or the local binary pattern histogram of an intensity
% image I. The LBP codes are computed using N sampling points on a
% circle of radius R and using mapping table defined by MAPPING.
% See the getmapping function for different mappings and use 0 for
% no mapping. Possible values for MODE are
% 'h' or 'hist' to get a histogram of LBP codes
% 'nh' to get a normalized histogram
% Otherwise an LBP code image is returned.
%
% J = LBP(I) returns the original (basic) LBP histogram of image I
%
% J = LBP(I,SP,MAPPING,MODE) computes the LBP codes using n sampling
% points defined in (n * 2) matrix SP. The sampling points should be
% defined around the origin (coordinates (0,0)).
%
% Examples
% --------
% I=imread('rice.png');
% mapping=getmapping(8,'u2');
% H1=LBP(I,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood
% %using uniform patterns
% subplot(2,1,1),stem(H1);
%
% H2=LBP(I);
% subplot(2,1,2),stem(H2);
%
% SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
% I2=LBP(I,SP,0,'i'); %LBP code image using sampling points in SP
% %and no mapping. Now H2 is equal to histogram
% %of I2.
function result = lbp(varargin) % image,radius,neighbors,mapping,mode)
% Version 0.3.2
% Authors: Marko Heikkil?and Timo Ahonen
% Changelog
% Version 0.3.2: A bug fix to enable using mappings together with a
% predefined spoints array
% Version 0.3.1: Changed MAPPING input to be a struct containing the mapping
% table and the number of bins to make the function run faster with high number
% of sampling points. Lauge Sorensen is acknowledged for spotting this problem.
% Check number of input arguments.
error(nargchk(1,5,nargin));
image=varargin{1};
d_image=double(image);
if nargin==1
spoints=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
neighbors=8;
mapping=0;
mode='h';
end
if (nargin == 2) && (length(varargin{2}) == 1)
error('Input arguments');
end
if (nargin > 2) && (length(varargin{2}) == 1)
radius=varargin{2};
neighbors=varargin{3};
spoints=zeros(neighbors,2);
% Angle step.
a = 2*pi/neighbors;
for i = 1:neighbors
spoints(i,1) = -radius*sin((i-1)*a);
spoints(i,2) = radius*cos((i-1)*a);
end
if(nargin >= 4)
mapping=varargin{4};
if(isstruct(mapping) && mapping.samples ~= neighbors)
error('Incompatible mapping');
end
else
mapping=0;
end
if(nargin >= 5)
mode=varargin{5};
else
mode='h';
end
end
if (nargin > 1) && (length(varargin{2}) > 1)
spoints=varargin{2};
neighbors=size(spoints,1);
if(nargin >= 3)
mapping=varargin{3};
if(isstruct(mapping) && mapping.samples ~= neighbors)
error('Incompatible mapping');
end
else
mapping=0;
end
if(nargin >= 4)
mode=varargin{4};
else
mode='h';
end
end
% Determine the dimensions of the input image.
[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
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 = (1 - tx) * (1 - ty);
w2 = tx * (1 - ty);
w3 = (1 - tx) * ty ;
w4 = tx * ty ;
% 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);
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
end
getmapping.m
% GETMAPPING returns a structure containing a mapping table for LBP codes.% MAPPING = GETMAPPING(SAMPLES,MAPPINGTYPE) returns a
% structure containing a mapping table for
% LBP codes in a neighbourhood of SAMPLES sampling
% points. Possible values for MAPPINGTYPE are
% 'u2' for uniform LBP
% 'ri' for rotation-invariant LBP
% 'riu2' for uniform rotation-invariant LBP.
%
% Example:
% I=imread('rice.tif');
% MAPPING=getmapping(16,'riu2');
% LBPHIST=lbp(I,2,16,MAPPING,'hist');
% Now LBPHIST contains a rotation-invariant uniform LBP
% histogram in a (16,2) neighbourhood.
%
function mapping = getmapping(samples,mappingtype)
% Version 0.1.1
% Authors: Marko Heikkil?and Timo Ahonen
% Changelog
% 0.1.1 Changed output to be a structure
% Fixed a bug causing out of memory errors when generating rotation
% invariant mappings with high number of sampling points.
% Lauge Sorensen is acknowledged for spotting this problem.
table = 0:2^samples-1;
newMax = 0; %number of patterns in the resulting LBP code
index = 0;
if strcmp(mappingtype,'u2') %Uniform 2
newMax = samples*(samples-1) + 3;
for i = 0:2^samples-1
j = bitset(bitshift(i,1,samples),1,bitget(i,samples)); %rotate left
numt = sum(bitget(bitxor(i,j),1:samples)); %number of 1->0 and
%0->1 transitions
%in binary string
%x is equal to the
%number of 1-bits in
%XOR(x,Rotate left(x))
if numt <= 2
table(i+1) = index;
index = index + 1;
else
table(i+1) = newMax - 1;
end
end
end
if strcmp(mappingtype,'ri') %Rotation invariant
tmpMap = zeros(2^samples,1) - 1;
for i = 0:2^samples-1
rm = i;
r = i;
for j = 1:samples-1
r = bitset(bitshift(r,1,samples),1,bitget(r,samples)); %rotate
%left
if r < rm
rm = r;
end
end
if tmpMap(rm+1) < 0
tmpMap(rm+1) = newMax;
newMax = newMax + 1;
end
table(i+1) = tmpMap(rm+1);
end
end
if strcmp(mappingtype,'riu2') %Uniform & Rotation invariant
newMax = samples + 2;
for i = 0:2^samples - 1
j = bitset(bitshift(i,1,samples),1,bitget(i,samples)); %rotate left
numt = sum(bitget(bitxor(i,j),1:samples));
if numt <= 2
table(i+1) = sum(bitget(i,1:samples));
else
table(i+1) = samples+1;
end
end
end
mapping.table=table;
mapping.samples=samples;
mapping.num=newMax;
lbptest.m
%%%%%%%%%%%%%%%%%%%LBP变换后的图像%%%%%%%%%%%%%%%%%%I11 = imread('test1.bmp');
SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
I12=LBP(I11,SP,0,'i');
I21 = imread('test2.bmp');
SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
I22=LBP(I21,SP,0,'i');
re1 = abs(I22-I12);
% re1 = 255 - re1;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
I31 = imread('test3.bmp');
SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
I32=LBP(I31,SP,0,'i');
I41 = imread('test4.bmp');
SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1];
I42=LBP(I41,SP,0,'i');
re2 = abs(I32-I42);
% re2 = 255 - re2;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
re3 = abs(I32 - I12);
% re3 = 255 - abs(I32 - I12);
re4 = abs(I32 - I22);
re5 = abs(I42 - I12);
re6 = abs(I42 - I22);
%%%%%%%%%%%%%%%%%%%%%%直方图%%%%%%%%%%%%%%%%%%%%%%%%
mapping=getmapping(8,'u2');
H11=LBP(I11,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood using uniform patterns
subplot(2,2,1),stem(H11);
H12=LBP(I11);
subplot(2,2,2),stem(H12);
H21=LBP(I21,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood using uniform patterns
subplot(2,2,3),stem(H21);
H22=LBP(I21);
subplot(2,2,4),stem(H22);
H31=LBP(I31,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood using uniform patterns
subplot(2,2,3),stem(H31);
H32=LBP(I31);
subplot(2,2,4),stem(H32);
%%%%%%%%%%%%%%%%%%%%%%%无聊瞎写%%%%%%%%%%%%%%%%%%%%%%%%
[row, col] = size(I11);
It1 = ones(row, col);
for i = 2 : col
It1(:, i - 1) = abs(I11(:, i) - I11(:, i - 1));
end
It2 = ones(row, col);
for i = 2 : row
It2(i-1,:) = abs(I11(i,:) - I11(i-1,:));
end
测试结果截图与分析
运行完lbptest后，可以输入imshow(I11)查看第一幅图像的原图，如下（昆虫翅膀）
输入imshow(I12)查看第一幅图像经LBP处理后的图像
输入imshow(re1)查看第一幅与第二幅图像做差值之后的图像，黑色越多，表示两个翅膀越相近
这是原图经过LBP处理之后的比较，lbptest.m后半部分的代码是比较两幅图的直方图，我个人建议用直方图，虽然不如显示原图的方法直观，但却有平移不变性，效果如图：
直方图越相似，代表两幅图像越相近。不过这种方法老让我想起哈希变换与RSA加密……不知道有没有人跟我同感。