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.
Related
I have to make a sinusoidal curve in an image to output an equal straight line in the resulting image.
An example of input sinusoidal image:
What I think is one solution should be:
Placing down the origin of x and y coordinates at the start of the curve, so we will have y=0 at the starting point. Then points on the upper limit will be counted as such that y= y-(delta_y) and for lower limits, y=y+(delta_y)
So to make upper points a straight line, our resulting image will be:
O[x,y-delta_y]= I[x,y];
But how to calculate deltaY for each y on horizontal x axis (it is showing the distance of curve points from horizontal axis)
Another solution could be, to save all information of the curve in a variable and to plot it as a straight line, but how to do it?
Since the curve is blue we can use information from the blue and red channels to extract the curve. Simply subtraction of red channel from blue channel will highlight the curve:
a= imread('kCiTx.jpg');
D=a(:,:,3)-a(:,:,1);
In each column of the image position of the curve is index of the row that it's value is the maximum of that column
[~,im]=max(D);
so we can use row position to shift each column so to create a horizontal line. shifting each column potentially increases size of the image so it is required to increase size of the image by padding the image from top and bottom by the size of the original image so the padded image have the row size of 3 times of the original image and padding value is 255 or white color
pd = padarray(a,[size(a,1) 0 0], 255);
finally for each channel cirshift each column with value of im
for ch = 1:3
for col = 1: size(a,2)
pd(:,col,ch) = circshift(pd(:,col,ch),-im(col));
end
end
So the result will be created with this code:
a= imread('kCiTx.jpg');
D=a(:,:,3)-a(:,:,1);
%position of curve found
[~,im]=max(D);
%pad image
pd = padarray(a,[size(a,1) 0 0], 255);
%shift columns to create a flat curve
for ch = 1:3
for col = 1: size(a,2)
pd(:,col,ch) = circshift(pd(:,col,ch),-im(col));
end
end
figure,imshow(pd,[])
If you are sure you have a sinusoid in your initial image, rather than calculating piece-meal offsets, you may as well estimate the sinusoidal equation:
amplitude = (yMax - yMin)/2
offset = (yMax + yMin)/2
(xValley needs to be the valley immediately after xPeak, alternately you could do peak to peak, but it changes the equation, so this is the more compact version (ie you need to see less of the sinusoid))
frequencyScale = π / (xValley - xPeak)
frequencyShift = xFirstZeroCrossRising
If you are able to calculate all of those, this is then your equation:
y = offset + amplitude * sin(frequencyScale * (x + frequencyShift))
This equation should be all you need to store from one image to be able to shift any other image, it can also be used to generate your offsets to exactly cancel your sinusoid in your image.
All of these terms should be able to be estimated from the image with relatively little difficulty. If you can not figure out how to get any of the terms from the image, let me know and I will explain that one in particular.
If you are looking for some type of distance plot:
Take your first point on the curvy line and copy it into your output image, then measure the distance between that point and the next point on the (curvy) line. Use that distance to offset (from the first point) that next point into the output image only along the x axis. You may want to do it pixel by pixel, or grab clumps of pixels through averaging or jumping (as pixel by pixel will give you some weird digital noise)
If you want to be cleaner, you will want to set up a step size which was sufficiently small to approximately match the maximum sinusoidal curvature without too much error. Then, estimate the distance as stated above to set up bounds and then interpolate each pixel between the start and end point into the image averaging into bins based on approximate position. IE if a pixel from the original image would fall between two bins in the output image, you would split it and add its weighted parts to those two bins.
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');
I have a 2d image, I have locations where local minimas occurs.
I want to measure the width of the valleys "leading" to those minimas.
I need either the radii of the circles or ellipses fitted to these valley.
An example attached here, dark red lines on the peaks contours is what I wish to find.
Thanks.
I am partially extending the answer of #Lucas.
Given a threshold t I would consider the points P_m that are below t and closer to a certain point m of minimum of your f (given a characteristic scale length r).
(You said your data are noisy; to distinguish minima and talk about wells, you need to estimate such r. In your example it can be for instance r=4, i.e. half the distance between the minima).
Then you have to consider a metric for each well region P_m, say for example
metric(P_m) = .5 * mean{ maximum vertical diameter of P_m ,
maximum horizontal diameter of P_m}.
In your picture metric(P_m) = 2 for both wells.
On the whole, in terms of pseudo-code you may consider
M := set of local minima of f
for_each(minimum m in M){
P_m += {p : d(p,m) < r and f(r)<t} % say that += is the push operation in a Stack
}
radius_of_region_around(m) = metric(P_m); %
I would suggest making a list of points that describe the values at the edge of your ellipse, perhaps by finding all the points where it crosses a threshold.
above = data > threshold
apply a simple edge detector
edges = EdgeDetector(above)
find coordinates of edges
[row,col] = find(edges)
Then apply this ellipse fitter http://www.mathworks.com/matlabcentral/fileexchange/3215-fitellipse
I'm assuming here you have access to the x, y and z data and are not processing a given JPG (or so) image. Then, you can use the function contourc to your advantage:
% plot some example function
figure(1), clf, hold on
[x,y,z] = peaks;
surf(x,y,z+10,'edgecolor', 'none')
grid on, view(44,24)
% generate contour matrix. The last entry is a 2-element vector, the last
% element of which is to ensure the right algorithm gets called (so leave
% it untouched), and the first element is your threshold.
C = contourc(x(1,:), y(:,1), z, [-4 max(z(:))+1]);
% plot the selected points
plot(C(1,2:end), C(2,2:end), 'r.')
Then use this superfast ellipse fitting tool to fit an ellipse through those points and find all the parameters of the ellipse you desire.
I suggest you read help contourc and doc contourc to find out why the above works, and what else you can use it for.
Using matlab's FILL function creates a filled region confined by a polygon with straight edges:
Unfortunately this leaves a small white region in the figure above, because the boundary of the region I want filled in is not a straight-edged polygon, but rather has a curved boundary on the left side. I have a curve (nearly parabolic but not exactly), and I want to fill in the region between two horizontal lines AND the curve itself. I also looked into the MATLAB function IMFILL, but with no luck.
What you need to do is make a polygon with more corners, so that it fits the curve more smoothly:
%# create a parabola and two straight lines
x = -3:0.1:3;
y = x.^2/4;
plot(x,y)
hold on, plot([-3 3],[1 1],'r',[-3 3],[2 2],'r')
%# create a polygon that hugs the parabola
%# note that we need to interpolate separately
%# for positive and negative x
x1 = interp1(y(x<0),x(x<0),1:0.1:2);
%# interpolate in reverse so that the corners are properly ordered
x2 = interp1(y(x>0),x(x>0),2:-0.1:1);
%# fill the area bounded by the three lines
fill([x1,x2],[1:0.1:2,2:-0.1:1],'g')
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!