Eliminate the pixel that have a lower intensity?

2 次查看（过去 30 天）
valerio auricchio
valerio auricchio 2020-11-9
评论： Ameer Hamza 2020-11-10
I read the immage in this way:
m=max(Imm,[],'all');
%Threshold
thr=10;
for i=1:2048
for j=1:2448
if Imm(i,j)>(m-thr)
pos(k,1:2)=[i,j];
k=k+1;
end
end
end
"Pos" contain all the pixel of my intrest, now i want to elimintate all the coordinates of the pixel that have intensity lower than a value

请先登录，再回答此问题。

采纳的回答

Ameer Hamza
Ameer Hamza 2020-11-9
You don't need a for-loop for this
m=max(Imm,[],'all');
%Threshold
thr=10;
[r,c] = find(Imm>(m-thr))
pos = [r c];
By elimination, if you mean to set them to zero, then try this
m=max(Imm,[],'all');
%Threshold
thr=10;
Imm(Imm<(m-thr)) = 0;
  4 个评论
显示 2更早的评论
valerio auricchio
valerio auricchio 2020-11-10
the fact is that after i find the point in fist immage and save them in this way:
m=max(Imm,[],'all');
%Threshold
thr=10;
[r,c] = find(Imm>(m-thr))
pos = [r c];
I want to use this point in the second immage and want to eliminate the point saved in "pos" where the intensity is lower than a value
Ameer Hamza
Ameer Hamza 2020-11-10
I didn't understand the question. Can you take 2 small 3x3 matrices and show with an example of what you want?

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Blocked Images 的更多信息

标签

产品


版本

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by