identify valid and invalid image

i execute a code and get two types of output.... one is valid and the other is invalid.... the valid output image has content content in it.... whereas the invalid image is totally black....
based on this information what can i do to display a message that the image is valid or invalid.... some condition check so that i can display the message correctly.... please can someone reply.....

 採用された回答

Walter Roberson
Walter Roberson 2013 年 1 月 5 日

1 投票

if ~any(TheOutputImage(:))
msgdlg('Outside of a dog, a book is man''s best friend. Inside of a dog, it is too dark to read.');
else
......
end

4 件のコメント

Elysi Cochin
Elysi Cochin 2013 年 1 月 5 日
編集済み: Elysi Cochin 2013 年 1 月 5 日
sir i got it but if it is invalid image i do not want to display it.... but that is also getting displayed for me when i do like this.... please tell me where i should make the change for not getting the invalid image displayed...
if ~any(YourImage(:))
msgbox('Invalid Image');
else
figure(1),
subplot(10,6,i);
imshow(YourImage);
axis image off
end
Walter Roberson
Walter Roberson 2013 年 1 月 5 日
The first thing you need to do is stop using "image" as a variable name: doing so makes it difficult to use the function image()
Second, if your code is not giving you the message box, then the output is not "totally black" as you claimed. Find out what
max(YourImage(:))
mean(YourImage(:))
std(YourImage(:))
are for the invalid images.
Elysi Cochin
Elysi Cochin 2013 年 1 月 5 日
編集済み: Elysi Cochin 2013 年 1 月 5 日
sir i got it... thank u sir... i wrote the display outside the for also... thank u sir...
Walter Roberson
Walter Roberson 2013 年 1 月 5 日
Put a breakpoint in at the "if ~any" and run your code again with something that creates an invalid image. Single-step through the code. I think you will find that the black image is not displayed by this section of code. If you continue single-stepping you will find the place that is displaying it.

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