matlab function

1 次查看（过去 30 天）
LIU WEIHUA
LIU WEIHUA 2012-4-16
I have write a code to find the first pixel that not equal to zero and store the pixel position to M and N respectively as follow:
j=1;
while (j>=1&&j<=size(I,1) )
for i=1:size(I,2)
if I(i,j) >= 0
M=j
N=i
break;
else
end
end
j=j+1;
end
where I is the image;
when the code find the first pixel that not equal to zeros,however , the loop can not stop, the break seems do not help to stop the loop;
Can you please help me on this?

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2012-4-16
The break is breaking out of the "for", but then you are still within the "while". You need to set a flag before you break and test it after the "for" loop code, using break when the flag is set.
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2012-4-16
j=1;
breakout = false;
while (j>=1&&j<=size(I,1) )
for i=1:size(I,2)
if I(i,j) >= 0
M=j
N=i
breakout = true;
break;
else
end
end
if breakout; break; end
j=j+1;
end
Image Analyst
Image Analyst 2012-4-17
I think you meant
while (j >= 1 && j <= size(I, 2) )
for i = 1 : size(I, 1)

请先登录，再进行评论。

更多回答（2 个）

Richard Brown
Richard Brown 2012-4-16
Why not just
[M, N] = find(I, 1, 'first')
?
Image Analyst
Image Analyst 2012-4-17
Um, did you notice that your "i" variable is the first dimension (row) in the matrix but you are going to the second dimension max (the max column, not row)? Similarly for your j variable. Thus your code will bomb for anything but square matrices.
And what do you define as the "first" pixel? Starting at the upper left pixel, do you want to go across columns, then down to the next row, or go down rows then over to the next column, or do you want to find the pixel that is closest to the upper left? Those might (probably) give three different pixels as the answer.

类别

在 Help Center 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

尚未输入任何标签。

Community Treasure Hunt

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

Start Hunting!

Translated by