I want to compute histogram for a tif image. But the regular histogram code does not accept my input. so please give me a solution

[Edit - format code. AU]

originalImage = imread('3-mar-08.tif'); 
subplot(3, 3, 1);
imshow(originalImage);
>> set(gcf, 'Position', get(0, 'ScreenSize'));
>> drawnow;
>> [pixelCount grayLevels] = imhist(originalImage);
subplot(3, 3, 2); 
bar(pixelCount); title('Histogram of original image');
xlim([0 grayLevels(end)]);
??? Error using ==> iptcheckinput
Function IMHIST expected its first input, I or X, to be
two-dimensional.

Error in ==> imhist>parse_inputs at 275
iptcheckinput(a,
{'double','uint8','logical','uint16','int16','single'}, ...

Error in ==> imhist at 57
[a, n, isScaled, top, map] = parse_inputs(varargin{:});

Answer 1

[Image Analyst]

Try this:

originalImage = imread('3-mar-08.tif'); 
[rows, columns, numberOfColorBands] = size(originalImage);
if numberOfColorBands > 1
  % It's not really gray scale like we expected - it's color.
  % Convert it to gray scale by taking only the green channel.
  grayImage = originalImage(:, :, 2); % Take green channel.
  % Alternatively you can call rgb2gray().  Do one or the other, not both.
  %grayImage = rgb2gray(originalImage);
else
  % It's already gray scale.
  grayImage = originalImage;
end
% Now use grayImage everywhere after this that you were using originalImage.

Or you can try my RGB histogram demo, attached below.

[Jos (10584)]

@Image Analyst, for some images that have no green in it, the first option would not work, is it?

You could also take the mean along the 3rd dimension:

grayImage = mean(originalImage,3)

which will also work for 2D images, so you do not have to check the sizes

[Image Analyst]

Why would it not work? Even if original image was multispectral and had 7 or more bands, it will still have a second band (the one I extract).

[Jos (10584)]

I meant: What if all image information is in the other channels and the second channel has no variation in it? For instance, because the image is filtered or so.

X = imread('board.tif') ; X(:,:,2) = 50 ; % create a test image
subplot(2,2,1) ; image(X) % it is still an image!
subplot(2,2,2) ; imhist(X(:,:,2)) % but no variation in the 2nd channel

[Image Analyst]

Well, yeah. You have to know your image and take the approach that makes sense. Taking one color channel is faster than calling rgb2gray() but it doesn't make sense if there's no information there - you'd be better off calling rgb2gray() even though it's slower because it has to do a weighted average of all the extracted color channels.

Answer 2

[Jos (10584)]

Apparently your array originalImage is not 2D. You can probably see that

ndims(originalImage)

will return 3, suggesting that the image is in RGB format. You might be able to use RGB2IND to convert it.

[X, MAP] = rgb2ind(originalImage)
imhist(X, MAP)

[sandhya chezhian]

[X, MAP] = rgb2ind(originalImage) imhist(X, MAP) ??? Error using ==> rgb2ind>parse_inputs at 131 RGB2IND(RGB) is a deprecated syntax. Specify number of colors, tolerance, or colormap.

Error in ==> rgb2ind at 50 [RGB,m,dith] = parse_inputs(varargin{:});

"im getting this error"..what shud i do??

[Jos (10584)]

a clear case of RTFM ...

doc rgb2ind