Image with accordion effect - image

I have read in an image file to MATLAB and I am trying to stretch it in one direction, but a variable amount (sinusoidal). This would create an accordion affect on the image. I have toyed around with imresize, however that only resizes the image linearly. I would like the amount of "stretch" to vary for each image line. I tried to convey this with the following code:
periods = 10; % Number of "stretch" cycles
sz = size(original_image,2)/periods;
s = 0;
x = 0;
for index = 1:periods
B = original_image(:,round(s+1:s+sz));
if mod(index,2) == 0
amp = 1.5;
else
amp = 0.75;
end
xi = size(B,2)*amp;
new_image(:,x+1:x+xi) = imresize(B, [size(B,1) size(B,2)*amp]);
s = s + sz;
x = x+xi;
end
You can see that segments of the image are stretched, then compressed, then stretched, etc, like an accordion. However, each segment has a uniform amount of stretch, whereas I'd like it to be increasing then decreasing as you move along the image.
I have also looked at MATLAB's example of Applying a Sinusoidal Transformation to a Checkerboard which seems very applicable to my problem, however I have been trying and I cannot get this to produce the desired result for my image.
Any help is much appreciated.
UPDATE:
Thank you for Answer #1. I was unable to get it to work for me, but also realized it would resulted in loss of data, as the code only called for certian lines in the original image, and other lines would have been ignored.
After experimenting further, I developed the code below. I used a checkerboard as an example. While combersome, it does get the job done. However, upon trying the script with an actual high-resolution image, it was extremely slow and ended up failing due to running out of memory. I believe this is because of the excessive number of "imresize" commands that are used in loop.
I = checkerboard(10,50);
I = imrotate(I,90);
[X Y] = size(I);
k = 4; % Number of "cycles"
k = k*2;
x = 1;
y = 2;
z = 2;
F = [];
i = 1;
t = 0;
s = 0;
for j = 1:k/2
t = t + 1;
for inc = round(s+1):round(Y/k*t)
Yi = i + 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z + 1;
i = i + 1;
end
y = y - 2;
z = z - 4;
for inc = round(Y/k*t+1):round(Y/k*(t+1))
Yi = i - 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z - 1;
i = i - 1;
end
y = y + 2;
z = z + 4;
s = Y/k*(t+1);
t = t + 1;
end
Fn = imresize(F, [X Y]);
imshow(Fn);
Does anyone know of a simpler way to achieve this? If you run the code above, you can see the effect I am trying to achieve. Unfortunately, my method above does not allow me to adjust the amplitude of the "stretch" either, only the number of "cycles," or frequency. Help on this would also be appreciated. Much thanks!

Here is how I would approach it:
Determine how the coordinate of each point in your Final image F maps into your Initial image I of size (M,N)
Since you want to stretch horizontally only, given a point (xF,yF) in your final image, that point would be (xI,yI) in your initial image where xI and yI can be obtained as follows:
yI = yF;
xI = xF + Lsin(xFK);
Notes:
these equations do not guarantee that xI remains within the range [1:N] so cropping needs to be added
K controls the how many wrinkles you want to have in your accordion effect. For example, if you only want one wrinkle, K would be 2*pi/N
L controls how much stretching you want to apply
Then simply express your image F from image I with the transforms you have in 1.
Putting it all together, the code below creates a sample image I and generates the image F as follows:
% Generate a sample input image
N=500;
xF=1:N;
I=(1:4)'*xF/N*50;
% Set the parameters for your accordion transform
K=2*pi/N;
L=100;
% Apply the transform
F=I(:, round(min(N*ones(1,N), max(ones(1,N), (xF + L*sin(xF*K))))) );
% Display the input and output images side by side
image(I);
figure;
image(F);
If you run this exact code you get:
As you can see, the final image on the right stretches the center part of the image on the left, giving you an accordion effect with one wrinkle.
You can fiddle with K and L and adjust the formula to get the exact effect you want, but note how by expressing the transform in a matrix form MATLAB executes the code in a fraction of second. If there is one take away for you is that you should stay away from for loops and complex processing whenever you can.
Have fun!

Related

Interp2 of image with transformed coordinates

I have 2 greyscale images that i am trying to align using scalar scaling 1 , rotation matrix [2,2] and translation vector [2,1]. I can calculate image1's transformed coordinates as
y = s*R*x + t;
Below the resulting images are shown.
The first image is image1 before transformation,
the second image is image1 (red) with attempted interpolation using interp2 shown on top of image2 (green)
The third image is when i manually insert the pixel values from image1 into an empty array (that has the same size as image2) using the transformed coordinates.
From this we can see that the coordinate transformation must have been successful, as the images are aligned although not perfectly (which is to be expected since only 2 coordinates were used in calculating s, R and t) .
How come interp2 is not producing a result more similar to when i manually insert pixel values?
Below the code for doing this is included:
Interpolation code
function [transformed_image] = interpolate_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
% doesn't help if i use get_grid that the other function is using here
[~, grid_xr, grid_yr] = get_ipgrid(im_r);
[x_t, grid_xt, grid_yt] = get_ipgrid(im_t);
y = s*R*x_t + t;
yx = reshape(y(1,:), m,n);
yy = reshape(y(2,:), m,n);
transformed_image = interp2(grid_xr, grid_yr, im_r, yx, yy, 'nearest');
end
function [x, grid_x, grid_y] = get_ipgrid(image)
[m,n] = size(image);
[grid_x,grid_y] = meshgrid(1:n,1:m);
x = [reshape(grid_x, 1, []); reshape(grid_y, 1, [])]; % X is [2xM*N] coordinate pairs
end
The manual code
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(round(yx(i,j)),round(yy(i,j))) = im_r(i,j);
end
end
end
end
function [x, grid_x, grid_y] = get_grid(image)
[m,n] = size(image);
[grid_y,grid_x] = meshgrid(1:n,1:m);
x = [grid_x(:) grid_y(:)]'; % X is [2xM*N] coordinate pairs
end
Can anyone see what i'm doing wrong with interp2? I feel like i have tried everything
Turns out i got interpolation all wrong.
In my question i calculate the coordinates of im1 in im2.
However the way interpolation works is that i need to calculate the coordinates of im2 in im1 such that i can map the image as shown below.
This means that i also calculated the wrong s,R and t since they were used to transform im1 -> im2, where as i needed im2 -> im1. (this is also called the inverse transform). Below is the manual code, that is basically the same as interp2 with nearest neighbour interpolation
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(i,j) = im_r(round(yx(i,j)),round(yy(i,j)));
end
end
end
end

How to blur an image in one specific direction in Matlab?

I have an image and I would like to blur it in one specific direction and distance using Matlab.
I found out there is a filter called fspecial('motion',len,theta).
Here there is an example:
I = imread('cameraman.tif');
imshow(I);
H = fspecial('motion',20,45);
MotionBlur = imfilter(I,H,'replicate');
imshow(MotionBlur);
However the blurred picture is blurred in 2 directions! In this case 225 and 45 degrees.
What should it do in order to blur it just in a specific direction (e.g. 45) and not both?
I think you want what's called a "comet" kernel. I'm not sure what kernel is used for the "motion" blur, but I'd guess that it's symmetrical based on the image you provided.
Here is some code to play with that applies the comet kernel in one direction. You'll have to change things around if you want an arbitrary angle. You can see from the output that it's smearing in one direction, since there is a black band on only one side (due to the lack of pixels there).
L = 5; % kernel width
sigma=0.2; % kernel smoothness
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
H1 = exp((-sigma.*X.^2)+(-sigma.*Y.^2));
kernel = H1/sum((H1(:)));
Hflag = double((X>0));
comet_kernel = Hflag.*H1;
comet_kernel=comet_kernel/sum(comet_kernel(:));
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Updated code: This will apply an arbitrary rotation to the comet kernel. Note also the difference between sigma in the previous example and sx and sy here, which control the length and width parameters of the kernel, as suggested by Andras in the comments.
L = 5; % kernel width
sx=3;
sy=10;
theta=0;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)))
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Based on Anger Density's answer I wrote this code that solves my problem completely:
L = 10; % kernel width
sx=0.1;
sy=100;
THETA = ([0,45,90,135,180,225,270,320,360])*pi/180;
for i=1:length(THETA)
theta=(THETA(i)+pi)*-1;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)));
smearedImage = conv2(double(I),comet_kernel,'same');
% Fix edges
smearedImage(:,[1:L, end-L:end]) = I(:,[1:L, end-L:end]); % Left/Right edge
smearedImage([1:L, end-L:end], :) = I([1:L, end-L:end], :); % Top/bottom edge
% Keep only inner blur
smearedImage(L:end-L,L:end-L) = min(smearedImage(L:end-L,L:end-L),double(I(L:end-L,L:end-L)));
figure
imshow(smearedImage,[]);
title(num2str(THETA(i)*180/pi))
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
end

Check for pixel values in a neighborhood

I'm trying to write a MATLAB script that does the following:
Given: pixel coordinates(x,y) for a .jpg image
Goal: Check, within a 5 pixel radius of given coordinates, if there is a pixel of a certain value.
For example, let's say I'm given the coordinates (100,100), then I want to check the neighborhood of (100,100) within my image for any pixels that are black (0,0,0). So perhaps, pixel (103, 100) and (104,100) might have the value (0,0,0).
Current code:
x_coord = uint32(coord(:,1));
y_coord = uint32(coord(:,2));
count = 0;
for i = 1:length(x_coord)
%(img(x,y) returns pixel value at that (x,y)
%Note 0 = black. Indicating that, at that position, the image is just
% black
if img(x_coord(i),y_coord(i)) == 0
count = count + 1;
end
end
It currently only checks at an exact location. Not in a local neighborhood. How to could I extend this?
EDIT: Also note, as long as there as at least one pixel in the neighborhood with the value, I increment count. I'm not trying to enumerate how many pixels in the neighborhood have that value, just trying to find evidence of at least one pixel that has that value.
EDIT:
Even though I am unable to identify an error with the code, I am not able to get the exact results I want. Here is the code I am using.
val = 0; %pixel value to check
N = 50; % neighbourhood radius
%2D grid of coordinates surrounding center coordinate
[R, C] = ndgrid(1 : size(img, 1), 1 : size(img, 2));
for kk = 1 : size(coord, 1)
r = coord(kk, 1); c = coord(kk, 2); % Get pixel locations
% mask of valid locations within the neighbourhood (avoid boundary problems)
mask = (R - r).^2 + (C - c).^2 <= N*N;
pix = img(mask); % Get the valid pixels
valid = any(pix(:) ~= val);
% Add either 0 or 1 depending if we have found any matching pixels
if(valid == 1)
img = insertMarker(img, [r c], 'x', 'color', 'red', 'size', 10);
imwrite(img, images(i).name,'tiff');
end
count = count + valid;
end
An easier way to do this would be to use indexing to grab a neighbourhood, then to check to see if any of the pixels in the neighbourhood have the value that you're looking for, use any on a flattened version of this neighbourhood. The trick with grabbing the right neighbourhood is to first generate a 2D grid of coordinates that span the entire dimensions of your image, then simply use the equation of a circle with the centre of it being each coordinate you are looking at and determine those locations that satisfy the following equation:
(x - a)^2 + (y - b)^2 <= N^2
N is the radius of the observation window, (a, b) is a coordinate of interest while (x, y) is a coordinate in the image. Use meshgrid to generate the coordinates.
You would use the above equation to create a logical mask, index into your image to pull the locations that are valid within the mask and check how many pixels match the one you want. Another added benefit with the above approach is that you are not subject to any out of bounds errors. Because you are pre-generating the list of all valid coordinates in your image, generating the mask will confine you within the boundaries of the image so you never have to check for out of boundaries conditions.... even when you specify coordinates to search that are out of bounds.
Specifically, assuming your image is stored in img, you would do:
count = 0; % Remembers total count of pixels matching a value
val = 0; % Value to match
N = 50; % Radius of neighbourhood
% Generate 2D grid of coordinates
[x, y] = meshgrid(1 : size(img, 2), 1 : size(img, 1));
% For each coordinate to check...
for kk = 1 : size(coord, 1)
a = coord(kk, 1); b = coord(kk, 2); % Get the pixel locations
mask = (x - a).^2 + (y - b).^2 <= N*N; % Get a mask of valid locations
% within the neighbourhood
pix = img(mask); % Get the valid pixels
count = count + any(pix(:) == val); % Add either 0 or 1 depending if
% we have found any matching pixels
end
The proposed solution:
fc = repmat(-5:5,11,1);
I = (fc.^2+fc'.^2)<=25;
fc_x = fc(I);
fc_y = fc'; fc_y = fc_y(I);
for i = 1:length(x_coord)
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
count = sum(img(x_toCheck(:),y_toCheck(:)) == 0);
end
If your img function can only check one pixel at a time, just add a for loop:
for i = 1:length(x_coord)
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
for j = 1:length(x_toCheck)
count = count + (img(x_toCheck(j),y_toCheck(j)) == 0);
end
end
Step-by-step:
You first need to get all the coordinates within 5 pixels range of the given coordinate.
We start by building a square of 11 pixels in length/width.
fc = repmat(-5:5,11,1);
fc_x = fc;
fc_y = fc';
plot(fc_x,fc_y,'.');
We now need to build a filter to get rid of those points outside the 5-pixel radius.
I = (fc.^2+fc'.^2)<=25;
Apply the filter, so we can get a circle of 5-pixel radius.
fc_x = fc_x(I);
fc_y = fc_y(I);
Next translate the centre of the circle to the given coordinate:
x_toCheck = fc_x + x_coord(i);
y_toCheck = fc_y + y_coord(i);
You need to check whether part of the circle is outside the range of your image:
I = x_toCheck>0 & x_toCheck<=yourImageWidth;
I = I.*(y_toCheck>0 & y_toCheck<=yourImageHeight);
x_toCheck = x_toCheck(logical(I));
y_toCheck = y_toCheck(logical(I));
Finally count the pixels:
count = sum(img(x_toCheck,y_toCheck) == 0);

Skipping some axis labels in a plot with imagesc

I have created a big heat map using matlab's imagesc command. It plots the error output for each combination of the values in x and y axes. As can be seen in the figure there are too many axes labels. This might become even denser as I plan to increase the number of points in both x and y axes - which means I will get more outputs on a finer grid.
I want to be flexible with the labels, and skip some of them. I want to do this for both X and Y. I also want to be flexible with the "ticks" and draw either all of them or maybe skip some of them. Keep in mind that both the X and Y values are not increasing in order, at first the increment is 0.01 for 9 points, then 0.1, then 1 or 3 or whatever. I will change these increments too.
I tried to show what I want the graph look like in the second image. I want roughly the labels shown in red boxes only. As I said these are not set values, and I will make the increments smaller which will lead to denser plot.
Thank you for your help.
OS: Windows 7, 8 (64 bit)
Matlab version: Matlab 2014 a
You can manipulate the ticks and labels like this:
ticksarray=[1 33 41 100 ...] % edit these to whatever you want
tickslabels={'1', '33', '41', '100'; ...} % match the size of both arrays
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
The same thing applies to the y-axis.
Small working example:
x=1:100;
y=2*x.^2-3*x+2;
plot(x,y)
ticksarray=[1 33 41 100];
tickslabels={'1', '33', '41', '100'};
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
Example:
figure(1)
load clown
subplot(211)
imagesc(X);
subplot(212)
imagesc(X);
h = gca;
Now you can either set a maximum number of labels per axis:
%// define maximum number of labels
maxLabel = 3;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxLabel);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxLabel);
or define how many labels should be skipped:
%// define number of labels to skip
skipLabel = 2;
h.XTick = h.XTick(1:skipLabel:end);
h.YTick = h.YTick(1:skipLabel:end)
You can also get a different number of ticks and labels, more complicated though:
maxLabel = 3;
maxTicks = 6;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxTicks);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxTicks);
h.XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
h.YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
If you use a prior version of Matlab 2014b, then you will need the set command to set all properties:
%// define maximum number of labels
maxLabel = 3;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxLabel));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxLabel));
%// or define number of labels to skip
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
%// or combined
maxLabel = 3;
maxTicks = 6;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxTicks));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxTicks));
XTickLabel = cellstr(get(h,'XTickLabel'));
YTickLabel = cellstr(get(h,'YTickLabel'));
XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
set(h,'XTickLabel',XTickLabel);
set(h,'YTickLabel',YTickLabel);
After applying the second method proposed by #thewaywewalk I got the second figure below. Apparently the labels need to be structured as well, because they only take the first so many labels.
Then I tried to manipulate the labels as shown below, and got the third image.
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
XTickLabel = get(h,'XTickLabel');
labelsX = cell( length(1: skipLabel:length(XTick)) , 1);
j = 1;
for i = 1: skipLabel:length(XTick)
labelsX{j} = XTickLabel(i, :);
j = j + 1;
end
set(h,'XTickLabel', labelsX);
YTickLabel = get(h,'YTickLabel');
labelsY = cell( length(1: skipLabel:length(YTick)) , 1);
j = 1;
for i = 1: skipLabel:length(YTick)
labelsY{j} = YTickLabel(i, :);
j = j + 1;
end
set(h,'YTickLabel', labelsY);
The Y axis labels seem to be in place as before (right next to tick), however the X axis labels seem to be shifted to the left a little. How can I correct this?
Another note: How can I change the scientific values into normal numbers? Also, probably there is a better approach at manipulating the labels.

matlab: efficient computation of local histograms within circular neighboorhoods

I've an image over which I would like to compute a local histogram within a circular neighborhood. The size of the neighborhood is given by a radius. Although the code below does the job, it's computationally expensive. I run the profiler and the way I'm accessing to the pixels within the circular neighborhoods is already expensive.
Is there any sort of improvement/optimization based maybe on vectorization? Or for instance, storing the neighborhoods as columns?
I found a similar question in this post and the proposed solution is quite in the spirit of the code below, however the solution is still not appropriate to my case. Any ideas are really welcomed :-) Imagine for the moment, the image is binary, but the method should also ideally work with gray-level images :-)
[rows,cols] = size(img);
hist_img = zeros(rows, cols, 2);
[XX, YY] = meshgrid(1:cols, 1:rows);
for rr=1:rows
for cc=1:cols
distance = sqrt( (YY-rr).^2 + (XX-cc).^2 );
mask_radii = (distance <= radius);
bwresponses = img(mask_radii);
[nelems, ~] = histc(double(bwresponses),0:255);
% do some processing over the histogram
...
end
end
EDIT 1 Given the received feedback, I tried to update the solution. However, it's not yet correct
radius = sqrt(2.0);
disk = diskfilter(radius);
fun = #(x) histc( x(disk>0), min(x(:)):max(x(:)) );
output = im2col(im, size(disk), fun);
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
circfilter = (dist <= radius);
end
Following on the technique I described in my answer to a similar question you could try to do the following:
compute the index offsets from a particular voxel that get you to all the neighbors within a radius
Determine which voxels have all neighbors at least radius away from the edge
Compute the neighbors for all these voxels
Generate your histograms for each neighborhood
It is not hard to vectorize this, but note that
It will be slow when the neighborhood is large
It involves generating an intermediate matrix that is NxM (N = voxels in image, M = voxels in neighborhood) which could get very large
Here is the code:
% generate histograms for neighborhood within radius r
A = rand(200,200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy zz] = meshgrid(1:sz(2), 1:sz(1), 1:sz(3));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 + (zz - center(3)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2), center(3));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius & ...
zz > radius + 1 & zz < sz(3) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
A 2D version of the same code (which might be all you need, and is considerably faster):
% generate histograms for neighborhood within radius r
A = rand(200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy] = meshgrid(1:sz(2), 1:sz(1));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
You're right, I don't think that colfilt can be used as you're not applying a filter. You'll have to check the correctness, but here's my attempt using im2col and your diskfilter function (I did remove the conversion to double so it now output logicals):
function circhist
% Example data
im = randi(256,20)-1;
% Ranges - I do this globally for the whole image rather than for each neighborhood
mini = min(im(:));
maxi = max(im(:));
edges = linspace(mini,maxi,20);
% Disk filter
radius = sqrt(2.0);
disk = diskfilter(radius); % Returns logical matrix
% Pad array with -1
im_pad = padarray(im, (size(disk)-1)/2, -1);
% Convert sliding neighborhoods to columns
B = im2col(im_pad, size(disk), 'sliding');
% Get elements from each column that correspond to disk (logical indexing)
C = B(disk(:), :);
% Apply histogram across columns to count number of elements
out = histc(C, edges)
% Display output
figure
imagesc(out)
h = colorbar;
ylabel(h,'Counts');
xlabel('Neighborhood #')
ylabel('Bins')
axis xy
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
disk = (dist <= radius);
If you want to set your ranges (edges) based on each neighborhood then you'll need to make sure that the vector is always the same length if you want to build a big matrix (and then the rows of that matrix won't correspond to each other).
You should note that the shape of the disk returned by fspecial is not as circular as what you were using. It's meant to be used a smoothing/averaging filter so the edges are fuzzy (anti-aliased). Thus when you use ~=0 it will grab more pixels. It'd stick with your own function, which is faster anyways.
You could try processing with an opposite logic (as briefly explained in the comment)
hist = zeros(W+2*R, H+2*R, Q);
for i = 1:R+1;
for j = 1:R+1;
if ((i-R-1)^2+(j-R-1)^2 < R*R)
for q = 0:1:Q-1;
hist(i:i+W-1,j:j+H-1,q+1) += (image == q);
end
end
end
end

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