Hi sir, please find the attached code,and correct it,i need to find likelihood ratio for 'n 4x4 block', in 'n frames'.i dontno to explain clearly,if u see the code u can able to understand what i want.please look after it

1 次查看（过去 30 天）

kaavya subramani 2015-8-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/235837-hi-sir-please-find-the-attached-code-and-correct-it-i-need-to-find-likelihood-ratio-for-n-4x4-bloc

评论： kaavya subramani 2015-8-25

for p=1:num of frames
i1 = cc(1:96,1:96);
i2 = cc(1:96,97:192);
i3 = cc(97:192,1:96);
i4 = cc(97:192,97:192);
y = {i1, i2, i3, i4};
for i=1:length(y)
  ui = edge(y{i},'canny');
  count(i) = length(find(ui(:)==1));
end
[~, o] = max(count);
% finding likelihood ratio
for o=1:length(o)
lr=(((var(mean(o))+var(mean(o+1))/2)+((mean(mean(o))-mean(mean(o+1))/2)^2)^2)/(var(mean(o))*var(mean(o+1)));
end
end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Walter Roberson 2015-8-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/235837-hi-sir-please-find-the-attached-code-and-correct-it-i-need-to-find-likelihood-ratio-for-n-4x4-bloc#answer_190220

You have no 4 x 4 blocks.

That code will count the number of canny == 1 results in each y, so count will be a vector the same length as the cell array y. [~, o] = max(count); will then find the index in count where the value is largest. o is going to be a scalar. Then for o=1:length(o) is going to be for o=1:1 (because length() of a scalar is 1) so inside the "for" loop, the "o" you calculated is going to be overwritten with the value 1.

Your "lr" formula does not look correct to me.

7 个评论
显示 5更早的评论隐藏 5更早的评论

kaavya subramani 2015-8-25

sir its for variance,it means max quarter of 1st and 2nd,1st and 3rd so on,then 2nd and 3rd,2nd and 4th and so on..likewise for 3rd and 4th frames,3rd and 5th frames ......instead of k-1,please consider k+1,by mistake i typed wrongly

kaavya subramani 2015-8-25

i1 = cc(1:96,1:96); i2 = cc(1:96,97:192); i3 = cc(97:192,1:96); i4 = cc(97:192,97:192); y = {i1, i2, i3, i4}; for i=1:length(y) ui = edge(y{i},'canny'); count(i) = length(find(ui(:)==1)); end [~, o] = max(count);%This 'o' contains the index of the block which contain maximum edge pixels. % finding likelihood ratio for o=1:length(o) lr=(((var(mean(o))+var(mean(o+1))/2)+((mean(mean(o))-mean(mean(o+1))/2)^2)^2)/(var(mean(o))*var(mean(o+1)));

end end I want to find like this for all frames.then need to perform likelihood ratio,for eg:if 1st frames maxcount block index'1',2nd frames maxcount block index is '3'.then find likelihood ratio for 1st frames '1' block matrix and 2nd frames '3' block matrix.

请先登录，再进行评论。

请先登录，再回答此问题。