I have a task that is compute the mean value of a sub-image that extract from input image I. Let explain my task. I have a image I (i.e, 9x9), and a window (i.e size 3x3). The window will be run from top-left to bottom-right of image. Hence, it will extract the input image into many subimage. I want to compute the mean value of these sub-images. Could you suggest to me some matlab code to compute it.
This is my solution. But it does not work.
First, I defined a window as Gaussian
Second, the Gaussian function will run from top-left to bottom-right using convolution function. (Note that, it must be use Gaussian Kernel)
Compute the mean value of each sub-window
%% Given Image I,Defined a Gaussian Kernel
sigma=3;
K=fspecial('gaussian',round(2*sigma)*2+1,sigma);
KI=conv2(I,K,'same');
%% mean value
mean(KI)
The problem in here is that mean value off all sub-image will have size similar image I. Because each pixel in image will made a sub-image. But my code returns only a value. What is problem?
If it is your desire to compute the average value in each sub-image once you filter your image with a Gaussian kernel, simply convolve your image with a mean or average filter. This will collect sub-images within your original image and for each output location, you will compute the average value.
Going with your initial assumption that the mask size is 3 x 3, simply use conv2 in conjunction with a 3 x 3 mask that has all 1/9 coefficients. In other words:
%// Your code
%% Given Image I,Defined a Gaussian Kernel
sigma=3;
K=fspecial('gaussian',round(2*sigma)*2+1,sigma);
KI=conv2(I,K,'same');
%// New code
mask = (1/9)*ones(3,3);
out = conv2(KI, mask, 'same');
Each location in out will give you what the average value was for each 3 x 3 sub-image in your Gaussian filtered result.
You can also create the averaging mask by using fspecial with the flag average and specifying the size / width of your mask. Given that you are already using it in your code, you already know of its existence. As such, you can also do:
mask = fspecial('average', 3);
The above code assumes the width and height of the mask are the same, and so it'll create a 3 x 3 mask of all 1/9 coefficients.
Aside
conv2 is designed for general 2D signals. If you are looking to filter an image, I recommend you use imfilter instead. You should have access to it, since fspecial is part of the Image Processing Toolbox, and so is imfilter. imfilter is known to be much more efficient than conv2, and also makes use of Intel Integrated Performance Primitives (Intel IPP) if available (basically if you are running MATLAB on a computer that has an Intel processor that supports IPP). Therefore, you should really perform your filtering this way:
%// Your code
%% Given Image I,Defined a Gaussian Kernel
sigma=3;
K=fspecial('gaussian',round(2*sigma)*2+1,sigma);
KI=imfilter(I,K,'replicate'); %// CHANGE
%// New code
mask = fspecial('average', 3);
out = imfilter(KI, mask, 'replicate'); %// CHANGE
The replicate flag is for handling the boundary conditions. When your mask goes out of bounds of the original image, replicate simply replicates the border of each side of your image so that the mask can fit comfortably within the image when performing your filtering.
Edit
Given your comment, you want to extract the subimages that are seen in KI. You can use the very powerful im2col function that's part of the Image Processing Toolbox. You call it like so:
B = im2col(A,[m n]);
A will be your input image, and B will be a matrix that is of size mn x L where L would be the total number of possible sub-images that exist in your image and m, n are the height and width of each sub-image respectively. How im2col works is that for each sub-image that exists in your image, it warps them so that it fits into a single column in B. Therefore, each column in B produces a single sub-image that is warped into a column. You can then use each column in B for your GMM modelling.
However, im2col only returns valid sub-images that don't go out of bounds. If you want to handle the edge and corner cases, you'll need to pad the image first. Use padarray to facilitate this padding. Therefore, to do what you're asking, we simply do:
Apad = padarray(KI, [1 1], 'replicate');
B = im2col(Apad, [3 3]);
The first line of code will pad the image so that you have a 1 pixel border that surrounds the image. This will allow you to extract 3 x 3 sub-images at the border locations. I use the replicate flag so that you can simply duplicate the border pixels. Next, we use im2col so that you get 3 x 3 sub-images that are then stored in B. As such, B will become a 9 x L matrix where each column gives you a 3 x 3 sub-image.
Be mindful that im2col warps these columns in column-major format. That means that for each sub-image that you have, it takes each column in the sub-image and stacks them on top of each other giving you a 9 x 1 column. You will have L total sub-images, and these are concatenated horizontally to produce a 9 x L matrix. Also, keep in mind that the sub-images are read top-to-bottom, then left-to-right as this is the nature of MATLAB operating in column-major order.
Related
function out = findWaldo(im, filter)
% convert image (and filter) to grayscale
im_input = im;
im = rgb2gray(im);
im = double(im);
filter = rgb2gray(filter);
filter = double(filter);
filter = filter/sqrt(sum(sum(filter.^2)));
out = normxcorr2(filter, im);
Question1: Why we first do rgb2gray on im and filter?
Question2: What does the last second line actually do? Namely,
filter = filter/sqrt(sum(sum(filter.^2)));
Question 1: Why apply rgb2gray first?
normxcorr2 standing for "Normalized 2-D cross-correlation" works on a 2D signal (see doc). A RGB image is a 3D signal: width x height x color (e.g. 1024 x 1024 x 3, 3 since it's three colors). That is why you flatten it first to one color channel. Applying the filter to the image on each color separately would be the alternative, but then you also need to process three correlations (average them or something...).
Question 2: What does filter = filter/sqrt(sum(sum(filter.^2)));do?
It squares the filter image, then sums over rows and columns (basically all squared gray values of the filters) to get a single number that the squareroot is applied to and then is used to divide all filter image values.
I'd say it is some sort of normalization to handle specific input signals, maybe an attempt to get values from 0 - 1. But since normalized cross correlation (normxcorr2) does normalization itself, this step is definitely not needed for it. Unless you don't do something other than cross correlation with the filter variable, I'd consider this an artifact that should be deleted.
General explanation of the function
This function receives two inputs: an image file and a template.
For example, the image file may be a large scene of findings Waldo game and the template can be a picture of Waldo himself.
The output is a matrix called 'out' with same size as the image file. s.t. each pixel holds a "matching results". The higher the value - the higher the chances that the patch centered around the pixel holds a similar pattern such as the template.
The maximal value should be on the pixel in which Waldo is.
Question 1
rgb2gray function receives an rgb image with 3 channels and transorm it into gray image.
It is done on im and on filter because normxcorr2 function only works with grayscale images.
Question 2
The last perform normalization of the pattern: it divides it by it's norm, thus changing it to 1. In fact, this line is not required and should be deleted. Normalization stage is already performed inside normxcorr2 function.
I am learning wavelet theory for image processing. To understand the theory, I write one Matlab program to decompose one black-white image. The program is as follows
Image = zeros(256, 256, 'uint8');
Image(101:200, 101:200) = 255;
figure; imshow(Image);
[cA1, cH1, cV1, cD1] = dwt2(Image, 'db1');
Image1 = [cA1, cH1; cV1, cD1];
figure; imshow(Image1, []);
[cA1, cH1, cV1, cD1] = dwt2(Image, 'db2');
Image1 = [cA1, cH1; cV1, cD1];
figure; imshow(Image1, []);
The first decomposition using the argument db1 produces zeros for all wavelet coefficients. The black-white image has the transition from 0 to 255 along horizontal and vertical directions and should have high-frequency component. Why are zero wavelet coefficients generated? If I change the argument from db1 to db2, the result will show horizontal and vertical lines in the subbands.
If you recall, db1 is the Haar Wavelet. The Haar Wavelet takes either an average of pixels within local windows for the approximation coefficients (or the LL band), or a difference of pixels within local windows for the detail coefficients (or the LH, HL and HH bands).
Be advised that the input image that you specified only consists of two intensities: 0 and 255. Also, you set a square grid within this image to be 255 and it is uniformly shaped.
For self-containment, this is what your test image looks like:
This uniformly shaped object within a square image is important as part of the reasoning why you are not getting any output for the detail images (HL, LH, and HH).
The best way to describe why you're seeing an output for db2 and not db1 can be shown visually.
This slide is from the University of Toronto's CS 320: Introduction to Visual Computing course, specifically, the Discrete Wavelet Transform lecture:
You are well aware that when you take the 2D DWT, you produce 4 sub-images that are half the resolution of the original image. The first output of dwt2 are the approximation coefficients where each output pixel is an average of a 2 x 2 window. The other outputs (second, third and fourth) are detail windows that take two pixels within the window and subtracts these with two other pixels in the window.
As such, the reason why you are not getting an output with db1 is because all of your calculations for the detail images will cancel out. Specifically, you will get 2 x 2 windows of either completely 0, or completely 255, and when you calculate the detail images, you will get 0s for the output regardless. You would take add two 0 values, or two 255 values and you would subtract these two 0 values, or 255 values respectively, thus causing the output to be 0 regardless.
The db2 wavelet is a more complicated transform which is a weighted sum of non-uniform coefficients and so you will certainly get outputs for the detail images rather than just a simple differencing of the 2 x 2 windows.
I would like to stress that if you have a more complicated shape that is non-uniform, db1 will certainly not give you a zero output. Try it on any test image that comes with MATLAB, like with cameraman.tif.
Hope this helps!
I have 2 images im1 and im2 shown below. Theim2 picture is the same as im1, but the only difference between them is the colors. im1 has RGB ranges of (0-255, 0-255, 0-255) for each color channel while im2 has RGB ranges of (201-255, 126-255, 140-255). My exercise is to reverse the added effects so I can restore im2 to im1 as closely as I can. I have 2 thoughts in mind. The first is to match their histograms so they both have the same colors. I tried it using histeq but it restores only a portion of the image. Is there any way to change im2's histogram to be exactly the same as im1? The second approach was just to copy each pixel value from im1 to im2 but this is wrong since it doesn't restore the original image state. Are there any suggestions to restore the image?
#sepdek below pretty much suggested the method that #NKN alluded to, but I will provide another approach. One more alternative I can suggest is to perform a colour correction based on a least mean squared solution. What this alludes to is that we can assume that transforming a pixel from im2 to im1 requires a linear combination of weights. In other words, given a RGB pixel where its red, green and blue components are shaped into a 3 x 1 vector from the corrupted image (im2), there exists some linear transformation to get its equivalent pixel in the clean image (im1). In other words, we have this relationship:
[R_im1] [R_im2]
[G_im1] = A * [G_im2]
[B_im1] [B_im2]
Y = A * X
A in this case would be a 3 x 3 matrix. This is essentially performing a matrix multiplication to get your output corrected pixel. The input RGB pixel from im2 would be X and the output RGB pixel from im1 would be Y. We can extend this to as many pixels as we want, where pairs of pixels from im1 and im2 would establish columns along Y and X. In general, this would further extend X and Y to 3 x N matrices. To find the matrix A, you would find the least mean squared error solution. I won't get into it, but to find the optimal matrix of A, this requires finding the pseudo-inverse. In our case here, A would thus equal to:
Once you find this matrix A, you would need to take each pixel in your image, shape it so that it becomes a 3 x 1 vector, then multiply A with this vector like the approach above. One thing you're probably asking yourself is what kinds of pixels do I need to grab from both images to make the above approach work? One guideline you must adhere to is that you need to make sure that you're sampling from the same spatial location between the two images. As such, if we were to grab a pixel at... say... row 4, column 9, you need to make sure that both pixels from im1 and im2 come from this same row and same column, and they are placed in the same corresponding columns in X and Y.
Another small caveat with this approach is that you need to be sure that you sample a lot of pixels in the image to get a good solution, and you also need to make sure the spread of your sampling is over the entire image. If we localize the sampling to be within a small area, then you're not getting a good enough distribution of the colours and so the output will not look very nice. It's up to you on how many pixels you choose for the problem, but from experience, you get to a point where the output starts to plateau and you don't see any difference. For demonstration purposes, I chose 2000 pixels in random positions throughout the image.
As such, this is what the code would look like. I use randperm to generate a random permutation from 1 to M where M is the total number of pixels in the image. These generate linear indices so that we can sample from the images and construct our matrices. We then apply the above equation to find A, then take each pixel and apply a matrix multiplication with A to get the output. Without further ado:
close all;
clear all;
im1 = imread('http://i.stack.imgur.com/GtgHU.jpg');
im2 = imread('http://i.stack.imgur.com/wHW50.jpg');
rng(123); %// Set seed for reproducibility
num_colours = 2000;
ind = randperm(numel(im1) / size(im1,3), num_colours);
%// Grab colours from original image
red_out = im1(:,:,1);
green_out = im1(:,:,2);
blue_out = im1(:,:,3);
%// Grab colours from corrupted image
red_in = im2(:,:,1);
green_in = im2(:,:,2);
blue_in = im2(:,:,3);
%// Create 3 x N matrices
X = double([red_in(ind); green_in(ind); blue_in(ind)]);
Y = double([red_out(ind); green_out(ind); blue_out(ind)]);
%// Find A
A = Y*(X.')/(X*X.');
%// Cast im2 to double for precision
im2_double = double(im2);
%// Apply matrix multiplication
out = cast(reshape((A*reshape(permute(im2_double, [3 1 2]), 3, [])).', ...
[size(im2_double,1) size(im2_double,2), 3]), class(im2));
Let's go through this code slowly. I am reading your images directly from StackOverflow. After, I use rng to set the seed so that you can reproduce the same results on your end. Setting the seed is useful because it allows you to reproduce the random pixel selection that I did. We generate those linear indices, then create our 3 x N matrices for both im1 and im2. Finding A is exactly how I described, but you're probably not used to the rdivide / / operator. rdivide finds the inverse on the right side of the operator, then multiplies it with whatever is on the left side. This is a more efficient way of doing the calculation, rather than calculating the inverse of the right side separately, then multiplying with the left when you're done. In fact, MATLAB will give you a warning stating to avoid calculating the inverse separately and that you should the divide operators instead. Next, I cast im2 to double to ensure precision as A will most likely be floating point valued, then go through the multiplication of each pixel with A to compute the result. That last line of code looks pretty intimidating, but if you want to figure out how I derived this, I used this to create vintage style photos which also require a matrix multiplication much like this approach and you can read up about it here: How do I create vintage images in MATLAB? . out stores our final image. After running this code and showing what out looks like, this is what we get:
Now, the output looks completely scrambled, but the colour distribution more or less mimics what the input original image looks like. I have a few explanations on why this is the case:
There is quantization noise. If you take a look at the final image, there is various white spotting all over. This is probably due to the quantization error that is introduced when compressing your image. Pixels that should map to the same colours between the images will have slight variations due to quantization which gives us that spotting
There is more than one colour from im2 that maps to im1. If there is more than one colour from im2 that maps to im1, it is impossible for a linear multiplication with the matrix A to be able to generate more than one kind of colour for im1 given a single pixel in im2. Instead, the least mean-squared solution will try and generate a colour that minimizes the error and give you the best colour possible instead. This is probably way the face and other fine details of the image are obscured because of this exact reason.
The image is noisy. Your im2 is not completely clean. I can also see various spots of salt and pepper noise across all of the channels. One bad thing about this method is that if your image is subject to noise, then this method will not faithfully reconstruct the original image properly. Your image can only be corrupted by a wrong mapping of colours. Should there be any other type of image noise introduced, then this method will definitely not work as you are trying to reconstruct the original image based on a noisy image. There are pixels in the noisy image that were never present in the original image, so you'll have no luck getting it back to the way it was before!
If you want to take a look at the histograms of each channel between the original image and the output image, this is what we get:
The code I used to generate the above figure was:
names = {'Red', 'Green', 'Blue'};
figure;
for idx = 1 : 3
subplot(3,2,2*idx - 1);
imhist(im1(:,:,idx));
title([names{idx} ': Image 1']);
end
for idx = 1 : 3
subplot(3,2,2*idx);
imhist(out(:,:,idx));
title([names{idx} ': Output']);
end
The left side shows the red, green and blue histograms for the original image while the right side shows the same histograms for the reconstructed image. You can see that the general shape more or less mimics the original image, but there are some spikes throughout - most likely attributed to quantization noise and the non-unique mapping between colours of both images.
All in all, this is the best that I could do, but I think that was the whole point of the exercise.... to show that it isn't possible.
For more information on how to perform colour correction, check out Richard Alan Peters' II Digital Image Processing slides on colour correction. This was what I started with, and the derivation of how to calculate A can be found in his slides. Perhaps you can use some of what he talks about in your future work.
Good luck!
It seems that you need a scaling function to map the values of im2 to the values of im1.
This is fairly simple and you could write a scaling function to have it available for any such case.
A basic scaling mapping would work as follows:
out_value = min_output + (in_value - min_input) * (outrange / inrange)
given that there is an input value in_value that is within a range of values inrange=max_input-min_input and the mapping results an output value out_value within a range outrange=max_output-min_output. We also need to take into account the minimum input and output range bounds (min_input and min_output) to have a correct mapping.
See for example the following code for a scaling function:
%
% scale the values of a matrix using a set of limits
% possible ways to use:
% y = scale( x, in_range, out_range) --> ex. y = scale( x, [8 230], [0 255])
% y = scale( x, out_range) --> ex. y = scale( x, [0 1])
%
function y = scale( x, varargin );
if nargin<2,
error([upper(mfilename),':: Syntax: y=',mfilename,'(x[,in_range],out_range)']);
end;
if nargin==2,
inrange=[min(x(:)) max(x(:))]; % compute the limits of the input variable
outrange=varargin{1}; % get the output limits from the arguments
else
inrange=varargin{1}; % get the input limits from the arguments
outrange=varargin{2}; % get the output limits from the arguments
end;
if diff(inrange)==0, % row or column vector matrix or scalar
% just do a clipping...
if x>=outrange(2),
y=outrange(2);
elseif x<=outrange(1),
y=outrange(1);
else
y=x;
end;
else
% actually scale the data
% using: out = min_output + (x-min_input) * (outrange / inrange)
y = outrange(1) + (x-inrange(1))*abs(diff(outrange))/abs(diff(inrange));
end;
This function gets a matrix of values and scales them to a desired range.
In your case it could be used as following (variable img is the scaled im2):
for i=1:size(im1,3), % for each of the input/output image channels
output_range = [min(min(im1(:,:,i))) max(max(im1(:,:,i)))];
img(:,:,i) = scale( im2(:,:,i), output_range);
end;
This way im2 is scaled to the range of values of im1 one channel at a time. Output variable img should be the desired one.
I have the following code in MATLAB:
I=imread(image);
h=fspecial('gaussian',si,sigma);
I=im2double(I);
I=imfilter(I,h,'conv');
figure,imagesc(I),impixelinfo,title('Original Image after Convolving with gaussian'),colormap('gray');
How can I define and apply a Gaussian filter to an image without imfilter, fspecial and conv2?
It's really unfortunate that you can't use the some of the built-in methods from the Image Processing Toolbox to help you do this task. However, we can still do what you're asking, though it will be a bit more difficult. I'm still going to use some functions from the IPT to help us do what you're asking. Also, I'm going to assume that your image is grayscale. I'll leave it to you if you want to do this for colour images.
Create Gaussian Mask
What you can do is create a grid of 2D spatial co-ordinates using meshgrid that is the same size as the Gaussian filter mask you are creating. I'm going to assume that N is odd to make my life easier. This will allow for the spatial co-ordinates to be symmetric all around the mask.
If you recall, the 2D Gaussian can be defined as:
The scaling factor in front of the exponential is primarily concerned with ensuring that the area underneath the Gaussian is 1. We will deal with this normalization in another way, where we generate the Gaussian coefficients without the scaling factor, then simply sum up all of the coefficients in the mask and divide every element by this sum to ensure a unit area.
Assuming that you want to create a N x N filter, and with a given standard deviation sigma, the code would look something like this, with h representing your Gaussian filter.
%// Generate horizontal and vertical co-ordinates, where
%// the origin is in the middle
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
%// Create Gaussian Mask
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
%// Normalize so that total area (sum of all weights) is 1
h = h / sum(h(:));
If you check this with fspecial, for odd values of N, you'll see that the masks match.
Filter the image
The basics behind filtering an image is for each pixel in your input image, you take a pixel neighbourhood that surrounds this pixel that is the same size as your Gaussian mask. You perform an element-by-element multiplication with this pixel neighbourhood with the Gaussian mask and sum up all of the elements together. The resultant sum is what the output pixel would be at the corresponding spatial location in the output image. I'm going to use the im2col that will take pixel neighbourhoods and turn them into columns. im2col will take each of these columns and create a matrix where each column represents one pixel neighbourhood.
What we can do next is take our Gaussian mask and convert this into a column vector. Next, we would take this column vector, and replicate this for as many columns as we have from the result of im2col to create... let's call this a Gaussian matrix for a lack of a better term. With this Gaussian matrix, we will do an element-by-element multiplication with this matrix and with the output of im2col. Once we do this, we can sum over all of the rows for each column. The best way to do this element-by-element multiplication is through bsxfun, and I'll show you how to use it soon.
The result of this will be your filtered image, but it will be a single vector. You would need to reshape this vector back into matrix form with col2im to get our filtered image. However, a slight problem with this approach is that it doesn't filter pixels where the spatial mask extends beyond the dimensions of the image. As such, you'll actually need to pad the border of your image with zeroes so that we can properly do our filter. We can do this with padarray.
Therefore, our code will look something like this, going with your variables you have defined above:
N = 5; %// Define size of Gaussian mask
sigma = 2; %// Define sigma here
%// Generate Gaussian mask
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
h = h / sum(h(:));
%// Convert filter into a column vector
h = h(:);
%// Filter our image
I = imread(image);
I = im2double(I);
I_pad = padarray(I, [floor(N/2) floor(N/2)]);
C = im2col(I_pad, [N N], 'sliding');
C_filter = sum(bsxfun(#times, C, h), 1);
out = col2im(C_filter, [N N], size(I_pad), 'sliding');
out contains the filtered image after applying a Gaussian filtering mask to your input image I. As an example, let's say N = 9, sigma = 4. Let's also use cameraman.tif that is an image that's part of the MATLAB system path. By using the above parameters, as well as the image, this is the input and output image we get:
a process of mine produces 256 binary (logical) matrices, one for each level of a grayscale source image.
Here is the code :
so = imread('bio_sd.bmp');
co = rgb2gray(so);
for l = 1:256
bw = (co == l); % Binary image from level l of original image
be = ordfilt2(bw, 1, ones(3, 3)); % Convolution filter
bl(int16(l)) = {bwlabel(be, 8)}; % Component labelling
end
I obtain a cell array of 256 binary images. Such a binary image contains 1s if the source-image pixel at that location has the same level as the index of the binary image.
ie. the binary image bl{12} contains 1s where the source image has pixels with the level 12.
I'd like to create new image by combining the 256 binary matrices back to a grayscale image.
But i'm very new to Matlab and i wonder if someone can help me to code it :)
ps : i'm using matlab R2010a student edition.
this whole answer only applies to the original form of the question.
Lets assume you can get all your binary matrices together into a big n-by-m-by-256 matrix binaryimage(x,y,greyvalue). Then you can calculate your final image as
newimage=sum(bsxfun(#times,binaryimage,reshape(0:255,1,1,[])),3)
The magic here is done by bsxfun, which multiplies the 3D (n x m x 256) binaryimage with the 1 x 1 x 256 vector containing the grey values 0...255. This produces a 3D image where for fixed x and y, the vector (y,x,:) contains many zeros and (for the one grey value G where the binary image contained a 1) it contains the value G. So now you only need to sum over this third dimension to get a n x m image.
Update
To test that this works correctly, lets go the other way first:
fullimage=floor(rand(100,200)*256);
figure;imshow(fullimage,[0 255]);
is a random greyscale image. You can calculate the 256 binary matrices like this:
binaryimage=false([size(fullimage) 256]);
for i=1:size(fullimage,1)
for j=1:size(fullimage,2)
binaryimage(i,j,fullimage(i,j)+1)=true;
end
end
We can now apply the solution I gave above
newimage=sum(bsxfun(#times,binaryimage,reshape(0:255,1,1,[])),3);
and verify that I returns the original image:
all(newimage(:)==fullimage(:))
which gives 1 (true) :-).
Update 2
You now mention that your binary images are in a cell array, I assume binimg{1:256}, with each cell containing an n x m binary array. If you can it probably makes sense to change the code that produces this data to create the 3D binary array I use above - cells are mostly usefull if different cells contain data of different types, shapes or sizes.
If there are good reasons to stick with a cell array, you can convert it to a 3D array using
binaryimage = reshape(cell2mat(reshape(binimg,1,256)),n,m,256);
with n and m as used above. The inner reshape is not necessary if you already have size(binimg)==[1 256]. So to sum it up, you need to use your cell array binimg to calculate the 3D matrix binaryimage, which you can then use to calculate the newimage that you are interested in using the code at the very beginning of my answer.
Hope this helps...
What your code does...
I thought it may be best to first go through what the code you posted is actually doing, since there are a couple of inconsistencies. I'll go through each line in your loop:
bw = (co == l);
This simply creates a binary matrix bw with ones where your grayscale image co has a pixel intensity equal to the loop value l. I notice that you loop from 1 to 256, and this strikes me as odd. Typically, images loaded into MATLAB will be an unsigned 8-bit integer type, meaning that the grayscale values will span the range 0 to 255. In such a case, the last binary matrix bw that you compute when l = 256 will always contain all zeroes. Also, you don't do any processing for pixels with a grayscale level of 0. From your subsequent processing, I'm guessing you purposefully want to ignore grayscale values of 0, in which case you probably only need to loop from 1 to 255.
be = ordfilt2(bw, 1, ones(3, 3));
What you are essentially doing here with ORDFILT2 is performing a binary erosion operation. Any values of 1 in bw that have a 0 as one of their 8 neighbors will be set to 0, causing islands of ones to erode (i.e. shrink in size). Small islands of ones will disappear, leaving only the larger clusters of contiguous pixels with the same grayscale level.
bl(int16(l)) = {bwlabel(be, 8)};
Here's where you may be having some misunderstandings. Firstly, the matrices in bl are not logical matrices. In your example, the function BWLABEL will find clusters of 8-connected ones. The first cluster found will have its elements labeled as 1 in the output image, the second cluster found will have its elements labeled as 2, etc. The matrices will therefore contain positive integer values, with 0 representing the background.
Secondly, are you going to use these labeled clusters for anything? There may be further processing you do for which you need to identify separate clusters at a given grayscale intensity level, but with regard to creating a grayscale image from the elements in bl, the specific label value is unnecessary. You only need to identify zero versus non-zero values, so if you aren't using bl for anything else I would suggest that you just save the individual values of be in a cell array and use them to recreate a grayscale image.
Now, onto the answer...
A very simple solution is to concatenate your cell array of images into a 3-D matrix using the function CAT, then use the function MAX to find the indices where the non-zero values occur along the third dimension (which corresponds to the grayscale value from the original image). For a given pixel, if there is no non-zero value found along the third dimension (i.e. it is all zeroes) then we can assume the pixel value should be 0. However, the index for that pixel returned by MAX will default to 1, so you have to use the maximum value as a logical index to set the pixel to 0:
[maxValue,grayImage] = max(cat(3,bl{:}),[],3);
grayImage(~maxValue) = 0;
Note that for the purposes of displaying or saving the image you may want to change the type of the resulting image grayImage to an unsigned 8-bit integer type, like so:
grayImage = uint8(grayImage);
The simplest solution would be to iterate through each of your logical matrices in turn, multiply it by its corresponding weight, and accumulate into an output matrix which will represent your final image.