我试图用方向梯度直方图来实现对拷贝移动伪造的纸张检测。
算法是:
我被第三步困住了,不能继续了。
我实现的代码是:
clc;
clear all;
close all;
%read image
img = imread('006_F.png');
img=rgb2gray(img);
img=imresize(img, 1/4);
figure(1);
imshow(img);
b=16; %block size
nrc=5; %no. of rows to check
td=416; %threshold
[r, c]=size(img);%Rows and columns;
column=(r-b+1)*(c-b+1);
M= zeros(column,4);
Mi = zeros(1,2);
i=1;
disp('starting extraction of features');
for r1 = 1:r-b+1
for c1 = 1:c-b+1
% Extract each block
B = img(r1:r1+b-1,c1:c1+b-1);
features = extractHOGFeatures(B);%extracting features
M(i, :) = features;
Mi(i,:) = [r1 c1];
i=i+1;
end
end
[S, index] = sortrows(M , [ 1 2 3 4]);
P= zeros(1,6);
b2=r-b+1;
disp('Finding Duplicates');
for i = 1:column
iv = index(i);
xi=mod(iv,b2) + 1;
yi=ceil(iv/b2);
j = i+1;
while j < column && abs(i - j) < 5
jv=index(j);
xj=mod(jv,b2) + 1;
yj=ceil(jv/b2);
z=sqrt(power(xi-xj,2) + power(yi-yj,2));
% only process those whose size is above Nd
if z > 16
offset = [xi-xj yi-yj];
P = [P;[xi yi xj yj xi-xj yi-yj]];
end
j = j + 1;
end
end
rows = size(P,1);
P(:,6) = P(:,6) - min(P(:,6));
P(:,5) = P(:,5) - min(P(:,5));
maxValP = max(P(:,6)) + 1;
P(:,5) = maxValP .* P(:,5) + P(:,6);
mostfrequentval = mode(P(:,5));
disp('Creating Image');
idx = 2;
% Create a copy of the image and mask it
RI = img;
while idx < rows
x1 = P(idx,1);
y1 = P(idx,2);
x2 = P(idx,3);
y2 = P(idx,4);
if (P(idx,5) == mostfrequentval)
RI(y1:y1,x1:x1) = 0;
RI(y2:y2,x2:x2) = 0;
end
idx = idx + 1;
end;发布于 2016-10-18 20:12:42
在阅读了论文中提到的一些参考文献之后(参考文献)。8和20):
字典排序相当于字母排序,对于数字,即:1 1 1<1 1 2 1<2 3 4 5<2 4 4 5。
因此,在您的例子中,您用以下方式使用函数sortrows():
A = [1 1 1 1;1 1 1 2;1 1 1 4;1 2 2 2; 1 2 2 1; 1 4 6 3; 2 3 4 5; 2 3 6 6]; % sample matrix
[B,idx] = sortrows(A,[1 2 3 4]); % Explicit notation but it is the Matlab default setting so equivalent to sortrows(A)它的意思是:首先查看第一列,然后在相等的情况下,查看第二列,对A行进行排序,等等。
如果您正在寻找相反的顺序,则在列号之前指定“-”。
因此,最后,您的代码是好的,如果结果不是预期的,它必须来自另一个步骤的实现.
编辑:参数idx记录排序行的原始索引。
https://stackoverflow.com/questions/40073025
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