I have a visualization output of gabor filter with 12 different orientations.I want to superimpose the vizualization image on my image of retina for vessel extraction.How do i do it?I have tried the below method.is there any other method to perform superimposition of images in matlab.
here is my code
I = getimage();
I=I(:,:,2);
lambda = 8;
theta = 0;
psi = [0 pi/2];
gamma = 0.5;
bw = 1;
N = 2;
img_in = im2double(I);
%img_in(:,:,2:3) = []; % discard redundant channels, it's gray anyway
img_out = zeros(size(img_in,1), size(img_in,2), N);
for n=1:N
gb = gabor_fn(bw,gamma,psi(1),lambda,theta)...
+ 1i * gabor_fn(bw,gamma,psi(2),lambda,theta);
% gb is the n-th gabor filter
img_out(:,:,n) = imfilter(img_in, gb, 'symmetric');
% filter output to the n-th channel
%theta = theta + 2*pi/N
%figure;
%imshow(img_out(:,:,n));
imshow(img_in); hold on;
h = imagesc(img_out(:,:,n)); % here i am getting error saying CDATA must be size[M*N]
set( h, 'AlphaData', .5 ); % .5 transparency
figure;
imshow(h);
theta = 15 * n; % next orientation
end
this is my original image
this is my visualized image got by gabor filter using orientation
this is the kind/type of image i have to get with respect to visualisation .i.e i have to impose visualized image on my original image and i have to get this type of image
With the information you have provided, my understanding is you want the third/final image to be an overlay on top of the first/initial image. I do things like this when using segmentation to detect hemorrhaging in MRI images of the brain.
First, let's set up some defintions:
I_src = source/original image
I_out = output/final image
Now, make a copy of I_src and make it a color image rather than grayscale.
I_hybrid = I_src
colorIm = gray2rgb(I_src)
Let's assume both I_src and I_out are the same visual dimensions (ie: width, height), and that I_out is strictly black-and-white (ie: monochrome). Now, we can use I_out as a mask template for alpha channel adjustments in the resulting image. This is where it gets fun.
BLACK=0;
WHITE=1;
[length width] = size(I_out);
for i = 1:1:length
for j = 1:1:width
if (I_out(i,j) == WHITE)
I_hybrid(i,j) = I_hybrid(i,j) + [0.25 0 0]a;
end
end
This will result in you getting your original image with the blood vessels in the eye being slightly brighter and tinted red. You now have a beautiful composite of your original image with the desired features highlighted, but not overwritten (ie: you can undo the highlighting by subtracting the original color vector).
I will include an example of what the output would look like, but it's noisy because I had to create it in GIMP as I don't have Matlab installed right now. The results will be similar, but yours would be much cleaner and prettier.
Please let me know how this goes.
References
"Converting Images from Grayscale to Color" http://blogs.mathworks.com/pick/2012/11/25/converting-images-from-grayscale-to-color/
Related
I would like to use OpenCV to detect which rectangles in an image have a majority of pixels close to a given color.
Here's an example of an image I would like to process using this to identify rectangular regions that contain mostly gray pixels (possibly roads):
More precisely, given:
dimensions h x w (height and weight of candidate rectangles)
a distance function dist for colors (for example, the norm of the vector difference between the color vector, which could be RGB or any other representation)
a color vector C
a maximum distance d for colors to be from C
a minimum percentage rate r of pixels in a given rectangle to be within distance d from C for the rectangle to be of interest,
return a mask M in which each pixel P is 1 if the rectangle of size h x w left-cornered by P contains at least r % of its pixels within distance d from C when measured with dist.
In pseudo-code, pixel P in the mask is 1 if and only if:
def rectangle_left_cornered_at_P_is_of_interest(P):
n_pixels_near_C = size([P' for P' in rectangle(P, P + (h,w)) if dist(P',C) < d])
return n_pixels_near_C / (h * w) > r
I imagine there may already exist a filter/kernel that does just that (or can be used to do that) in OpenCV, but I am still learning about it and could not identify one by looking at the documentation. Is there such a thing?
You can use HSV for this . you may have to play with the values a bit for the mask but it will get the job done.
img = cv2.imread(img)
hsv = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
lower_gray = np.array([0, 5, 50], np.uint8)
upper_gray = np.array([350, 50, 255], np.uint8)
mask = cv2.inRange(hsv, lower_gray, upper_gray)
img_res = cv2.bitwise_and(img, img, mask = mask)
cv2.imwrite('gray.png',img_res)
You should also refer to this post. Its a good post on the use of HSV.
Basicly all you will need for this job will be :
HSV masks,
Otsu thresholding , blurs and may be erosion and dilation.
Use them in some combition that fits your requirement best.
I have tried to extract the dark line inside very noisy images without success. Some tips?
My current steps for the first example:
1) Clahe: with clip_limit = 10 and grid_size = (8,8)
2) Box Filter: with size = (5,5)
3) Inverted Image: 255 - image
4) Threshold: when inverted_image < 64
UPDATE
I have performed some preprocessing steps to improve the quality of tested images. I adjusted my ROI mask to crop top and down (because they are low intensities) and added a illumination correction to see better the line. Follow below the current images:
Even though the images are noisy, you are only looking for straight lines towards the north of image. So, why don't use some kind of matched filter with morphological operations?
EDIT: I have modified it.
1) Use median filter along the x and y axis, and normalize the images.
2) Matched filter with all possible orientations of lines.
% im=imread('JwXON.png');
% im=imread('Fiy72.png');
% im=imread('Ya9AN.png');
im=imread('OcgaIt8.png');
imOrig=im;
matchesx = fl(im, 1);
matchesy = fl(im, 0);
matches = matchesx + matchesy;
[x, y] = find(matches);
figure(1);
imagesc(imOrig), axis image
hold on, plot(y, x, 'r.', 'MarkerSize',5)
colormap gray
%----------
function matches = fl(im, direc)
if size(im,3)~=1
im = double(rgb2gray(im));
else
im=double(im);
end
[n, m] = size(im);
mask = bwmorph(imfill(im>0,'holes'),'thin',10);
indNaN=find(im==0); im=255-im; im(indNaN)=0;
N = n - numel(find(im(:,ceil(m/2))==0));
N = ceil(N*0.8); % possible line length
% Normalize the image with median filter
if direc
background= medfilt2(im,[1,30],'symmetric');
thetas = 31:149;
else
background= medfilt2(im,[30,1],'symmetric');
thetas = [1:30 150:179];
end
normIm = im - background;
normIm(normIm<0)=0;
% initialize matched filter result
matches=im*0;
% search for different angles of lines
for theta=thetas
normIm2 = imclose(normIm>0,strel('line',5,theta));
normIm3 = imopen(normIm2>0,strel('line',N,theta));
matches = matches + normIm3;
end
% eliminate false alarms
matches = imclose(matches,strel('disk',2));
matches = matches>3 & mask;
matches = bwareaopen(matches,100);
I am looking to analyze the most dominant color in a UIImage on iOS (color present in the most pixels) and I stumbled upon Core Image's filter based API, particularly CIAreaHistogram.
It seems like this filter could probably help me but I am struggling to understand the API. Firstly it says the output of the filter is a one-dimensional image which is the length of your input-bins and one pixel in height. How do I read this data? I basically want to figure out the color-value with the highest frequency so I am expecting the data to contain some kind of frequency count for each color, its not clear to me how this one-dimensional image would represent that because it does not really explain the data I can expect inside this 1-d image. And if its truly a histogram why would it not return a data-structure representing that like a dictionary
Second, in the API it asks for a number of bins? What should that input be? If I want an exact analysis would the input bin parameter be the color-space of my image? What does making the bin value smaller do, I would imagine it just approximates nearby colors via Euclidean distance to the nearest bin. If this is the case will that not yield exact histogram results, why would anyone want to do that?
Any input on the above two questions from an API perspective would help me greatly
Ian Ollmann's idea of calculating the histogram just for the hue is really neat and can be done with a simple color kernel. This kernel returns a monochrome image of just the hue of an image (based on this original work)
let shaderString = "kernel vec4 kernelFunc(__sample c)" +
"{" +
" vec4 K = vec4(0.0, -1.0 / 3.0, 2.0 / 3.0, -1.0);" +
" vec4 p = mix(vec4(c.bg, K.wz), vec4(c.gb, K.xy), step(c.b, c.g));" +
" vec4 q = mix(vec4(p.xyw, c.r), vec4(c.r, p.yzx), step(p.x, c.r));" +
" float d = q.x - min(q.w, q.y);" +
" float e = 1.0e-10;" +
" vec3 hsv = vec3(abs(q.z + (q.w - q.y) / (6.0 * d + e)), d / (q.x + e), q.x);" +
" return vec4(vec3(hsv.r), 1.0);" +
"}"
let colorKernel = CIColorKernel(string: shaderString)
If I get the hue of an image of a blue sky, the resulting histogram looks like this:
...while a warm sunset gives a histogram like this:
So, that looks like a good technique to get the dominant hue of an image.
Simon
CIAreaHistogram returns an image where the reg, green, blue and alpha values of each of the pixels indicates the frequency of that tone in the image. You can render that image to an array of UInt8 to look at the histogram data. There's also an undocumented outputData value:
let filter = CIFilter(
name: "CIAreaHistogram",
withInputParameters: [kCIInputImageKey: image])!
let histogramData = filter.valueForKey("outputData")
However, I've found vImage to be a better framework for working with histograms. First off, you need to create a vImage image format:
var format = vImage_CGImageFormat(
bitsPerComponent: 8,
bitsPerPixel: 32,
colorSpace: nil,
bitmapInfo: CGBitmapInfo(
rawValue: CGImageAlphaInfo.PremultipliedLast.rawValue),
version: 0,
decode: nil,
renderingIntent: .RenderingIntentDefault)
vImage works with image buffers that can be created from CGImage rather than CIImage instances (you can create one with the createCGImage method of CIContext. vImageBuffer_InitWithCGImage will create an image buffer:
var inBuffer: vImage_Buffer = vImage_Buffer()
vImageBuffer_InitWithCGImage(
&inBuffer,
&format,
nil,
imageRef,
UInt32(kvImageNoFlags))
Now to create arrays of Uint which will hold the histogram values for the four channels:
let red = [UInt](count: 256, repeatedValue: 0)
let green = [UInt](count: 256, repeatedValue: 0)
let blue = [UInt](count: 256, repeatedValue: 0)
let alpha = [UInt](count: 256, repeatedValue: 0)
let redPtr = UnsafeMutablePointer<vImagePixelCount>(red)
let greenPtr = UnsafeMutablePointer<vImagePixelCount>(green)
let bluePtr = UnsafeMutablePointer<vImagePixelCount>(blue)
let alphaPtr = UnsafeMutablePointer<vImagePixelCount>(alpha)
let rgba = [redPtr, greenPtr, bluePtr, alphaPtr]
let histogram = UnsafeMutablePointer<UnsafeMutablePointer<vImagePixelCount>>(rgba)
The final step is to perform the calculation, which will populate the four arrays, and free the buffer's data:
vImageHistogramCalculation_ARGB8888(&inBuffer, histogram, UInt32(kvImageNoFlags))
free(inBuffer.data)
A quick check of the alpha array of an opaque image should yield 255 zeros with the final value corresponding to the number of pixels in the image:
print(alpha) // [0, 0, 0, 0, 0 ... 409600]
A histogram won't give you the dominant color from a visual perspective: an image which is half yellow {1,1,0} and half black {0,0,0} will give the same results as an image which is half red {1,0,0} and held green {0,1,0}.
Hope this helps,
Simon
One problem with the histogram approach is that you lose correlation between the color channels. That is, half your image could be magenta and half yellow. You will find a red histogram that is all in the 1.0 bin, but the blue and green bins would be evenly split between 0.0 and 1.0 with nothing in between. Even though you can be quite sure that red is bright, you won't be able to say much about what the blue and green component should be for the "predominant color"
You could use a 3D histogram with 2**(8+8+8) bins, but this is quite large and you will find the signal is quite sparse. By happenstance three pixels might land in one bin and have no two the same elsewhere, even though many users could tell you that there is a predominant color and it has nothing to do with that pixel.
You could make the 3D histogram a lot lower resolution and have (for example) just 16 bins per color channel. It is much more likely that bins will have a statistically meaningful population count this way. This should give you a starting point to find a mean for a local population of pixels in that bin. If each bin had a count and a {R,G,B} sum, then you could quickly find the mean color for pixels in that bin once you had identified the most popular bins. This method is still subject to some influence from the histogram grid. You will be more likely to identify colors in the middle of a grid cell than at the edges. Populations may span multiple grid cells. Something like kmeans might be another method.
If you just want predominant hue, then conversion to a color space like HSV followed by a histogram of hue would work.
I'm not aware of any filters in vImage, CI or MetalPerformanceShaders to do these things for you. You can certainly write code in either the CPU or Metal to do it without a lot of trouble.
I have set of images with four possible color R, G, B and Y. In front of camera I have sequence of four images with any possible combination of color. For ex. R,R,G,B or R,G,B,Y etc. In order to find the correct sequence which algorithm or approach is best?
I have added an example image.
The code should return correct sequence as RGBG.
As I mentioned before, convert the image to HSV plane. HSV plane is more better to choose a specific color. (Code is in Python)
import numpy as np
import cv2
img = cv2.imread('sofqn.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
hsv = cv2.cvtColor(gray,cv2.COLOR_BGR2HSV)
Then binarize the image. You can use threshold() function. But I don't know how your brightness would be. So I for edge detection with Canny(). And find contours in it.
edges = cv2.Canny(img,50,150)
contours,hierarchy = cv2.findContours(edges,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
Then for each contour (you can consider contour as an object in your image for now), find its area. If it is small, it is noise, avoid it. Else, we find a bounding rectangle for it, which gives us its topleft corners(x,y), width(w),height(h). From that, we find center point of the square. Check its color in HSV image, and check if it is R,G,B,Y. We put them, ie centroids (cx,cy) and color in a list, (or array). Finally we sort them as per x coordinate, so that first row corresponds to first square and so on.
res = []
for cnt in contours:
if cv2.contourArea(cnt) > 100:
x,y,w,h = cv2.boundingRect(cnt)
cx,cy = x+w/2, y+h/2
color = hsv[cy,cx,0]
if (color < 10 or color > 170):
res.append([cx,cy,'R'])
elif(50 < color < 70):
res.append([cx,cy,'G'])
elif(20 < color <40):
res.append([cx,cy,'Y'])
elif(110 < color < 130):
res.append([cx,cy,'B'])
res = sorted(res,key = lambda res : res[0])
colors = [x[2] for x in res]
print colors
This gives me the result : ['R', 'G', 'B', 'G']
This is a formula for LoG filtering:
(source: ed.ac.uk)
Also in applications with LoG filtering I see that function is called with only one parameter:
sigma(σ).
I want to try LoG filtering using that formula (previous attempt was by gaussian filter and then laplacian filter with some filter-window size )
But looking at that formula I can't understand how the size of filter is connected with this formula, does it mean that the filter size is fixed?
Can you explain how to use it?
As you've probably figured out by now from the other answers and links, LoG filter detects edges and lines in the image. What is still missing is an explanation of what σ is.
σ is the scale of the filter. Is a one-pixel-wide line a line or noise? Is a line 6 pixels wide a line or an object with two distinct parallel edges? Is a gradient that changes from black to white across 6 or 8 pixels an edge or just a gradient? It's something you have to decide, and the value of σ reflects your decision — the larger σ is the wider are the lines, the smoother the edges, and more noise is ignored.
Do not get confused between the scale of the filter (σ) and the size of the discrete approximation (usually called stencil). In Paul's link σ=1.4 and the stencil size is 9. While it is usually reasonable to use stencil size of 4σ to 6σ, these two quantities are quite independent. A larger stencil provides better approximation of the filter, but in most cases you don't need a very good approximation.
This was something that confused me too, and it wasn't until I had to do the same as you for a uni project that I understood what you were supposed to do with the formula!
You can use this formula to generate a discrete LoG filter. If you write a bit of code to implement that formula, you can then to generate a filter for use in image convolution. To generate, say a 5x5 template, simply call the code with x and y ranging from -2 to +2.
This will generate the values to use in a LoG template. If you graph the values this produces you should see the "mexican hat" shape typical of this filter, like so:
(source: ed.ac.uk)
You can fine tune the template by changing how wide it is (the size) and the sigma value (how broad the peak is). The wider and broader the template the less affected by noise the result will be because it will operate over a wider area.
Once you have the filter, you can apply it to the image by convolving the template with the image. If you've not done this before, check out these few tutorials.
java applet tutorials more mathsy.
Essentially, at each pixel location, you "place" your convolution template, centred at that pixel. You then multiply the surrounding pixel values by the corresponding "pixel" in the template and add up the result. This is then the new pixel value at that location (typically you also have to normalise (scale) the output to bring it back into the correct value range).
The code below gives a rough idea of how you might implement this. Please forgive any mistakes / typos etc. as it hasn't been tested.
I hope this helps.
private float LoG(float x, float y, float sigma)
{
// implement formula here
return (1 / (Math.PI * sigma*sigma*sigma*sigma)) * //etc etc - also, can't remember the code for "to the power of" off hand
}
private void GenerateTemplate(int templateSize, float sigma)
{
// Make sure it's an odd number for convenience
if(templateSize % 2 == 1)
{
// Create the data array
float[][] template = new float[templateSize][templatesize];
// Work out the "min and max" values. Log is centered around 0, 0
// so, for a size 5 template (say) we want to get the values from
// -2 to +2, ie: -2, -1, 0, +1, +2 and feed those into the formula.
int min = Math.Ceil(-templateSize / 2) - 1;
int max = Math.Floor(templateSize / 2) + 1;
// We also need a count to index into the data array...
int xCount = 0;
int yCount = 0;
for(int x = min; x <= max; ++x)
{
for(int y = min; y <= max; ++y)
{
// Get the LoG value for this (x,y) pair
template[xCount][yCount] = LoG(x, y, sigma);
++yCount;
}
++xCount;
}
}
}
Just for visualization purposes, here is a simple Matlab 3D colored plot of the Laplacian of Gaussian (Mexican Hat) wavelet. You can change the sigma(σ) parameter and see its effect on the shape of the graph:
sigmaSq = 0.5 % Square of σ parameter
[x y] = meshgrid(linspace(-3,3), linspace(-3,3));
z = (-1/(pi*(sigmaSq^2))) .* (1-((x.^2+y.^2)/(2*sigmaSq))) .*exp(-(x.^2+y.^2)/(2*sigmaSq));
surf(x,y,z)
You could also compare the effects of the sigma parameter on the Mexican Hat doing the following:
t = -5:0.01:5;
sigma = 0.5;
mexhat05 = exp(-t.*t/(2*sigma*sigma)) * 2 .*(t.*t/(sigma*sigma) - 1) / (pi^(1/4)*sqrt(3*sigma));
sigma = 1;
mexhat1 = exp(-t.*t/(2*sigma*sigma)) * 2 .*(t.*t/(sigma*sigma) - 1) / (pi^(1/4)*sqrt(3*sigma));
sigma = 2;
mexhat2 = exp(-t.*t/(2*sigma*sigma)) * 2 .*(t.*t/(sigma*sigma) - 1) / (pi^(1/4)*sqrt(3*sigma));
plot(t, mexhat05, 'r', ...
t, mexhat1, 'b', ...
t, mexhat2, 'g');
Or simply use the Wavelet toolbox provided by Matlab as follows:
lb = -5; ub = 5; n = 1000;
[psi,x] = mexihat(lb,ub,n);
plot(x,psi), title('Mexican hat wavelet')
I found this useful when implementing this for edge detection in computer vision. Although not the exact answer, hope this helps.
It appears to be a continuous circular filter whose radius is sqrt(2) * sigma. If you want to implement this for image processing you'll need to approximate it.
There's an example for sigma = 1.4 here: http://homepages.inf.ed.ac.uk/rbf/HIPR2/log.htm