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Hi I am trying to extract x,y coordinates from gray scale photo (example.jpg).
I mark the points (with red dots) where I want the x, y coordinates (example1.jpg)
Then I extract the red dots area with below codes
A=imread('example.jpg');
B=imread('example1.jpg');
Size=size(A)
C=zeros(Size);
for j=1:Size(2)
for i=1:Size(1)
if A(i,j)==B(i,j);
C(i,j)=1;
else C(i,j)=255;
end
end
end
K=mat2gray(C);
imshow(K)
By doing this, I can extract the dots (I am interested) and got below image but the dot points bigger than my real marking in the photo (dots.png). And it did not capture all dots (8 out of 10 dots)
Then to get the x,y coordinates of these dot points, I use below codes:
X=imread('dots.png');
[I,J] = find(X(:,:,1) == 255); %// Change
scatter(I,J)
Then I got x and y coordinates in terms of I,J and the picture to counter check if the coordinates show the correct positions as in the photo.
But here are the problems I need solve:
1) How can I get the smaller dot points in figure 3 (as the dot I mark in gray photos are small but when it is generated in figure 3, it become bigger. As a consequence, when I extract x,y coordinates (in terms of I,J here), I got multiple x and y for one dot as each dot is so big. Instead, I want to get one x and one y for each dot.
2) How to make to capture all dots that I mark
3) And when you look at the figure 3 and 4, figure 3 shows the real orientation of pumpkin but in the figure 4, it is rotated. How this happen and how can I correct it?
4) And I thought there can be an easier way to do this extraction than my method. Can you please advice?
Thank you
Here's a more complete example, for the sake of an answer:
I = imread('cameraman.tif');
figure, imshow(I)
[y, x] = ginput(5); % This chooses the five points interactively.
close
color = I; % making a RGB color version of the image.
color(:,:,2) = I;
color(:,:,3) = I;
for idx = 1:numel(x)
% Change the color of the points to red one by one.
xpoint = x(idx);
ypoint = y(idx);
color(xpoint, ypoint, 1) = 255;
color(xpoint, ypoint, 2) = 0;
color(xpoint, ypoint, 3) = 0;
end
figure,
imshow(color) % display the colored dot image.
This should select five points in the image, interactively, and then color those points red.
In terms of the blobs I suggest that you have saved the image with the red dots using a compressed file format.
The blobs seem to be 24x24 macroblocks - not sure what compression algorithm uses that (not jpeg)...
Here it is one up close:
Regardless, an easy way to see why it is happening is to list the values of A(i,j) and B(i,j) when A(i,j) != B(i,j) - a basic bug finding process you can do yourself.
I'm making an image processing project and I have stuck in one the project's steps. Here is the situation;
This is my mask:
and I want to detect the maximum-sized rectangle that can fit into this mask like this.
I'm using MATLAB for my project. Do you know any fast way to accomplish this aim. Any code sample, approach or technique would be great.
EDIT 1 : The two algorithms below are works with lot's of the cases. But both of them give wrong results in some difficult cases. I'am using both of them in my project.
This approach starts with the entire image and shrinks each border in turn pixel-by-pixel until it finds an acceptable rectangle.
It takes ~0.02 seconds to run on the example image, so it's reasonably fast.
EDIT: I should clarify that this isn't meant to be a universal solution. This algorithm relies on the rectangle being centered and having roughly the same aspect ratio as the image itself. However, in the cases where it is appropriate, it is fast. #DanielHsH offered a solution which they claim works in all cases.
The code:
clear; clc;
tic;
%% // read image
imrgb= imread('box.png');
im = im2bw(rgb2gray(imrgb)); %// binarize image
im = 1-im; %// convert "empty" regions to 0 intensity
[rows,cols] = size(im);
%% // set up initial parameters
ULrow = 1; %// upper-left row (param #1)
ULcol = 1; %// upper-left column (param #2)
BRrow = rows; %// bottom-right row (param #3)
BRcol = cols; %// bottom-right column (param #4)
parameters = 1:4; %// parameters left to be updated
pidx = 0; %// index of parameter currently being updated
%% // shrink region until acceptable
while ~isempty(parameters); %// update until all parameters reach bounds
%// 1. update parameter number
pidx = pidx+1;
pidx = mod( pidx-1, length(parameters) ) + 1;
p = parameters(pidx); %// current parameter number
%// 2. update current parameter
if p==1; ULrow = ULrow+1; end;
if p==2; ULcol = ULcol+1; end;
if p==3; BRrow = BRrow-1; end;
if p==4; BRcol = BRcol-1; end;
%// 3. grab newest part of region (row or column)
if p==1; region = im(ULrow,ULcol:BRcol); end;
if p==2; region = im(ULrow:BRrow,ULcol); end;
if p==3; region = im(BRrow,ULcol:BRcol); end;
if p==4; region = im(ULrow:BRrow,BRcol); end;
%// 4. if the new region has only zeros, stop shrinking the current parameter
if isempty(find(region,1))
parameters(pidx) = [];
end
end
toc;
params = [ULrow ULcol BRrow BRcol]
area = (BRrow-ULrow)*(BRcol-ULcol)
The results for this image:
Elapsed time is 0.027032 seconds.
params =
10 25 457 471
area =
199362
Code to visualize results:
imrgb(params(1):params(3),params(2):params(4),1) = 0;
imrgb(params(1):params(3),params(2):params(4),2) = 255;
imrgb(params(1):params(3),params(2):params(4),3) = 255;
imshow(imrgb);
Another example image:
Here is a correct answer.
You must use dynamic programming! Other methods of direct calculation (like cutting 1 pixel from each edge) might produce sub-optimal results. My method guarantees that it selects the largest possible rectangle that fits in the mask. I assume that the mask has 1 big convex white blob of any shape with black background around it.
I wrote 2 methods. findRect() which finds the best possible square (starting on x,y with length l). The second method LargestInscribedImage() is an example of how to find any rectangle (of any aspect ratio). The trick is to resize the mask image, find a square and resize it back.
In my example the method finds the larges rectangle that can be fit in the mask having the same aspect ration as the mask image. For example if the mask image is of size 100x200 pixels than the algorithm will find the largest rectangle having aspect ratio 1:2.
% ----------------------------------------------------------
function LargestInscribedImage()
% ----------------------------------------------------------
close all
im = double(imread('aa.bmp')); % Balck and white image of your mask
im = im(:,:,1); % If it is colored RGB take only one of the channels
b = imresize(im,[size(im,1) size(im,1)]); Make the mask square by resizing it by its aspect ratio.
SC = 1; % Put 2..4 to scale down the image an speed up the algorithm
[x1,y1,l1] = findRect(b,SC); % Lunch the dyn prog algorithm
[x2,y2,l2] = findRect(rot90(b),SC); % rotate the image by 90deg and solve
% Rotate back: x2,y2 according to rot90
tmp = x2;
x2 = size(im,1)/SC-y2-l2;
y2 = tmp;
% Select the best solution of the above (for the original image and for the rotated by 90degrees
if (l1>=l2)
corn = sqCorn(x1,y1,l1);
else
corn = sqCorn(x2,y2,l2);
end
b = imresize(b,1/SC);
figure;imshow(b>0); hold on;
plot(corn(1,:),corn(2,:),'O')
corn = corn*SC;
corn(1,:) = corn(1,:)*size(im,2)/size(im,1);
figure;imshow(im); hold on;
plot(corn(1,:),corn(2,:),'O')
end
function corn = sqCorn(x,y,l)
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
end
% ----------------------------------------------------------
function [x,y,l] = findRect(b,SC)
b = imresize(b,1/SC);
res = zeros(size(b,1),size(b,2),3);
% initialize first col
for i = 1:1:size(b,1)
if (b(i,1) > 0)
res(i,1,:) = [i,1,0];
end
end
% initialize first row
for i = 1:1:size(b,2)
if (b(1,i) > 0)
res(1,i,:) = [1,i,0];
end
end
% DynProg
for i = 2:1:size(b,1)
for j = 2:1:size(b,2)
isWhite = b(i,j) > 0;
if (~isWhite)
res(i,j,:)=res(i-1,j-1,:); % copy
else
if (b(i-1,j-1)>0) % continuous line
lineBeg = [res(i-1,j-1,1),res(i-1,j-1,2)];
lineLenght = res(i-1,j-1,3);
if ((b(lineBeg(1),j)>0)&&(b(i,lineBeg(2))>0)) % if second diag is good
res(i,j,:) = [lineBeg,lineLenght+1];
else
res(i,j,:)=res(i-1,j-1,:); % copy since line has ended
end
else
res(i,j,:) = [i,j,0]; % Line start
end
end
end
end
% check last col
[maxValCol,WhereCol] = max(res(:,end,3));
% check last row
[maxValRow,WhereRow] = max(res(end,:,3));
% Find max
x= 0; y = 0; l = 0;
if (maxValCol>maxValRow)
y = res(WhereCol,end,1);
x = res(WhereCol,end,2);
l = maxValCol;
else
y = res(end,WhereRow,1);
x = res(end,WhereRow,2);
l = maxValRow;
end
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
% figure;imshow(b>0); hold on;
% plot(corn(1,:),corn(2,:),'O')
return;
end
The black boundaries in your image are curved and not closed. For example, in the top right corner, the black boundaries won't meet and form a closed contour. Therefore, a simple strategy in one of my comments will not work.
I am now providing you with a skeleton of a code which you can play with and add conditions as per your need. My idea is as follows:
To find left-side x-coordinate of the rectangle, first count the white pixels each column of the image contains:
%I assume that the image has already been converted to binary.
whitePixels=sum(img,1);
Then find the rate of change:
diffWhitePixels=diff(whitePixels);
If you see the bar plot of diffWhitePixels then you will observe various large entries (which indicate that the white region is still not in a straight line, and it is not a proper place to put the rectangles left vertical edge). Small entries (in your image, less than 5) indicate you can put the rectangle edge there.
You can do similar things to determine right, top and bottom edge positions of the rectangles.
Discussion:
First of all, the problem is ill-posed in my opinion. What do you mean by maximum-sized rectangle? Is it maximum area or length of side? In all possible cases, I don't think above method can get the correct answer. I can think of two or three cases right now where above method would fail, but it will at least give you the right answer on images similar to the given image, provided you adjust the values.
You can put some constraints once you know how your images are going to look. For example, if the black boundary curves inside, you can say that you don't want a column such as [0;0;...0;1;1;...0;0;...;0;1;1;...;1] i.e. zeros surrounded by ones. Another constraint could be how many black pixels do you want to allow? You can also crop the image till to remove extra black pixels. In your image, you can crop the image (programmatically) from the left and the bottom edge. Cropping an image is probably necessary, and definitely the better thing to do.
I'm making an image processing project and I have stuck in one the project's steps. Here is the situation;
This is my mask:
and I want to detect the maximum-sized rectangle that can fit into this mask like this.
I'm using MATLAB for my project. Do you know any fast way to accomplish this aim. Any code sample, approach or technique would be great.
EDIT 1 : The two algorithms below are works with lot's of the cases. But both of them give wrong results in some difficult cases. I'am using both of them in my project.
This approach starts with the entire image and shrinks each border in turn pixel-by-pixel until it finds an acceptable rectangle.
It takes ~0.02 seconds to run on the example image, so it's reasonably fast.
EDIT: I should clarify that this isn't meant to be a universal solution. This algorithm relies on the rectangle being centered and having roughly the same aspect ratio as the image itself. However, in the cases where it is appropriate, it is fast. #DanielHsH offered a solution which they claim works in all cases.
The code:
clear; clc;
tic;
%% // read image
imrgb= imread('box.png');
im = im2bw(rgb2gray(imrgb)); %// binarize image
im = 1-im; %// convert "empty" regions to 0 intensity
[rows,cols] = size(im);
%% // set up initial parameters
ULrow = 1; %// upper-left row (param #1)
ULcol = 1; %// upper-left column (param #2)
BRrow = rows; %// bottom-right row (param #3)
BRcol = cols; %// bottom-right column (param #4)
parameters = 1:4; %// parameters left to be updated
pidx = 0; %// index of parameter currently being updated
%% // shrink region until acceptable
while ~isempty(parameters); %// update until all parameters reach bounds
%// 1. update parameter number
pidx = pidx+1;
pidx = mod( pidx-1, length(parameters) ) + 1;
p = parameters(pidx); %// current parameter number
%// 2. update current parameter
if p==1; ULrow = ULrow+1; end;
if p==2; ULcol = ULcol+1; end;
if p==3; BRrow = BRrow-1; end;
if p==4; BRcol = BRcol-1; end;
%// 3. grab newest part of region (row or column)
if p==1; region = im(ULrow,ULcol:BRcol); end;
if p==2; region = im(ULrow:BRrow,ULcol); end;
if p==3; region = im(BRrow,ULcol:BRcol); end;
if p==4; region = im(ULrow:BRrow,BRcol); end;
%// 4. if the new region has only zeros, stop shrinking the current parameter
if isempty(find(region,1))
parameters(pidx) = [];
end
end
toc;
params = [ULrow ULcol BRrow BRcol]
area = (BRrow-ULrow)*(BRcol-ULcol)
The results for this image:
Elapsed time is 0.027032 seconds.
params =
10 25 457 471
area =
199362
Code to visualize results:
imrgb(params(1):params(3),params(2):params(4),1) = 0;
imrgb(params(1):params(3),params(2):params(4),2) = 255;
imrgb(params(1):params(3),params(2):params(4),3) = 255;
imshow(imrgb);
Another example image:
Here is a correct answer.
You must use dynamic programming! Other methods of direct calculation (like cutting 1 pixel from each edge) might produce sub-optimal results. My method guarantees that it selects the largest possible rectangle that fits in the mask. I assume that the mask has 1 big convex white blob of any shape with black background around it.
I wrote 2 methods. findRect() which finds the best possible square (starting on x,y with length l). The second method LargestInscribedImage() is an example of how to find any rectangle (of any aspect ratio). The trick is to resize the mask image, find a square and resize it back.
In my example the method finds the larges rectangle that can be fit in the mask having the same aspect ration as the mask image. For example if the mask image is of size 100x200 pixels than the algorithm will find the largest rectangle having aspect ratio 1:2.
% ----------------------------------------------------------
function LargestInscribedImage()
% ----------------------------------------------------------
close all
im = double(imread('aa.bmp')); % Balck and white image of your mask
im = im(:,:,1); % If it is colored RGB take only one of the channels
b = imresize(im,[size(im,1) size(im,1)]); Make the mask square by resizing it by its aspect ratio.
SC = 1; % Put 2..4 to scale down the image an speed up the algorithm
[x1,y1,l1] = findRect(b,SC); % Lunch the dyn prog algorithm
[x2,y2,l2] = findRect(rot90(b),SC); % rotate the image by 90deg and solve
% Rotate back: x2,y2 according to rot90
tmp = x2;
x2 = size(im,1)/SC-y2-l2;
y2 = tmp;
% Select the best solution of the above (for the original image and for the rotated by 90degrees
if (l1>=l2)
corn = sqCorn(x1,y1,l1);
else
corn = sqCorn(x2,y2,l2);
end
b = imresize(b,1/SC);
figure;imshow(b>0); hold on;
plot(corn(1,:),corn(2,:),'O')
corn = corn*SC;
corn(1,:) = corn(1,:)*size(im,2)/size(im,1);
figure;imshow(im); hold on;
plot(corn(1,:),corn(2,:),'O')
end
function corn = sqCorn(x,y,l)
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
end
% ----------------------------------------------------------
function [x,y,l] = findRect(b,SC)
b = imresize(b,1/SC);
res = zeros(size(b,1),size(b,2),3);
% initialize first col
for i = 1:1:size(b,1)
if (b(i,1) > 0)
res(i,1,:) = [i,1,0];
end
end
% initialize first row
for i = 1:1:size(b,2)
if (b(1,i) > 0)
res(1,i,:) = [1,i,0];
end
end
% DynProg
for i = 2:1:size(b,1)
for j = 2:1:size(b,2)
isWhite = b(i,j) > 0;
if (~isWhite)
res(i,j,:)=res(i-1,j-1,:); % copy
else
if (b(i-1,j-1)>0) % continuous line
lineBeg = [res(i-1,j-1,1),res(i-1,j-1,2)];
lineLenght = res(i-1,j-1,3);
if ((b(lineBeg(1),j)>0)&&(b(i,lineBeg(2))>0)) % if second diag is good
res(i,j,:) = [lineBeg,lineLenght+1];
else
res(i,j,:)=res(i-1,j-1,:); % copy since line has ended
end
else
res(i,j,:) = [i,j,0]; % Line start
end
end
end
end
% check last col
[maxValCol,WhereCol] = max(res(:,end,3));
% check last row
[maxValRow,WhereRow] = max(res(end,:,3));
% Find max
x= 0; y = 0; l = 0;
if (maxValCol>maxValRow)
y = res(WhereCol,end,1);
x = res(WhereCol,end,2);
l = maxValCol;
else
y = res(end,WhereRow,1);
x = res(end,WhereRow,2);
l = maxValRow;
end
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
% figure;imshow(b>0); hold on;
% plot(corn(1,:),corn(2,:),'O')
return;
end
The black boundaries in your image are curved and not closed. For example, in the top right corner, the black boundaries won't meet and form a closed contour. Therefore, a simple strategy in one of my comments will not work.
I am now providing you with a skeleton of a code which you can play with and add conditions as per your need. My idea is as follows:
To find left-side x-coordinate of the rectangle, first count the white pixels each column of the image contains:
%I assume that the image has already been converted to binary.
whitePixels=sum(img,1);
Then find the rate of change:
diffWhitePixels=diff(whitePixels);
If you see the bar plot of diffWhitePixels then you will observe various large entries (which indicate that the white region is still not in a straight line, and it is not a proper place to put the rectangles left vertical edge). Small entries (in your image, less than 5) indicate you can put the rectangle edge there.
You can do similar things to determine right, top and bottom edge positions of the rectangles.
Discussion:
First of all, the problem is ill-posed in my opinion. What do you mean by maximum-sized rectangle? Is it maximum area or length of side? In all possible cases, I don't think above method can get the correct answer. I can think of two or three cases right now where above method would fail, but it will at least give you the right answer on images similar to the given image, provided you adjust the values.
You can put some constraints once you know how your images are going to look. For example, if the black boundary curves inside, you can say that you don't want a column such as [0;0;...0;1;1;...0;0;...;0;1;1;...;1] i.e. zeros surrounded by ones. Another constraint could be how many black pixels do you want to allow? You can also crop the image till to remove extra black pixels. In your image, you can crop the image (programmatically) from the left and the bottom edge. Cropping an image is probably necessary, and definitely the better thing to do.
I'm trying to measure the areas of each particle shown in this image:
I managed to get the general shape of each particle using MSER shown here:
but I'm having trouble removing the background. I tried using MATLAB's imfill, but it doesn't fill all the particles because some are cut off at the edges. Any tips on how to get rid of the background or find the areas of the particles some other way?
Cheers.
Edit: This is what imfill looks like:
Edit 2: Here is the code used to get the outline. I used this for the MSER.
%Compute region seeds and elliptial frames.
%MinDiversity = how similar to its parent MSER the region is
%MaxVariation = stability of the region
%BrightOnDark is used as the void is primarily dark. It also prevents dark
%patches in the void being detected.
[r,f] = vl_mser(I,'MinDiversity',0.7,...
'MaxVariation',0.2,...
'Delta',10,...
'BrightOnDark',1,'DarkOnBright',0) ;
%Plot region frames, but not used right now
%f = vl_ertr(f) ;
%vl_plotframe(f) ;
%Plot MSERs
M = zeros(size(I)) ; %M = no of overlapping extremal regions
for x=r'
s = vl_erfill(I,x) ;
M(s) = M(s) + 1;
end
%Display region boundaries
figure(1) ;
clf ; imagesc(I) ; hold on ; axis equal off; colormap gray ;
%Create contour plot using the values
%0:max(M(:))+.5 is the no of contour levels. Only level 0 is needed so
%[0 0] is used.
[c,h]=contour(M,[0 0]) ;;
set(h,'color','r','linewidth',1) ;
%Retrieve the image data from the contour image
f = getframe;
I2 = f.cdata;
%Convert the image into binary; the red outlines are while while the rest
%is black.
I2 = all(bsxfun(#eq,I2,reshape([255 0 0],[1 1 3])),3);
I2 = imcrop(I2,[20 1 395 343]);
imshow(~I2);
Proposed solution / trick and code
It seems you can work with M here. One trick that you can employ here would be to pad zeros all across the boundaries of the image M and then fill its holes. This would take care of filling the blobs that were touching the boundaries before, as now there won't be any blob touching the boundaries because of the zeros padding.
Thus, after you have M, you can add this code -
%// Get a binary version of M
M_bw = im2bw(M);
%// Pad zeros all across the grayscale image
padlen = 2; %// length of zeros padding
M_pad = padarray(M_bw,[padlen padlen],0);
%// Fill the holes
M_pad_filled = imfill(M_pad,'holes');
%// Get the background mask after the holes are gone
background_mask = ~M_pad_filled(padlen+1:end-padlen,padlen+1:end-padlen);
%// Overlay the background mask on the original image to show that you have
%// a working background mask for use
I(background_mask) = 0;
figure,imshow(I)
Results
Input image -
Foreground mask (this would be ~background_mask) -
Output image -
I have a binary image that represents a number in MATLAB:
I'd like to fill all the digits. The desired result is:
The only thing I found was the imfill function, but that wasn't really helpfull since I've lost my inner data (the 9's inner circle for example).
Another possibility is to use the BWBOUNDARIES function, which:
traces the exterior boundaries of objects, as well as boundaries of
holes inside these objects
That information is contained in the fourth output A, an adjacency matrix that represents the parent-child-hole dependencies.
%# read binary image
bw = imread('SUvif.png');
%# find all boundaries
[B,L,N,A] = bwboundaries(bw, 8, 'holes');
%# exclude inner holes
[r,~] = find(A(:,N+1:end)); %# find inner boundaries that enclose stuff
[rr,~] = find(A(:,r)); %# stuff they enclose
idx = setdiff(1:numel(B), [r(:);rr(:)]); %# exclude both
bw2 = ismember(L,idx); %# filled image
%# compare results
subplot(311), imshow(bw), title('original')
subplot(312), imshow( imfill(bw,'holes') ), title('imfill')
subplot(313), imshow(bw2), title('bwboundaries')
The problem is how to distinguish the holes from the digits. A possible ad hoc solution is filtering them by the area of the pixels inside.
function SolveSoProblem()
I = imread('http://i.stack.imgur.com/SUvif.png');
%Fill all the holes
F = imfill(I,'holes');
%Find all the small ones,and mark their edges in the image
bw = bwlabel(I);
rp = regionprops(bw,'FilledArea','PixelIdxList');
indexesOfHoles = [rp.FilledArea]<150;
pixelsNotToFill = vertcat(rp(indexesOfHoles).PixelIdxList);
F(pixelsNotToFill) = 0;
figure;imshow(F);
%Remove the inner area
bw1 = bwlabel(F,4);
rp = regionprops(bw1,'FilledArea','PixelIdxList');
indexesOfHoles1 = [rp.FilledArea]<150;
pixelListToRemove = vertcat(rp(indexesOfHoles1).PixelIdxList);
F(pixelListToRemove) = 0;
figure;imshow(F);
end
After step(1):
After step(2):
Assuming the top-left pixel is always outside the regions to filled:
Work across the top line, copying pixels to the output image
When you come to a white pixel followed by a black pixel in the input image, start setting white pixels in the output image, until you come to black pixel followed by a white pixel.