I am trying to find a way to crop from a circle object (Image A) the largest square that can fit inside it.
Can someone please explain/show me how to find the biggest square fit parameters of the white space inside the circle (Image I) and based on them crop the square in the original image (Image A).
Script:
A = imread('E:/CirTest/Test.jpg');
%imshow(A)
level = graythresh(A);
BW = im2bw(A,level);
%imshow(BW)
I = imfill(BW, 'holes');
imshow(I)
d = imdistline;
[centers, radii, metric] = imfindcircles(A,[1 500]);
imageCrop=imcrop(A, [BoxBottomX BoxBottomY NewX NewY]);
I have a solution for you but it requires a bit of extra work. What I would do first is use imfill but directly on the grayscale image. This way, noisy pixels in uniform areas get inpainted with the same intensities so that thresholding is easier. You can still use graythresh or Otsu's thresholding and do this on the inpainted image.
Here's some code to get you started:
figure; % Open up a new figure
% Read in image and convert to grayscale
A = rgb2gray(imread('http://i.stack.imgur.com/vNECg.jpg'));
subplot(1,3,1); imshow(A);
title('Original Image');
% Find the optimum threshold via Otsu
level = graythresh(A);
% Inpaint noisy areas
I = imfill(A, 'holes');
subplot(1,3,2); imshow(I);
title('Inpainted image');
% Threshold the image
BW = im2bw(I, level);
subplot(1,3,3); imshow(BW);
title('Thresholded Image');
The above code does the three operations that I mentioned, and we see this figure:
Notice that the thresholded image has border pixels that need to be removed so we can concentrate on the circular object. You can use the imclearborder function to remove the border pixels. When we do that:
% Clear off the border pixels and leave only the circular object
BW2 = imclearborder(BW);
figure; imshow(BW2);
... we now get this image:
Unfortunately, there are some noisy pixels, but we can very easily use morphology, specifically the opening operation with a small circular disk structuring element to remove these noisy pixels. Using strel with the appropriate structuring element in addition to imopen should help do the trick:
% Clear out noisy pixels
SE = strel('disk', 3, 0);
out = imopen(BW2, SE);
figure; imshow(out);
We now get:
This mask contains the locations of the circular object we now need to use to crop our original image. The last part is to determine the row and column locations using this mask to locate the top left and bottom right corner of the original image and we thus crop it:
% Find row and column locations of circular object
[row,col] = find(out);
% Find top left and bottom right corners
top_row = min(row);
top_col = min(col);
bottom_row = max(row);
bottom_col = max(col);
% Crop the image
crop = A(top_row:bottom_row, top_col:bottom_col);
% Show the cropped image
figure; imshow(crop);
We now get:
It's not perfect, but it will of course get you started. If you want to copy and paste this in its entirety and run this on your computer, here we are:
figure; % Open up a new figure
% Read in image and convert to grayscale
A = rgb2gray(imread('http://i.stack.imgur.com/vNECg.jpg'));
subplot(2,3,1); imshow(A);
title('Original Image');
% Find the optimum threshold via Otsu
level = graythresh(A);
% Inpaint noisy areas
I = imfill(A, 'holes');
subplot(2,3,2); imshow(I);
title('Inpainted image');
% Threshold the image
BW = im2bw(I, level);
subplot(2,3,3); imshow(BW);
title('Thresholded Image');
% Clear off the border pixels and leave only the circular object
BW2 = imclearborder(BW);
subplot(2,3,4); imshow(BW2);
title('Cleared Border Pixels');
% Clear out noisy pixels
SE = strel('disk', 3, 0);
out = imopen(BW2, SE);
% Show the final mask
subplot(2,3,5); imshow(out);
title('Final Mask');
% Find row and column locations of circular object
[row,col] = find(out);
% Find top left and bottom right corners
top_row = min(row);
top_col = min(col);
bottom_row = max(row);
bottom_col = max(col);
% Crop the image
crop = A(top_row:bottom_row, top_col:bottom_col);
% Show the cropped image
subplot(2,3,6);
imshow(crop);
title('Cropped Image');
... and our final figure is:
You can use bwdist with L_inf distance (aka 'chessboard') to get the axis-aligned distance to the edges of the region, thus concluding the dimensions of the largest bounded box:
bw = imread('http://i.stack.imgur.com/7yCaD.png');
lb = bwlabel(bw);
reg = lb==2; %// pick largest area
d = bwdist(~reg,'chessboard'); %// compute the axis aligned distance from boundary inward
r = max(d(:)); %// find the largest distance to boundary
[cy cx] = find(d==r,1); %// find the location most distant
boundedBox = [cx-r, cy-r, 2*r, 2*r];
And the result is
figure;
imshow(bw,'border','tight');
hold on;
rectangle('Position', boundedBox, 'EdgeColor','r');
Once you have the bounding box, you can use imcrop to crop the original image
imageCrop = imcrop(A, boundedBox);
Alternatively, you can
imageCrop = A(cy + (-r:r-1), cx + (-r:r-1) );
Related
I'm implementing image mosaic in Matlab using SURF.the problem is
outputView = imref2d(size(img1)*2);
Ir = imwarp(img2,tform,'OutputView',outputView);
it produces
i want it something like this
if i change
outputView = imref2d(size(img1)*2);
to
outputView = imref2d(size(img1));
matlab crops the second image so it can fit in first image size after transforming.
Notice that when you warp the image with respect to the target plane, many of the pixels in this new plane are equal to 0. A very rudimentary algorithm is to simply threshold your image so that you find values above 0 then find the largest bounding box that encompasses the non-zero pixels... then crop:
[rows,cols] = find(Ir(:,:,1) > 0);
topLeftRow = min(rows);
topLeftCol = min(cols);
bottomRightRow = max(rows);
bottomRightCol = max(cols);
Ir_crop = Ir(topLeftRow:bottomRightRow, topLeftCol:bottomRightCol, :);
I'd like to take in an RGB image, find the points in the image that are white, and get the cartesian coordinates of those points in the image. I've gotten most of the way there, but when I try to plot the cartesian coordinates, I get a vertically tiled image (i.e. 5 overlapped copies of what I should see). Anyone know what could be causing this?
,
Code: (JPG comes in as 2448 x x3264 x 3 uint8)
I = imread('IMG_0245.JPG');
imshow(I); % display unaltered image
% Convert image to grayscale
I = rgb2gray(I);
% Convert image to binary (black/white)
I = im2bw(I, 0.9);
% Generate cartesian coordinates of image
imageSize = size(I);
[x, y] = meshgrid( 1:imageSize(1), 1:imageSize(2) );
PerspectiveImage = [x(:), y(:), I(:)];
% Get indices of white points only
whiteIndices = find(PerspectiveImage(:,3));
figure; plot( PerspectiveImage(whiteIndices, 1), PerspectiveImage(whiteIndices, 2),'.');
% Flip vertically to correct indexing vs. plotting issue
axis ij
Very simple. You're declaring your meshgrid wrong. It should be:
[x, y] = meshgrid( 1:imageSize(2), 1:imageSize(1) );
The first parameter denotes the horizontal extents of the 2D grid, and so you want to make this vary for as many columns as you have. Similarly, the second parameter denotes the vertical extents of the 2D grid, and so you want to make this for as many rows as you have.
I had to pre-process some of your image to get some good results because your original image had a large white border surrounding the image. I had to remove this border by removing all pure white pixels. I also read in the image directly from StackOverflow:
I = imread('http://s7.postimg.org/ovb53w4ff/Track_example.jpg');
mask = all(I == 255, 3);
I = bsxfun(#times, I, uint8(~mask));
This was the image I get after doing my pre-processing:
Once I do this and change your meshgrid call, I get this:
I'm trying to crop an image but not with a rectangle (like in imcrop()) but with a polygon that has four corners. I've searched a lot and discovered that I need to perform a homography to reajust the cropped polygon into a rectangle.
So I've used imcrop() to select a polygon in an image :
img = imread('pout.tif');
imshow(img);
h = impoly;
position = wait(h);
x1 = min(position(:, 1));
x2 = max(position(:, 1));
y1 = min(position(:, 2));
y2 = max(position(:, 2));
BW = createMask(h);
How could I use these two things to crop out an area in the shape of a polygon with four corners ?
First of all, it is a bad idea to transform the image for cropping. It will results in changing the content of the ROI with artifacts due to interpolation when applying the homography. In addition, if one day you want to turn into a ROI defined by more than 4 points, this approach doesn't apply anylonger.
Second, I make some minor changes to your script, like this:
img = imread('circuit.tif');
imshow(img);
h = impoly;
position = wait(h);
boundbox = [min(position(:,1)), ....
min(position(:,2)), ....
max(position(:,1))-min(position(:,1)), ....
max(position(:,2))-min(position(:,2))];
BW = createMask(h);
img = imcrop(uint8(BW).*img, boundbox);
imshow(img)
You were almost there ... just mask the ROI of the image you want and crop with the bounding box of the ROI. Here it puts 0 outside the mask; you can adapt differently if you want.
Try "impoly" function in MATLAB
refer http://www.mathworks.in/help/images/ref/impoly.html
After I did a 'imclearborder', there are still a bit of unwanted object around the barcode. How can I remove those objects to isolate the barcode? I have pasted my code for your reference.
rgb = imread('barcode2.jpg');
% Resize Image
rgb = imresize(rgb,0.33);
figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = double(rgb2gray(rgb));
% Calculate the Gradients
[dIx, dIy] = gradient(Igray);
B = abs(dIx) - abs(dIy);
% Low-Pass Filtering
H = fspecial('gaussian', 20, 10);
C = imfilter(B, H);
C = imclearborder(C);
figure(),imagesc(C);colorbar;
Well, i have already explained it in your previous question How to find the location of red region in an image using MATLAB? , but with a opencv code and output images.
Instead of asking for code, try to implement it yourself.
Below is what to do next.
1) convert image 'C' in your code to binary.
2) Apply some erosion to remove small noises.( this time, barcode region also shrinks)
3) Apply dilation to compensate previous erosion.(most of noise will have removed in previous erosion. So they won't come back)
4) Find contours in the image.
5) Find their area. Most probably, contour which has maximum area will be the barcode, because other things like letters, words etc will be small ( you can understand it in the grayscale image you have provided)
6) Select contour with max. area. Draw a bounding rectangle for it.
Its result is already provided in your previous question. It works very nice. Try to implement it yourself with help of MATLAB documentation. Come here only when you get an error which you don't understand.
%%hi, i am ading my code to yours at the end of your code%%%%
clear all;
rgb = imread('barcode.jpeg');
% Resize Image
rgb = imresize(rgb,0.33);
figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = double(rgb2gray(rgb));
Igrayc = Igray;
% Calculate the Gradients
[dIx, dIy] = gradient(Igray);
B = abs(dIx) - abs(dIy);
% Low-Pass Filtering
H = fspecial('gaussian', 10, 5);
C = imfilter(B, H);
C = imclearborder(C);
imshow(Igray,[]);
figure(),imagesc(C);colorbar;
%%%%%%%%%%%%%%%%%%%%%%%%from here my code starts%%%%%%%%%%%%%%%%
bw = im2bw(C);%%%binarising the image
% imshow(bw);
%%%%if there are letters or any other noise is present around the barcode
%%Note: the size of the noise and letters should be smaller than the
%%barcode size
labelImage = bwlabel(bw,8);
len=0;labe=0;
for i=1:max(max(labelImage))
a = find(labelImage==i);
if(len<length(a))
len=length(a);
labe=i;
end
end
imag = zeros(size(l));
imag(find(labelImage==labe))=255;
% imtool(imag);
%%%if Necessary do errossion
% se2 = strel('line',10,0);
% imag= imerode(imag,se2);
% imag= imerode(imag,se2);
[r c]= find(imag==255);
minr = min(r);
maxc = max(c);
minc = min(c);
maxr = max(r);
imag1 = zeros(size(l));
for i=minr:maxr
for j=minc:maxc
imag1(i,j)=255;
end
end
% figure,imtool(imag1);
varit = find(imag1==0);
Igrayc(varit)=0;
%%%%%result image having only barcode
imshow(Igrayc,[]);
%%%%%original image
figure(),imshow(Igray,[]);
Hope it is useful
I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)