I am trying to figuring out about correlation between two images that shifted by pixel each of other and measure the correlation between these images. I have images woods and rooster image like this rooster image and woods image
and then I doing some code in matlab like this
Im_Rooster = imread('rooster.jpg'); //read image file
Im_Woods = imread('woods.png');
Im_DRooster = im2double(rgb2gray(Im_Rooster)); //convert to gray image and double data type
Im_DWoods = im2double(Im_Woods);
for i = 0:1:30
Img_Rooster_shift = circshift(Im_DRooster,i,2); // shift image by 1 pixel
Img_Woods_shift = circshift(Im_DRooster,i,2);
Rooster_correlation_val(1,i+1) = corr2(Im_DRooster,Img_Rooster_shift ); // calculate correlation coefficient between original image and shifted image
Woods_correlation_val(1,i+1) = corr2(Im_DWoods,Img_Woods_shift );
end
x = 0:1:30;
figure(1),plot(x,Rooster_correlation_val,x,Woods_correlation_val) // plot the result graph
legend('rooster','woods')
and then I have plotted graph like this plot graph result
Can somebody explain the meaning of this graph result?
What is correlation coefficient connection between natural images?
The result imply that in natural images the probability that neighborhood pixel will share same color information is higher than the probability that it will share same light intensity/value. This is naturally explained by the fact that color in nature are result of light-matter interaction, and matter type is not distributed randomly, but usually we will find a collection of same matter in same place, while shadow depends on surrounding space so it can beahve more randomly.
When you circshift the image you actually compare neighbor pixels, this is auto-correlation but in space. When you ignore the color information than you stay with value/intensity value that can easily have higher frequencies because now we enter to light-shadow realm, that is depend on light angles and matter shape and neighbor shapes and location.
Related
I was working on my image processing problem with detecting coins.
I have some images like this one here:
and wanted to separate the falsely connected coins.
We already tried the watershed method as stated on the MATLAB-Homepage:
the-watershed-transform-strategies-for-image-segmentation.html
especially since the first example is exactly our problem.
But instead we get a somehow very messed up separation as you can see here:
We already extracted the area of the coin using the regionprops Extrema parameter and casting the watershed only on the needed area.
I'd appreciate any help with the problem or even another method of getting it separated.
If you have the Image Processing Toolbox, I can also suggest the Circular Hough Transform through imfindcircles. However, this requires at least version R2012a, so if you don't have it, this won't work.
For the sake of completeness, I'll assume you have it. This is a good method if you want to leave the image untouched. If you don't know what the Hough Transform is, it is a method for finding straight lines in an image. The circular Hough Transform is a special case that aims to find circles in the image.
The added advantage of the circular Hough Transform is that it is able to detect partial circles in an image. This means that those regions in your image that are connected, we can detect them as separate circles. How you'd call imfindcircles is in the following fashion:
[centers,radii] = imfindcircles(A, radiusRange);
A would be your binary image of objects, and radiusRange is a two-element array that specifies the minimum and maximum radii of the circles you want to detect in your image. The outputs are:
centers: A N x 2 array that tells you the (x,y) co-ordinates of each centre of a circle that is detected in the image - x being the column and y being the row.
radii: For each corresponding centre detected, this also gives the radius of each circle detected. This is a N x 1 array.
There are additional parameters to imfindcircles that you may find useful, such as the Sensitivity. A higher sensitivity means that it is able to detect circular shapes that are more non-uniform, such as what you are showing in your image. They aren't perfect circles, but they are round shapes. The default sensitivity is 0.85. I set it to 0.9 to get good results. Also, playing around with your image, I found that the radii ranged from 50 pixels to 150 pixels. Therefore, I did this:
im = im2bw(imread('http://dennlinger.bplaced.net/t06-4.jpg'));
[centers,radii] = imfindcircles(im, [50 150], 'Sensitivity', 0.9);
The first line of code reads in your image directly from StackOverflow. I also convert this to logical or true black and white as the image you uploaded is of type uint8. This image is stored in im. Next, we call imfindcircles in the method that we described.
Now, if we want to visualize the detected circles, simply use imshow to show your image, then use the viscircles to draw the circles in the image.
imshow(im);
viscircles(centers, radii, 'DrawBackgroundCircle', false);
viscircles by default draws the circles with a white background over the contour. I want to disable this because your image has white circles and I don't want to show false contouring. This is what I get with the above code:
Therefore, what you can take away from this is the centers and radii variables. centers will give you the centre of each detected circle while radii will tell you what the radii is for each circle.
Now, if you want to simulate what regionprops is doing, we can iterate through all of the detected circles and physically draw them onto a 2D map where each circle would be labeled by an ID number. As such, we can do something like this:
[X,Y] = meshgrid(1:size(im,2), 1:size(im,1));
IDs = zeros(size(im));
for idx = 1 : numel(radii)
r = radii(idx);
cen = centers(idx,:);
loc = (X - cen(1)).^2 + (Y - cen(2)).^2 <= r^2;
IDs(loc) = idx;
end
We first define a rectangular grid of points using meshgrid and initialize an IDs array of all zeroes that is the same size as the image. Next, for each pair of radii and centres for each circle, we define a circle that is centered at this point that extends out for the given radius. We then use these as locations into the IDs array and set it to a unique ID for that particular circle. The result of IDs will be that which resembles the output of bwlabel. As such, if you want to extract the locations of where the idx circle is, you would do:
cir = IDs == idx;
For demonstration purposes, this is what the IDs array looks like once we scale the IDs such that it fits within a [0-255] range for visibility:
imshow(IDs, []);
Therefore, each shaded circle of a different shade of gray denotes a unique circle that was detected with imfindcircles.
However, the shades of gray are probably a bit ambiguous for certain coins as this blends into the background. Another way that we could visualize this is to apply a different colour map to the IDs array. We can try using the cool colour map, with the total number of colours to be the number of unique circles + 1 for the background. Therefore, we can do something like this:
cmap = cool(numel(radii) + 1);
RGB = ind2rgb(IDs, cmap);
imshow(RGB);
The above code will create a colour map such that each circle gets mapped to a unique colour in the cool colour map. The next line applies a mapping where each ID gets associated with a colour with ind2rgb and we finally show the image.
This is what we get:
Edit: the following solution is more adequate to scenarios where one does not require fitting the exact circumferences, although simple heuristics could be used to approximate the radii of the coins in the original image based on the centers found in the eroded one.
Assuming you have access to the Image Processing toolbox, try imerode on your original black and white image. It will apply an erosion morphological operator to your image. In fact, the Matlab webpage with the documentation of that function has an example strikingly similar to your problem/image and they use a disk structure.
Run the following code (based on the example linked above) assuming the image you submitted is called ima.jpg and is local to the code:
ima=imread('ima.jpg');
se = strel('disk',50);
eroded = imerode(ima,se);
imshow(eroded)
and you will see the image that follows as output. After you do this, you can use bwlabel to label the connected components and compute whatever properties you may want, for example, count the number of coins or detect their centers.
I have 10 gray scale images<2559*3105>. These images are taken from X-ray reflectivity measurement. Each image has two spots except first, showing intensity of X-ray. First image has one highest intensity spot. From second to tenth image each has two spots first one is same as in the first image but second one differs with respect to the location and intensity value. I want to search and crop these spots. The problem is when i apply a condition that find() maximum intensity point in the image, it always points to the spot which is common in all images.
here's some basic image processing code that allows you to select the indices of the spots
%# read the image
im=rgb2gray(imread('a.jpg'));
%# select only relevant area
d=im(5:545,5:660);
%# set a threshold and filter
thres = (max([min(max(d,[],1)) min(max(d,[],2))])) ;
filt=fspecial('gaussian', 7,1);
% reduce noise threshold and smooth the image
d=medfilt2(d);
d=d.*uint8(d>thres);
d=conv2(double(d),filt,'same') ;
d=d.*(d>thres);
% find coonected objets 1
L = bwlabel(d);
%# or also
CC = bwconncomp(d);
Both L and CC have information about the indices of the 2 blobs, so you can now select only that part of the image using them
I am looking for some measurements to do to distinct between these two binary images (texte and noise).
Hough transform of the frequency domain don't tell me much (either in skeleton or in the original shape), as can be seen below !
in the spatial domain, I have try to measure, if a given pixel participate to line or curve, or participate to a random shape, and then measures the percentage of all pixels participating and not participating to normal shape (lines and curves) to distinguish between these images, but I didn't succeed, in implementation.
what do you think ?
I use matlab for test.
Thanks in advance
Looking at the skeleton images, one could notice how the noise image has lots of branches in it, as compared to the text image and this looks like one of the features that could be exploited. The experiment as code shown below soughts to verify the same, using the OP's images -
Experiment Code
%%// Experiment to research what features what might help us
%%// differentiate betwen noise and text images
%%// Read in the given images
img1 = imread('noise.png');
img2 = imread('text.png');
%%// Since the given images had the features as black and rest as white,
%%// we must invert them
img1 = ~im2bw(img1);
img2 = ~im2bw(img2);
%%// Remove the smaller blobs from both of the images which basically
%%// denote the actual noise in them
img1 = rmnoise(img1,60);
img2 = rmnoise(img2,60);
%// Get the skeleton images
img1 = bwmorph(img1,'skel',Inf);
img2 = bwmorph(img2,'skel',Inf);
%%// Find blobs branhpoints for each blob in both images
[L1, num1] = bwlabel(img1);
[L2, num2] = bwlabel(img2);
for k = 1:num1
img1_bpts_count(k) = nnz(bwmorph(L1==k,'branchpoints'));
end
for k = 1:num2
img2_bpts_count(k) = nnz(bwmorph(L2==k,'branchpoints'));
end
%%// Get the standard deviation of branch points count
img1_branchpts_std = std(img1_bpts_count)
img2_branchpts_std = std(img2_bpts_count)
Note: Above code uses a function - rmnoise shown below that is built based on the problem discussed at this link :
function NewImg = rmnoise(Img,threshold)
[L,num] = bwlabel( Img );
counts = sum(bsxfun(#eq,L(:),1:num));
B1 = bsxfun(#eq,L,permute(find(counts>threshold),[1 3 2]));
NewImg = sum(B1,3)>0;
return;
Output
img1_branchpts_std =
73.6230
img2_branchpts_std =
12.8417
One can see the big difference between the standard deviations of the two input images, suggesting this feature could be used.
Runs on some other samples
To make our theory a bit more concrete, let's use a pure text image and gradually add noise and see if the standard deviation of branch-points, naming it as check_value suggest anything on them.
(I) Pure text image
check_value = 1.7461
(II) Some added noise image
check_value = 30.1453
(III) Some more added noise image
check_value = 54.6446
Conclusion: As can be seen, this parameter provides quite a good indicator to decide on the nature of images.
Finalized Code
A script could be written to test for whether another input image would be a text or noise one, like this -
%%// Parameters
%%// 1. Decide this based on the typical image size and count of pixels
%%// in the biggest noise blob
rmnoise_threshold = 60;
%%// 2. Decide this based on the typical image size and how convoluted the
%%// noisy images are
branchpts_count_threshold = 50;
%%// Actual processing
%%// We are assuming input images as binary images with features as true
%%// and false in rest of the region
img1 = im2bw(imread(FILE));
img1 = rmnoise(img1,rmnoise_threshold);
img1 = bwmorph(img1,'skel',Inf);
[L1, num1] = bwlabel(img1);
for k = 1:num1
img1_bpts_count(k) = nnz(bwmorph(L1==k,'branchpoints'));
end
if std(img1_bpts_count) > branchpts_count_threshold
disp('This is a noise image');
else
disp('This is a text image');
end
And now what you suggest if we try to use the original shape instead of the skeleton, (to avoid the loss of information).
I try to measure for a given pixel, the elongation of the strokes (instead of straight branches) that past throughout that pixel, by counting the number of transitions from white to black in a clockwise.
I am thinking to use a circle with a radius, and for the origin the pixel in consideration, and store the pixels locating at the edge of the circle in an ordered list (clockwise) and then compute the number of transitions (black to white) from this list.
by increasing the radius of the circle we could trace the shape of elongated stokes and know his orientation.
this is a schema illustrating this.
the pixels that have a number of transitions equal to 0 or bigger than 2 (red ones) have to be classified as noise, and those that have 2 or 1 transition classified as normal.
What do you think of this approach !
I have two images – mannequin with and without garment.
Please refer sample images below. Ignore the jewels, footwear on the mannequin, imagine the second mannequin has only dress.
I want to extract only the garment from the two images for further processing.
The complexity is that there is slight displacement in the position of camera when taking the two pictures. Due to this simple subtraction to generate the garment mask will not work.
Can anyone tell me how to handle it?
I think I need to do registration between the two images so that I can extract only the garment from the image?
Any references to blogs, articles and codes is highly appreciated.
--
Thanks
Idea
This is an idea of how you could do it, I haven't tested it but my gut tells me it might work. I'm assuming that there will be slight differences in the pose of the manequin as well as the camera attitude.
Let the original image be A, and the clothed image be B.
Take the difference D = |A - B|, apply a median filter that is proportional to the largest deviation you expect from pose and camera attitude error: Dmedian = Median(D, kernelsize).
Quantize Dmedian into a binary mask Dmask = Q(Dmedian, threshold) using appropriate threshold values to obtain an approximate mask for the garment (this will be smaller than the garment itself due to the median filter). Reject any shapes in Dmedian that have too small area by setting their pixels to 0.
Expand the shape(s) in Dmask proportionally to the size of the median kernel into Emask=expand(Dmask, k*kernelsize). Then construct the difference in the masks Fmask=|Dmask - Emask| which now contains areas of pixels where the garment edge is expected to be. For every pixel in Fmask which is in this area, find the correlation Cxy between A and B using a small neighbourhood, store the correlations into an image C=1.0 - Corr(A,B, Fmask, n).
Your final garment mask will be M=C+Dmask.
Explanation
Since your image has nice and continuous swatches of colour, the difference between the two similar images will be thin lines and small gradients where the pose and camera attitude is different. When taking a median filter of the difference image over a sufficiently large kernel, these lines will be removed because they are in a minority of the pixels.
The garment on the other hand will (hopefully) have a significant difference from the colors in the unclothed version. And will generate a bigger difference. Thresholding the difference after the median filter should give you a rough mask of the garment that is undersized dues to some of the pixels on the edge being rejected due to their median values being too low. You could stop here if the approximation is good enough for you.
By expanding the mask we obtained above we get a probable region for the "true" edge. The above process has served to narrow our search region for the true edge considerably and we can apply a more costly correlation search between the images along this edge to find where the garment is. High correlation means no carment and low correlation means garment.
We use the inverted correlation as an alpha value together with the initially smaller mask to obtain a alpha valued mask of the garment that can be used for extracting it.
Clarification
Expand: What I mean by "expanding the mask" is to find the contour of the mask region and outsetting/growing/enlarging it to make it larger.
Corr(A,B,Fmask,n): Is just an arbitrarily chosen correlation function that gives correlation between pixels in A and B that are selected by the mask Fmask using a region of size n. The function returns 1.0 for perfect match and 0.0 for anti-match for each pixel tested. A good function is this pseudocode:
foreach px_pos in Fmask where Fmask[px_pos] == 1
Ap = subregion(A, px_pos, size) - mean(mean(A));
Bp = subregion(B, px_pos, size) - mean(mean(B))
Cxy = sum(sum(Ap .* Bp))*sum(sum(Ap .* Bp)) / (sum(sum(Ap.*Ap))*sum(sum(Bp.*Bp)))
C[px_pos] = 1.0 - Cxy;
end
where subregion selects a region of size size around the pixel with position px_pos.
You can see that if Ap == Bp then Cxy=1
Here with i have attached two consecutive frames captured by a cmos camera with IR Filter.The object checker board was stationary at the time of capturing images.But the difference between two images are nearly 31000 pixels.This could be affect my result.can u tell me What kind of noise is this?How can i remove it.please suggest me any algorithms or any function possible to remove those noises.
Thank you.Sorry for my poor English.
Image1 : [1]: http://i45.tinypic.com/2wptqxl.jpg
Image2: [2]: http://i45.tinypic.com/v8knjn.jpg
That noise appears to result from camera sensor (Bayer to RGB conversion). There's the checkerboard pattern still left.
Also lossy jpg contributes a lot to the process. You should first have an access to raw images.
From those particular images I'd first try to use edge detection filters (Sobel Horizontal and Vertical) to make a mask that selects between some median/local histogram equalization for the flat areas and to apply some checker board reducing filter to the edges. The point is that probably no single filter is able to do good for both jpeg ringing artifacts and to the jagged edges. Then the real question is: what other kind of images should be processed?
From the comments: if corner points are to be made exact, then the solution more likely is to search for features (corner points with subpixel resolution) and make a mapping from one set of points to the other images set of corners, and search for the best affine transformation matrix that converts these sets to each other. With this matrix one can then perform resampling of the other image.
One can fortunately estimate motion vectors with subpixel resolution without brute force searching all possible subpixel locations: when calculating a matched filter, one gets local maximums for potential candidates of exact matches. But this is not all there is. One can try to calculate a more precise approximation of the peak location by studying the matched filter outputs in the nearby pixels. For exact match the output should be symmetric. Otherwise the 'energies' of the matched filter are biased towards the second best location. (A 2nd degree polynomial fit + finding maximum can work.)
Looking closely at these images, I must agree with #Aki Suihkonen.
In my view, the main noise comes from the jpeg compression, that causes sharp edges to "ring". I'd try a "de-speckle" type of filter on the images, and see if this makes a difference. Some info that can help you implement this can be found in this link.
In a more quick and dirty fashion, you apply one of the many standard tools, for example, given the images are a and b:
(i) just smooth the image with a Gaussian filter, this can reduce noise differences between the images by an order of magnitude. For example:
h=fspecial('gaussian',15,2);
a=conv2(a,h,'same');
b=conv2(b,h,'same');
(ii) Reduce Noise By Adaptive Filtering
a = wiener2(a,[5 5]);
b = wiener2(b,[5 5]);
(iii) Adjust ntensity Values Using Histogram Equalization
a = histeq(a);
b = histeq(b);
(iv) Adjust Intensity Values to a Specified Range
a = imadjust(a,[0 0.2],[0.5 1]);
b = imadjust(b,[0 0.2],[0.5 1]);
If your images are supposed to be black and white but you have captured them in gray scale there could be difference due to noise.
You can convert the images to black and white by defining a threshold, any pixel with a value less than that threshold should be assigned 0 and anything larger than that threshold should be assigned 1, or whatever your gray scale range is (maybe 255).
Assume your image is I, to make it black and white assuming your gray scale image level is from 0 to 255, assume you choose a threshold of 100:
ind = find(I < 100);
I(ind) = 0;
ind = find(I >= 100);
I(ind) = 255;
Now you have a black and white image, do the same thing for the other image and you should get very small difference if the camera and the subject have note moved.