LBP源代码阅读

静下心来看了一下LBP的源代码，作者写得很漂亮。以前经常用，但没有注意过细节，写了一些注释与大家分享。

%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('test1.bmp');
%       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.% %% 检查参数的个数nargin，使其大于1小于5。如果不在此区间，就报错
error(nargchk(1,5,nargin));

image=varargin{1};% % %% 把第一个参数赋值给image
d_image=double(image);% % % 把图像从uint8转成double类型，以便以后计算

% % % 只有给出待处理的图像（一个参数）时，使用默认的设置。
% % % sp定义了中心点与它的近邻的相对位置
% % % neighbors定义近邻个数
% % % mapping定义的映射
% % % mode区别直方图的类型，'h' or 'hist'是直方图，nh是规一化的直方图
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

% % % 给出两个参数，并且第二个参数（代表近邻半径）的长度为1时
% % % 只给出了近邻的半径，没给出近邻的个数，报错
if (nargin == 2) && (length(varargin{2}) == 1)
    error('Input arguments');
end
% % % 如果给出两个以上的参数，并且第二个参数（代表近邻半径）的长度为1
% % % 半径设为第二个参数
% % % 近邻个数设为第三个参数
if (nargin > 2) && (length(varargin{2}) == 1)
    radius=varargin{2};
    neighbors=varargin{3};
    
    spoints=zeros(neighbors,2);
    
% % %     把360度均匀分成neighbors分，以计算近邻点与中心的相对坐标
    % Angle step.
    a = 2*pi/neighbors;
    
% % %       计算坐标，每一维代表y,第二维代表x
% % %       spoints的第i行代表第i个近邻
    for i = 1:neighbors
        spoints(i,1) = -radius*sin((i-1)*a);
        spoints(i,2) = radius*cos((i-1)*a);
    end
% % %     如果参数个数大于等于4，第四个参数赋值给映射mapping；否则，无映射。
    if(nargin >= 4)
        mapping=varargin{4};
        if(isstruct(mapping) && mapping.samples ~= neighbors)
            error('Incompatible mapping');
        end
    else
        mapping=0;
    end
% % %     第五个参数确定直方图的属性
    if(nargin >= 5)
        mode=varargin{5};
    else
        mode='h';
    end
end
% % % 如果参数个数大于1，并且第二个参数的长度大于1。则第二个参数给出近邻点与中心点的相对位置
if (nargin > 1) && (length(varargin{2}) > 1)
    spoints=varargin{2};
    neighbors=size(spoints,1);
% % %     如果还有第三个参数，把它赋值给映射mapping
    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.图像的大小，第一维是y，第二维是x
[ysize xsize] = size(image);

% % % 确定block的左上和右下两个点
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 
% % % block的大小
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
% % % 在block里中心点的坐标
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.
% % % 检查block和img的大小
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.
% % % 所有可以作为模板中心点的像素集合
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);
% % % 初始化结果矩阵


%Compute the LBP code image  这一段写得很漂亮！！！！
% % % 对于每一个neighbor，先使要比较的点与中心点对齐，然后利用D = N >= C比较它们的大小。
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.
% % %   更新结果矩阵
  v = 2^(i-1);
  result = result + v*D;
end

%Apply mapping if it is defined
% % % 如果mapping已经存在，那么利用这个mapping.
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
% % % 如果要参数列表指定了直方图的属性，计算直方图
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
% % %     如果没有指定直方图的属性，返回数值方阵
    if ((bins-1)<=intmax('uint8'))
        result=uint8(result);
    elseif ((bins-1)<=intmax('uint16'))
        result=uint16(result);
    else
        result=uint32(result);
    end
end

end

生成mapping的函数

%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是i循环左移的结果
    j = bitset(bitshift(i,1,'uint8'),1,bitget(i,samples)); %rotate left
% % %     计算跳变的次数
    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)) 
% % %     如果跳变次数不大于2，那么新建一个标记index;否则放入最后一类
    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;
% % %     计算所有rotate中最小的一个
    for j = 1:samples-1
      r = bitset(bitshift(r,1,'uint8'),1,bitget(r,samples)); %rotate
                                                             %left
      if r < rm
        rm = r;
      end
    end
% % %     同上
    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,'uint8'),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;

LBP函数的第五个参数，作者只给出了三种选择，'h'，‘hist’和‘uh’。如果没有设置，程度自动设置为mode='h'。如果想要看一下图像的原始LBP特征，而非直方图，可以随便设一个其它的值。

clear;clc;close all
I=imread('test1.bmp');
mapping=getmapping(8,'u2');

H1=lbp(I,1,8,mapping,'a');%%%如果想以图像形式显示lbp特征，第五个参数随便设一个值，但不能不设。H1=lbp(I,1,8,mapping,1);
figure;imshow(H1)


H1=lbp(I,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood
                                 %using uniform patterns
figure                                 
subplot(2,1,1),stem(H1);
H2=lbp(I);
subplot(2,1,2),stem(H2);