I have the following binary image:
http://www.4shared.com/download/BozvHQcHba/untitled2.jpg?lgfp=3000
I manually select the start and end points using the ruler in imtool to get the length. Is there a way to automatically get the length i.e the first white pixel to last white pixel (longest length) without doing it manually.
Code
%%// Get the binary image
img = imread(filename1);
BW = im2bw(img);
%%// Find biggest blob
[L,num] = bwlabel( BW );
count_pixels_per_obj = sum(bsxfun(#eq,L(:),1:num));
[~,ind] = max(count_pixels_per_obj);
biggest_blob = (L==ind);
%%// Find row and column info for all edge pixels
BW1 = edge(biggest_blob,'canny');
[row1,col1] = find(BW1);
%%// Get the distance matrix and thus find the largest length separating
%%// them which is the length of the object/blob
%dist_mat = pdist2([row1 col1],[row1 col1]);
dist_mat = dist2s([row1 col1],[row1 col1]); %// If you do not have pdist2
length_blob = max(dist_mat(:))
Associated function
function out = dist2s(pt1,pt2)
out = NaN(size(pt1,1),size(pt2,1));
for m = 1:size(pt1,1)
for n = 1:size(pt2,1)
if(m~=n)
out(m,n) = sqrt( (pt1(m,1)-pt2(n,1)).^2 + (pt1(m,2)-pt2(n,2)).^2 );
end
end
end
Related
I want to pad an object based on edge pixel to replace the zeros.
I am not sure if padarray is applicable to this, I am showing a sample code below to replicate my need. I am able to do it but I think this is not an efficient way as I am scanning each row at a time to find and pad the zeros.
%% Example code to recreate my need
image = imread('moon.tif');
[~, ncols] = size(image);
image(image <50) = 0;
image = fliplr(image(:,1:round(ncols/2)));
% figure, imshow(image,[])
BW = bwareafilt(logical(image),1);
% create bounding box
boxProps=regionprops(BW,'BoundingBox');
cords_BoundingBox = boxProps(1).BoundingBox;
% Extract sub_Image
sub_Image = imcrop(image, cords_BoundingBox);
% figure, imshow(sub_Image,[])
%% This is the part I want to use better or existing function for padding
duplicate_sub_Image = sub_Image;
[nrows, ~] = size(duplicate_sub_Image);
for nrow = 1:nrows
% current_row_inverted = fliplr(sub_Image(nrow,:));
[~,col,pad_value] = find(duplicate_sub_Image(nrow,:),1,'last');
duplicate_sub_Image(nrow,col+1:end) = pad_value;
end
figure,
subplot(131),imshow(image,[]), title('original image');
subplot(132),imshow(sub_Image,[]), title('bounding box image');
subplot(133),imshow(duplicate_sub_Image,[]), title('row padded image');
Any suggestions to improve this code or use of existing functions to address this issue?
Thanks
Here is a way without using loops:
[~,imin] = min(sub_Image, [], 2);
col = max(1, imin-1);
ind = sub2ind(size(sub_Image), (1:numel(col)).', col);
duplicate_sub_Image = sub_Image(ind) .* ~sub_Image + sub_Image;
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);
I performed some operations on an image of a cube and I obtained a binary image of the edges of the cube which are disconnected at some places.The image I obtained is shown below:
I want to join the sides to make it a closed figure.I have tried the following:
BW = im2bw(image,0.5);
BW = imdilate(BW,strel('square',5));
figure,imshow(BW);
But this only thickens the image.It does not connect the edges.I have also tried bwmorph() and various other functions, but it is not working.Can anyone please suggest any function or steps to connect the edges? Thank you
This could be one approach -
%// Read in the input image
img = im2bw(imread('http://i.imgur.com/Bl7zhcn.jpg'));
%// There seems to be white border, which seems to be non-intended and
%// therefore could be removed
img = img(5:end-4,5:end-4);
%// Thin input binary image, find the endpoints in it and connect them
im1 = bwmorph(img,'thin',Inf);
[x,y] = find(bwmorph(im1,'endpoints'));
for iter = 1:numel(x)-1
two_pts = [y(iter) x(iter) y(iter+1) x(iter+1)];
shapeInserter = vision.ShapeInserter('Shape', 'Lines', 'BorderColor', 'White');
rectangle = int32(two_pts);
im1 = step(shapeInserter, im1, rectangle);
end
figure,imshow(im1),title('Thinned endpoints connected image')
%// Dilate the output image a bit
se = strel('diamond', 1);
im2 = imdilate(im1,se);
figure,imshow(im2),title('Dilated Thinned endpoints connected image')
%// Get a convex shaped blob from the endpoints connected and dilate image
im3 = bwconvhull(im2,'objects',4);
figure,imshow(im3),title('Convex blob corresponding to previous output')
%// Detect the boundary of the convex shaped blob and
%// "attach" to the input image to get the final output
im4 = bwboundaries(im3);
idx = im4{:};
im5 = false(size(im3));
im5(sub2ind(size(im5),idx(:,1),idx(:,2))) = 1;
img_out = img;
img_out(im5==1 & img==0)=1;
figure,imshow(img_out),title('Final output')
Debug images -
I used the above code to write the following one.I haven't tested it on many images and it may not be as efficient as the one above but it executes faster comparatively.So I thought I would post it as a solution.
I = imread('image.jpg'); % your original image
I=im2bw(I);
figure,imshow(I)
I= I(5:end-4,5:end-4);
im1 = bwmorph(I,'thin',Inf);
[x,y] = find(bwmorph(im1,'endpoints'));
for iter = 1:numel(x)-1
im1=linept(im1, x(iter), y(iter), x(iter+1), y(iter+1));
end
im2=imfill(im1,'holes');
figure,imshow(im2);
BW = edge(im2);
figure,imshow(BW);
se = strel('diamond', 1);
im3 = imdilate(BW,se);
figure,imshow(im3);
The final result is this:
I got the "linept" function from here:http://in.mathworks.com/matlabcentral/fileexchange/4177-connect-two-pixels
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
I'm using normxcorr2 to find the area that exactly match with my pattern and i also want to find the other area(in the red rectangle) that is look like the pattern. I think it will be works if i can find the next maximum and so on and that value must not in the first maximum area or the first one that it has been detected but i can't do it. Or if you have any idea that using normxcorr2 to find the others area please advise me, I don't have any idea at all.
Here's my code. I modified from this one http://www.mathworks.com/products/demos/image/cross_correlation/imreg.html
onion = imread('pattern103.jpg'); %pattern image
peppers = imread('rsz_1jib-159.jpg'); %Original image
onion = rgb2gray(onion);
peppers = rgb2gray(peppers);
%imshow(onion)
%figure, imshow(peppers)
c = normxcorr2(onion,peppers);
figure, surf(c), shading flat
% offset found by correlation
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = ind2sub(size(c),imax(1));
corr_offset = [(xpeak-size(onion,2))
(size(onion,1)-ypeak)]; %size of window show of max value
offset = corr_offset;
xoffset = offset(1);
yoffset = offset(2);
xbegin = round(xoffset+1); fprintf(['xbegin = ',num2str(xbegin)]);fprintf('\n');
xend = round(xoffset+ size(onion,2));fprintf(['xend = ',num2str(xbegin)]);fprintf('\n');
ybegin = round(yoffset+1);fprintf(['ybegin = ',num2str(ybegin)]);fprintf('\n');
yend = round(yoffset+size(onion,1));fprintf(['yend = ',num2str(yend)]);fprintf('\n');
% extract region from peppers and compare to onion
extracted_onion = peppers(ybegin:yend,xbegin:xend,:);
if isequal(onion,extracted_onion)
disp('pattern103.jpg was extracted from rsz_org103.jpg')
end
recovered_onion = uint8(zeros(size(peppers)));
recovered_onion(ybegin:yend,xbegin:xend,:) = onion;
figure, imshow(recovered_onion)
[m,n,p] = size(peppers);
mask = ones(m,n);
i = find(recovered_onion(:,:,1)==0);
mask(i) = .2; % try experimenting with different levels of
% transparency
% overlay images with transparency
figure, imshow(peppers(:,:,1)) % show only red plane of peppers
hold on
h = imshow(recovered_onion); % overlay recovered_onion
set(h,'AlphaData',mask)