I'm trying to read colormap of an image using this code:
[X, map] = imread('D:\Eye.png');
But map is rescaled to [0,1] type double. How can I get the colormap in uint8 range [0,255]?
This can be solved by simply rescaling map and casting it to uint8:
uint8(255*map);
Optionally, you can round it before casting (the default rounding scheme, as above, is floor):
uint8(round(255*map));
Related
Log Transform for image is defined as this
s=T(r) = c*log(r+1)
where
s is the output image
r is the input image
c = 255/log(1+Maximum pixel value from the input image).
If the "Maximum pixel value from the input image" is 255 then the C value will be
105.886458025=255/log(256)
This does not make sense. 105.89 is a very high intensity value. Multiplying with this C value will yield a very bright image. Using the 'tire' image from matlab I get the output below:
I=imread('tire.tif');
I2=im2double(I);
J=105*log(1+I2);
imshow(I)
Is my understanding correct?
Image for demonstration
Alternate image
c is a scaling factor.
The following script works both on Matlab and Octave.
clc;
close all;
clearvars;
%imageFileName = 'https://www.forestfoliage.com/wp-content/original/2010_10/Fall-Foliage-mountain-sunset.jpg';
imageFileName = 'https://pixnio.com/free-images/2020/06/27/2020-06-27-11-50-57-1200x800.jpg';
I = imread(imageFileName);
inputMin = min(I(:))
inputMax = max(I(:))
I2 = double(I);
outputMax = 255
c = outputMax/log10(1 + double(inputMax))
J = uint8(double(c) * log10(1 + I2));
%I2(1:5,1:5,:)
%J(1:5,1:5,:)
imshow(I)
title("Raw")
pause(5)
imshow(J)
title("Processed")
im2double documentation:
im2double(I) converts the image I to double precision. I can be a grayscale intensity image, a truecolor image, or a binary image. im2double rescales the output from integer data types to the range [0, 1].
So, we will use double() instead. Thanks to #CrisLuengo for this. Earlier, I was rescaling the scaled output.
In Matlab,
Images stored as matrices of type double should have values in the range [0, 1].
Images stored as integer matrices should have values in the range [0, 255].
On line 12, we are casting the matrix to an integer matrix for this reason.
From the imshow() documentation,
If you specify a truecolor image of data type single or double, then values should be in the range [0, 1]. If pixel values are outside this range, then you can use the rescale function to scale pixel values to the range [0, 1]. The 'DisplayRange' argument has no effect when the input image is truecolor.
I already know uint8 contains intensity values between 0 and 255 (28-1) and single contains values between 0 and 1, it is used to hold larger values without upsetting the range error. But, apart from that, are there any other differences?
What is the difference between imagesc((I/64)*64) and imagesc((Is/64)*64), where I is uint8 and Is is single?
imagesc just calls image underneath. As for image, it behaves a bit differently if integers or floats are supplied, as can be learned from image's documentation:
If C is of type double, then an RGB triplet value of [0 0 0] corresponds to black and [1 1 1] corresponds to white.
If C is an integer type, then the image uses the full range of data to determine the color. For example, if C is of type uint8, then [0 0 0] corresponds to black and [255 255 255] corresponds to white. If CData is of type int8, then [-128 -128 -128] corresponds to black and [127 127 127] corresponds to white.
...
Converting Between Data Types
To convert indexed image data from an integer type to type double, add 1. For example, if X8 is indexed image data of type uint8, convert it to type double using:
X64 = double(X8) + 1;
To convert indexed image data from type double to an integer type, subtract 1 and use round to ensure that all the values are integers. For example, if X64 is indexed image data of type double, convert it to uint8 using:
X8 = uint8(round(X64 - 1));
To convert true color image data from an integer type to type double, rescale the data. For example, if RGB8 is true color image data of type uint8, convert it to double using:
RGB64 = double(RGB8)/255;
To convert true color image data from type double to an integer type, rescale the data and use round to ensure that all the values are integers. For example, if RGB64 is image data of type double, convert it to uint8 using:
RGB8 = uint8(round(RGB64*255));
I = uint8(255*rand(1e3));
Is = single(I)/255;
tmpI = (I/64)*64;
tmpIs = (Is/64)*64;
% plot for prosterity
% figure;
% subplot(211)
% imagesc(tmpI)
% subplot(212)
% imagesc(tmpIs)
numel(unique(tmpI(:))) % gives 5
numel(unique(tmpIs(:))) % gives 256
Dividing an integer basically means binning of values, and then it stretches the data back to the original extend for plotting. In this case, you get 256/64 = 4 bins, with 0 as well, thus 5 possible values for your uint8 image. However, using single you retain all unique numbers, since the precision is a lot higher.
If you'd do the same test with a lot (order 2^52) elements in the rand and use double you'd see that that again has 2^32 times the number of unique elements of single, just as uint16 will have 2^8 the number of unique elements of uint8.
The following code snippet results a double image.
f = imread('C:\Users\Administrator\Desktop\2.tif');
h = double(f);
figure;
imshow(h);
whereas, this other code snippet results a uint8 image.
f = imread('C:\Users\Administrator\Desktop\2.tif');
figure;
imshow(f);
While displaying these two figures, the displayed results of these two images using imshow are different, but what is the reason behind this difference?
Images of type double are assumed to have values between 0 and 1 and uint8 images are assumed to have values between 0 and 255. Since your double data contains values between 0 and 255 (since you simply cast it as a double and don't perform any scaling), it will appear as mostly white since most values are greater than 1.
You can use the second input to imshow to indicate that you would like to ignore this assumption and automatically scale the display to the dynamic range of the data
imshow(h, [])
Or you can normalize the double version using mat2gray prior to displaying the image
h = mat2gray(h);
imshow(h)
My goal is to make a vector of (1,250x250) for learning in a Neural Network. I am converting 250x250 jpg images to grayscale in matlab. The result is <250x250 uint8>. My question is how do I convert the uint8 vector into float32?
To convert a uint8 type array into 32 bit floating point you can simply:
for_learning = single( gray_imag );
for_learning is a single precision 250x250 image with pixel values in range [0..255]. If you want pixel values to range in the interval [0..1], you simply need to divide it
for_learning = single( gray_imag )/ 255.0;
See single and double for more information
I have an image I get it using this code
[path,user_cance]=imgetfile();
im=imread(path);
Now I don't know if this image is RGB or indexed or ...
How to convert im to RGB for example ?
If it is indexed image you can easily use ind2rgb function:
read the image:
[X,map] = imread('imagefile.tif');
Verify that the colormap, map, is not empty, and convert the data in X to RGB.
if ~isempty(map)
Im = ind2rgb(X,map);
end
finally you can View the size and class of X.
whos Im
ind2rgb converts the matrix X and corresponding colormap map to RGB (truecolor) format.
X can be of class uint8, uint16, single, or double. RGB is an m-by-n-by-3 array of class double.
here you can find the image formats that you can read using MATLAB.
According to Mathworks documentation (http://www.mathworks.fr/fr/help/matlab/ref/imread.html), imread can infer the data type from its content.
You could do test if the colormap exists or is empty :
[im, map] = imread(path_to_image);
if(isempty(map)) % image is RGB or grayscale
if(size(im, 3) == 1) % image is grayscale
im = cat(3, img, img, img);
end
else % image is indexed
im = ind2rgb(im, map);
end
% now 'im' is a RGB-image