I know that I can change the saturation of a RGB image with rgb2hsv, but not for grey-value images. I've already tried the neighbourhood function. Can you give me a hint?
As others have pointed out, the saturation of a gray-scale image is 0 by definition. If you are looking to improve contrast try imadjust or histeq.
As has already been mentioned, "saturation" is not a term that can apply to a grayscale image. Below is a suggestion that might approach the kind of effect you are imagining:
sat = 10;
imshow(img1,rgb2gray(map).^(exp(-0.1*sat)));
This assumes that your image img1 has a corresponding colormap in map. If you have no colormap you can replace rgb2gray(map) with gray(256).
In the example above, positive values of sat will produce brighter images, and negative values of sat will produce darker images. Really you can change the function that warps your colormap in whatever way you choose to get the desired effect.
Related
I have an image and I apply thresholding to it to apply binary mask.I draw histogram before and after the thresholding process.The histograms look like below.
The second figure which is after thresholding,doesn't show any peaks.Is that mean,my thresholding is wrong.Can anyone please explain these histograms.
Update
Image after thresholding
To summarize Sardar's comment, the horizontal range of your plot is tight. Simply loosen the range a bit so you can see the result better. Doing xlim([-0.5 1.5]); will certainly do that and we can see that in the last figure of your update. How you interpret the histogram... well, for black and white images, examining the histogram is never meaningful because there are only two possible intensities to examine - 0 and 1. Histograms usually give a glimpse as to the contrast of the image. If the histogram is spread out, this usually gives an indication that the image has high contrast. However, if the histogram is within a small range this usually means the image is poor contrast.
Remember that the histogram simply count the occurrence of instances in a data set. In this case, we are counting how many times we see 0 and 1 in the image. Referring to your last plot, this means that approximately 9000 pixels that are intensity 0 and approximately 4000 pixels that are intensity 1. This gives absolutely no indication as to the contrast or the spread of the intensities in your image. because there are only two possible intensities that are seen in the image. As such, to answer your question in such a long-winded way, the answer is that you can't really interpret anything.
The only thing I can possible suggest is that it tells you the ratio of object pixels to background pixels and could indicate a measure of quality. Usually when we determine what is an object and what are background pixels, we would expect that there would be more background pixels than object pixels to be able to discern this from the background. Therefore, the more black pixels you have the better it may be. That being said, I can't really say more unless you actually show us what your image looks like after you threshold it.
This question already has an answer here:
What does the index refer to when selecting a pixel on an image in Matlab?
(1 answer)
Closed 6 years ago.
I have gray scale image "lena.bmp". I want read this image in matlab using imread() function.
When i use code below to read and show image my image is dark (black).
img = imread('lena.bmp');
imshow(img);
But when i use code below, I have no problem to view.
[img map]= imread('lena.bmp');
imshow(img,map);
It seems that my first code doses not reading image in grayscale mode (like what rgb2gray function generate).
My image is as follows:
What can i do to solve this problem?
Your image is an "indexed" image. That means it contains integer values which act as "labels" more than anything, and each of those labels is mapped to a colour (i.e. an rgb triplet). Your map variable represents that mapping; at row 5 you have the rgb triplet that corresponds to 'label' "5", for instance.
To see what I mean, do unique(img) and you'll see that the values of your img array are in fact quite regular. The command rgbplot can demonstrate the actual colourmap graphically. Run rgbplot(map) on your map variable to see the mapping for each of the red green and blue colours.
Now, save and read the image below on your computer as img2 and compare the array values.
This image was generated by converting from the "indexed" image you linked to, to a "grayscale" one using photoediting software (the GIMP). The difference is that
in a grayscale image, the pixel values represent actual intensities, rather than integer 'labels'. Imread reads grayscale images as uint8 images by default, meaning it assigns intensity values to pixels ranging from 0 (black) to 255 (white). Since these values happen to be integers you could still cheat and treat them as 'labels' and force a colour-mapping on them. But if you assign a 'linear map' (i.e. value 1 = intensity 1, value 2 = intensity 2, etc) then your image will look as you would expect.
You'll see that the values from unique(img2) are quite different. If you imshow(img2) you'll see this displays as you'd expect. If you don't specify a colormap for imshow, it will assume that the map is a linear mapping from the lowest to the highest value in the image array, which explains why your indexed image looked weird, since its values were never supposed to correspond to intensities.
Also try imagesc(img2) which will show this but using the "current" colormap. imagesc causes the colormap to be "scaled", so that the lowest colour goes to the lowest value in the image, and similarly for the highest.
The default colormap is jet so you should see a psychedelic looking image but you should be able to make out lena clearly. If you try colormap gray you should see the gray version again. Also try colormap hot. Now to make sense of the colormaps, try the rgbplot command on them (e.g. rgbplot(gray), rgbplot(hot) etc).
So, going back to imshow, imshow basically allows you to display an indexed image, and specify what colormap you want to use to display it. If you don't specify the colormap, it will just use a linear interpolation from the lowest value to the highest as your map. Therefore imshow(img) will show the image pretty much in the same way as imagesc(img) with a gray colormap. And since the values in your first img represent evenly spaced 'labels' rather than actual intensities, you'll get a rubbish picture out.
EDIT: If you want to convert your indexed image to a grayscale image, matlab provides the ind2gray function, e.g.:
[img, map] = imread('lena.bmp');
img_gray = ind2gray(img, map);
This is probably what you need if you mean to process pixel values as intensities.
I'm trying to implement a morphological method for image colors from the article: "Probabilistic pseudo-morphology for grayscale and color images". At one point, we compute the PCA on the entire image, calculate a chebyschev inequality ( the equation 11 in the paper: http://perso.telecom-paristech.fr/~bloch/P6Image/Projets/pseudoMorphology/Caliman-PR2014.pdf) of each 3 components which gives us 3 pairs of vector. We next have to represent these vectors back in the RGB space. I don't understand how do we do that? Can someone help me?
Looking at the paper, I'm not sure which representation you're talking about. I'm guessing Fig. 16, but I'm not sure. There's a note in the caption of Fig. 16 that's helpful: "(For interpretation of the references to color in this figure caption, the reader is referred to the web version of this article.)"
Possible answer: if you have a matrix of size A = (y_pixels,x_pixels,3), then you can display this as an RGB image via:
A = rand(100,100,3);
figure()
imshow(A)
Note that your matrix must be scaled in the range [0..1].
It seems easy to map your your PCA scores for each pixel onto such a matrix, and simply display that as RGB via imshow. Does that solve your problem?
I have obtained the following figure using a 120x120 matrix and the surf function.
Is there a simple way to make the lines between different colors look smoother?
First of all, surf may not be the best way to display 2D-image - if you don't actually need the height information, imagesc will work just fine. Even better, it won't show the differently colored lines between hexagons, since it's not going through the colormap at intersections.
However, regardless of your approach, a low-resolution bitmap will not be automatically transformed into a "arbitrary"-resolution vector graphics - and you may not want that, anyway, if you use the figure to allow you to inspect at which combination of (x,y) you obtained at a given value.
There are three approaches to make your image prettier - (1) segment the hexagons, and use patch to create a vector-graphics image. (2) upsample the image with imresample. (3) create a RGB image and smoothen each color separately to get softer transitions:
%# assume img is your image
nColors = length(unique(img));
%# transform the image to rgb
rgb = ind2rgb((img+fliplr(img)),jet(nColors)); %# there are much better colormaps than jet
%# filter each color
for i=1:3,rgbf(:,:,i)=imfilter(rgb(:,:,i),fspecial('gaussian',9,3),'replicate');end
we are doing a mat lab based robotics project.which actually sorts objects based on its color so we need an algorithm to detect specific color from the image captured from a camera using mat lab.
it will be a great help if some one can help me with it.its the video of the project
In response to Amro's answer:
The five squares above all have the same Hue value in HSV space. Selecting by Hue is helpful, but you'll want to impose some constraints on Saturation and value as well.
HSV allows you to describe color in a more human-meaningful way, but you still need to look at all three values.
As a starting point, I would use the rgb space and the euclidian norm to detect if a pixel has a given color. Typically, you have 3 values for a pixel: [red green blue]. You also have also 3 values defining a target color: [255 0 0] for red. Compute the euclidian norm between those two vectors, and apply a decision threshold to classify the color of your pixel.
Eventually, you want to get rid of the luminance factor (i.e is it a bright red or a dark red?). You can switch to HSV space and use the same norm on the H value. Or you can use [red/green blue/green] vectors. Before that, apply a low pass filter to the images because divisions (also present in the hsv2rgb transform) tend to increase noise.
You probably want to convert to the HSV colorspace, and detect colors based on the Hue values. MATLAB offers the RGB2HSV function.
Here is an example submission on File Exchange that illustrate color detection based on hue.
For obtaining a single color mask, first of all convert the rgb image gray using rgb2gray. Also extract the desired color plane from the rgb image ,(eg for obtaining red plain give rgb_img(:,:,1)). Subtract the given plane from the gray image........