uint8 and double in case of image

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vaishali
vaishali am 27 Nov. 2014
Bearbeitet: Image Analyst am 27 Nov. 2014
Hello
In one of the image encryption method I need to convert an image from uint8 to double. But after converting the whole, image is losing some information only white colour with some coloured dots will appear in that double image. so after the required task how to convert it to original image? please guide..
Thanks

Akzeptierte Antwort

per isakson
per isakson am 27 Nov. 2014
Try
a=imread('lena.jpg');
imshow(a);
b=double(a);
imshow(b/255);

Weitere Antworten (1)

Image Analyst
Image Analyst am 27 Nov. 2014
Look at the variable in the workspace/variable editor and tell me if there is data loss there. Chances are you're just not displaying it correctly. For double, if it's not in the range o-1, you need to use [] as the second argument in imshow():
imshow(doubleImage, []);
  2 Kommentare
vaishali
vaishali am 27 Nov. 2014
Bearbeitet: per isakson am 27 Nov. 2014
in workspace the number of elements not changing.but while displaying
a=imread('lena.jpg');
imshow(a);
b=double(a);
imshow(b,[]);
after converting it to double it is showing blank white background with some dots. if this is the case how to get back original lena?
Image Analyst
Image Analyst am 27 Nov. 2014
Bearbeitet: Image Analyst am 27 Nov. 2014
The badly-named "a" must be a color image. You cannot display double RGB images unless they're in the range 0-1. You must cast to uint8 first. And since it's color, you won't need the [] - it will ignore them.
imshow(uint8(b));
Or you can divide your array by 255 (which is intmax('uint8')) like per showed.

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