Hello, I have an image as shown above. Is it possible for me to detect the center point of the cross and output the result using Matlab? Thanks.
Here you go. I'm assuming that you have the image toolbox because if you don't then you probably shouldn't be trying to do this sort of thing. However, all of these functions can be implemented with convolutions I believe. I did this process on the image you presented above and obtained the point (139,286) where 138 is the row and 268 is the column.
1.Convert the image to a binary image:
bw = bw2im(img, .25);
where img is the original image. Depending on the image you might have to adjust the second parameters (which ranges from 0 to 1) so that you only get the cross. Don't worry about the cross not being fully connected because we'll remedy that in the next step.
2.Dilate the image to join the parts. I had to do this twice because I had to set the threshold so low on the binary image conversion (some parts of your image were pretty dark). Dilation essentially just adds pixels around existing white pixels (I'll also be inverting the binary image as I send it into bwmorph because the operations are made to act on white pixels which are the ones that have a value of 1).
bw2 = bwmorph(~bw, 'dilate', 2);
The last parameter says how many times to do the dilation operation.
3.Shrink the image to a point.
bw3 = bwmorph(bw2, 'shrink',Inf);
Again, the last parameter says how many times to perform the operation. In this case I put in Inf which shrinks until there is only one pixel that is white (in other words a 1).
4.Find the pixel that is still a 1.
[i,j] = find(bw3);
Here, i is the row and j is the column of the pixel in bw3 such that bw3(i,j) is equal to 1. All the other pixels should be 0 in bw3.
There might be other ways to do this with bwmorph, but I think that this way works pretty well. You might have to adjust it depending on the picture too. I can include images of each step if desired.
I just encountered the same kind of problem, and I found other solutions that I would like to share:
Assume image file name is pict1.jpg.
1.Read input image, crop relevant part and covert to Gray-scale:
origI = imread('pict1.jpg'); %Read input image
I = origI(32:304, 83:532, :); %Crop relevant part
I = im2double(rgb2gray(I)); %Covert to Grayscale and to double (set pixel range [0, 1]).
2.Convert image to binary image in robust approach:
%Subtract from each pixel the median of its 21x21 neighbors
%Emphasize pixels that are deviated from surrounding neighbors
medD = abs(I - medfilt2(I, [21, 21], 'symmetric'));
%Set threshold to 5 sigma of medD
thresh = std2(medD(:))*5;
%Convert image to binary image using above threshold
BW = im2bw(medD, thresh);
BW Image:
3.Now I suggest two approaches for finding the center:
Find find centroid (find center of mass of the white cluster)
Find two lines using Hough transform, and find the intersection point
Both solutions return sub-pixel result.
3.1.Find cross center using regionprops (find centroid):
%Find centroid of the cross (centroid of the cluster)
s = regionprops(BW, 'centroid');
centroids = cat(1, s.Centroid);
figure;imshow(BW);
hold on, plot(centroids(:,1), centroids(:,2), 'b*', 'MarkerSize', 15), hold off
%Display cross center in original image
figure;imshow(origI), hold on, plot(82+centroids(:,1), 31+centroids(:,2), 'b*', 'MarkerSize', 15), hold off
Centroid result (BW image):
Centroid result (original image):
3.2 Find cross center by intersection of two lines (using Hough transform):
%Create the Hough transform using the binary image.
[H,T,R] = hough(BW);
%ind peaks in the Hough transform of the image.
P = houghpeaks(H,2,'threshold',ceil(0.3*max(H(:))));
x = T(P(:,2)); y = R(P(:,1));
%Find lines and plot them.
lines = houghlines(BW,T,R,P,'FillGap',5,'MinLength',7);
figure, imshow(BW), hold on
L = cell(1, length(lines));
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','green');
% Plot beginnings and ends of lines
plot(xy(1,1),xy(1,2),'x','LineWidth',2,'Color','yellow');
plot(xy(2,1),xy(2,2),'x','LineWidth',2,'Color','red');
%http://robotics.stanford.edu/~birch/projective/node4.html
%Find lines in homogeneous coordinates (using cross product):
L{k} = cross([xy(1,1); xy(1,2); 1], [xy(2,1); xy(2,2); 1]);
end
%https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection
%Lines intersection in homogeneous coordinates (using cross product):
p = cross(L{1}, L{2});
%Convert from homogeneous coordinate to euclidean coordinate (divide by last element).
p = p./p(end);
plot(p(1), p(2), 'x', 'LineWidth', 1, 'Color', 'white', 'MarkerSize', 15)
Hough transform result:
I think that there is a far simpler way of solving this. The lines which form the cross-hair are of equal length. Therefore it in will be symmetric in all orientations. So if we do a simple line scan horizontally as well as vertically, to find the extremities of the lines forming the cross-hair. the median of these values will give the x and y co-ordinates of the center. Simple geometry.
I just love these discussions of how to find something without defining first what that something is! But, if I had to guess, I’d suggest the center of mass of the original gray scale image.
What about this;
a) convert to binary just to make the algorithm faster.
b) Perform a find on the resulting array
c) choose the element which has either lowest/highest row/column index (you would have four points to choose from then
d) now keep searching neighbours
have a global criteria for search that if search does not result in more than a few iterations, the point selected is false and choose another extreme point
e) going along the neighbouring points, you will end up at a point where you have three possible neighbours.That is you intersection
I would start by using the grayscale image map. The darkest points are on the cross, so discriminating on the highest values is a starting point. After discrimination, set all the lower points to white and leave the rest as they are. This would maximize the contrast between points on the cross and points in the image. Next up is to come up with a filter for determining the position with the highest average values. I would step through the entire image with a NxM array and take the mean value at the center point. Create a new array of these means and you should have the highest mean at the intersection. I'm curious to see how someone else may try this!
Related
I have fitted a bounding box on image which have concerned area is human silhouette in BW, and found centroid according to concerned human silhouette?
now i have to plot red line by 45 degree angle from centroid.
from centroid to vertical, horizental and diagonal logic? Need code in MATLAB?
I have fitted a bounding box on image and found centroid according to concerned human silhouette?
% bounding box
labeledImage = bwlabel(Ibw);
blobMeasurements = regionprops(labeledImage, 'BoundingBox');
thisBlobsBoundingBox = blobMeasurements.BoundingBox;
subImage = imcrop(Ibw, thisBlobsBoundingBox);
figure, imshow(subImage);
imwrite(subImage,fullfile(cd, strcat('Croped By BoundingBox','.png')));
%centroid
Ibw = imread('Croped By BoundingBox.png');
Ibw = imfill(Ibw,'holes');
Ilabel = bwlabel(Ibw);
stat = regionprops(Ilabel,'centroid');
imshow(Ibw),hold on;
for x = 1: numel(stat)
plot(stat(x).Centroid(1),stat(x).Centroid(2),'ro');
[rows, cols] =ndgrid(1:size(Ibw, 1), 1:size(Ibw, 2));
centroidrowcol = mean([rows(:) .* Ibw(:), cols(:) .* Ibw(:)]);
hold on
end
figure, imshow(Ibw);
imwrite(Ibw,fullfile(cd, strcat('Centroid','.png')));
plot lines from centroid to vercally horizentally and diagonally by 45 degree from centroid.
i have to obtaon these results
enter image description here
I'll start with assuming centroid, a length-2 vector, is the coordinates of a centroid -- since you already have them.
As a reminder, 45 degree lines centered at (0,0) are the images of the functions f(x)=x and f(x)=-x. In order to shift the lines to intercept points of your choice, say (x0, y0), you need f(x-x0)+y0. In other words, f(x)=x-x0+y0 and f(x)=-x+x0+y0 respectively.
Knowing the functions (whose images are) the lines of your desire, there are a number of things you can do in order to complete the task. You can define Matlab functions as such and plot the usual way with plot, or for a quick display, with ezplot:
ezplot(#(x)x-centroid(1)+centroid(2));
line() in Matlab allows you to draw a line segment between two points in 2D, among other things. With line(), you actually don't have to know the equations explicitly. You just need two points. So, one possibility is that you use [centroid(1)+1, centroid(2)+1] as your first point and [centroid(1)-1, centroid(2)-1] as your second point for the shifted f(x)=x line.
Just read the documentation for the correct syntax. In the above example, you would put in
line([centroid(1)+1;centroid(1)-1], [centroid(2)+1;centroid(2)-1]);
This draws a 45 degree line segment alright. You also want the length of the line segment to be visually sensible. You can either autoscale the length, ie. change the 1 into a formula, based on the size of your image. OR, you can make sure your segment is longer than the extent of the image but confine your plot to a certain size by changing xlim and ylim.
I have this BW image:
And using the function RegionProps, it shows that some objetcs are connected:
So I used morphological operations like imerode to separte the objects to get their centroids:
Now I have all the centroids of each object separated, but to that I lost a lot of information when eroding the region, like you can see in picture 3 in comparison with picture 1.
So I was thinking if is there anyway to "dilate" the picture 3 till get closer to picture 1 but without connecting again the objects.
You might want to take a look at bwmorph(). With the 'thicken', inf name-value pair it will thicken the labels until they would overlap. It's a neat tool for segmentation. We can use it to create segmentation borders for the original image.
bw is the original image.
labels is the image of the eroded labels.
lines = bwmorph(labels, 'thicken', inf);
segmented_bw = bw & lines
You could also skip a few phases and achieve similiar results with a marker based watershed. Or even better, as the morphological seesaw has destroyed some information as seen on the poorly segmented cluster on the lower right.
You can assign each white pixel in the mask to the closest centroid and work with the resulting label map:
[y x]= find(bw); % get coordinates of mask pixels
D = pdist2([x(:), y(:)], [cx(:), cy(:)]); % assuming cx, cy are centers' coordinates
[~, lb] = min(D, [], 2); % find index of closest center
lb_map = 0*bw;
lb_map(bw) = lb; % should give you the map.
See pdist2 for more information.
Hello I am working with matlab. I am trying to generate a bounding box around a silhouette. The problem here is that the silhouette is fragmented
as shown here
The code i tried is
BW=bwconncomp(image);
STATS = regionprops(BW, 'FilledArea','BoundingBox');
which gives me a bounding box around a part of the silhouette. I cannot use dilate which is the preferred morphological operation in this case as it connects the silhouette with neighboring fragments.
Thanks in advance for the help.
Here is something to get you going with the image you posted. I used a line structuring element with an angle to dilate the image and amplify the signal from the small white chunks at the left of the silhouette. Then using regionprops its easier to identify objects individually and select the object with the largest area (i.e. the silhouette), calculated with the property FilledArea, and report back the bounding box on the original image. It might not be perfect but it's a start and it seems to give a pretty decent result.
Here is the code:
clear
clc
close all
BW = im2bw(imread('Silhouette.png'));
BW = imclearborder(BW);
%// Dilate with a line structuring element oriented at about 60 degrees to
%// amplify the elements at an angle that you don't want.
se = strel('line',5,60);
dilateddBW = imdilate(BW,se);
figure;
imshow(dilateddBW)
The dilated image looks like this:
Calling regionprops and displaying the output:
%// Get the region properties and select that with the largest area.
S = regionprops(dilateddBW,'BoundingBox','FilledArea','PixelIdxList');
boundingboxes = cat(1, S.BoundingBox);
FilledAreas = cat(1,S.FilledArea);
[~,MaxAreaIndex] = max(FilledAreas);
%// Get linear indices of the corresponding silhouette to display along
%// with its bounding box.
MaxIndices = S(MaxAreaIndex).PixelIdxList;
%// Create empty image to put the silhouette + box
NewIm = false(size(dilateddBW));
NewIm(MaxIndices) = 1;
figure;
imshow(BW)
rectangle('Position',boundingboxes(MaxAreaIndex,:),'EdgeColor','r')
Output:
Hope that helps somehow!
Since you have an array of vectors containing indice of pixels(bwconncomp() returns a struct which have a member named PixelIdxList), you can create a rectangle by finding pixels with min x, min y, max x, max y.
Here is a good example: 2D Minimal Bounding Box
In my progress work, I have to detect a parasite. I have found the parasite using HSV and later made it into a grey image. Now I have done edge detection too. I need some code which tells MATLAB to find the largest contour (parasite) and make the rest of the area as black pixels.
You can select the "largest" contour by filling in the holes that each contour surrounds, figure out which shape gives you the largest area, then use the locations of the largest area and copy that over to a final image. As what Benoit_11 suggested, use regionprops - specifically the Area and PixelList flags. Something like this:
im = imclearborder(im2bw(imread('http://i.stack.imgur.com/a5Yi7.jpg')));
im_fill = imfill(im, 'holes');
s = regionprops(im_fill, 'Area', 'PixelList');
[~,ind] = max([s.Area]);
pix = sub2ind(size(im), s(ind).PixelList(:,2), s(ind).PixelList(:,1));
out = zeros(size(im));
out(pix) = im(pix);
imshow(out);
The first line of code reads in your image from StackOverflow directly. The image is also a RGB image for some reason, and so I convert this into binary through im2bw. There is also a white border that surrounds the image. You most likely had this image open in a figure and saved the image from the figure. I got rid of this by using imclearborder to remove the white border.
Next, we need to fill in the areas that the contour surround, so use imfill with the holes flag. Next, use regionprops to analyze the different filled objects in the image - specifically the Area and which pixels belong to each object in the filled image. Once we obtain these attributes, find the filled contour that gives you the biggest area, then access the correct regionprops element, extract out the pixel locations that belong to the object, then use these and copy over the pixels to an output image and display the results.
We get:
Alternatively, you can use the Perimeter flag (as what Benoit_11) suggested, and simply find the maximum perimeter which will correspond to the largest contour. This should still give you what you want. As such, simply replace the Area flag with Perimeter in the third and fourth lines of code and you should still get the same results.
Since my answer was pretty much all written out I'll give it to you anyway, but the idea is similar to #rayryeng's answer.
Basically I use the Perimeter and PixelIdxList flags during the call to regionprops and therefore get the linear indices of the pixels forming the largest contour, once the image border has been removed using imclearborder.
Here is the code:
clc
clear
BW = imclearborder(im2bw(imread('http://i.stack.imgur.com/a5Yi7.jpg')));
S= regionprops(BW, 'Perimeter','PixelIdxList');
[~,idx] = max([S.Perimeter]);
Indices = S(idx).PixelIdxList;
NewIm = false(size(BW));
NewIm(Indices) = 1;
imshow(NewIm)
And the output:
As you see there are many ways to achieve the same result haha.
This could be one approach -
%// Read in image as binary
im = im2bw(imread('http://i.stack.imgur.com/a5Yi7.jpg'));
im = im(40:320,90:375); %// clear out the whitish border you have
figure, imshow(im), title('Original image')
%// Fill all contours to get us filled blobs and then select the biggest one
outer_blob = imfill(im,'holes');
figure, imshow(outer_blob), title('Filled Blobs')
%// Select the biggest blob that will correspond to the biggest contour
outer_blob = biggest_blob(outer_blob);
%// Get the biggest contour from the biggest filled blob
out = outer_blob & im;
figure, imshow(out), title('Final output: Biggest Contour')
The function biggest_blob that is based on bsxfun is an alternative to what other answers posted here perform with regionprops. From my experience, I have found out this bsxfun based technique to be faster than regionprops. Here are few benchmarks comparing these two techniques for runtime performances on one of my previous answers.
Associated function -
function out = biggest_blob(BW)
%// Find and labels blobs in the binary image BW
[L, num] = bwlabel(BW, 8);
%// Count of pixels in each blob, basically should give area of each blob
counts = sum(bsxfun(#eq,L(:),1:num));
%// Get the label(ind) cooresponding to blob with the maximum area
%// which would be the biggest blob
[~,ind] = max(counts);
%// Get only the logical mask of the biggest blob by comparing all labels
%// to the label(ind) of the biggest blob
out = (L==ind);
return;
Debug images -
How can i calculate the average of a certain area in an image using mat-lab?
For example, if i have an intensity image with an area that is more alight and i want to know what is the average of the intensity there- how do i calculate it?
I think i can find the coordinates of the alight area by using the 'impixelinfo' command.
If there is another more efficient way to find the coordinates i will also be glad to know.
After i know the coordinates how do i calculate the average of part of the image?
You could use one of the imroi type functions in Matlab such as imfreehand
I = imread('cameraman.tif');
h = imshow(I);
e = imfreehand;
% now select area on image - do not close image
% this makes a mask from the area you just drew
BW = createMask(e);
% this takes the mean of pixel values in that area
I_mean = mean(I(BW));
Alternatively, look into using regionprops, especially if there's likely to be more than one of these features in the image. Here, I'm finding points in the image above some threshold intensity and then using imdilate to pick out a small area around each of those points (presuming the points above the threshold are well separated, which may not be the case - if they are too close then imdilate will merge them into one area).
se = strel('disk',5);
BW = imdilate(I>thresh,se);
s = regionprops(BW, I, 'MeanIntensity');