how get centroid contour distance at every 10 degrees?

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Diah 2012-12-3

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评论： Muhammad Andi Yusran 2019-12-9

I have an image and I already get the boundary/edge and the weighted centroid of region of interest. I want to get the coordinate point of boundary at every 10 degrees. It starts from the fartest distance of center to edge. I have tried, but the result is not appropriate with my goal. Can anyone help me about this? Thx for the help

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Image Analyst 2016-1-8

shape_recognition_demo.m

See attached shape recognition demo. One of the ways it calculates the number of sides it to look at the centroid-to-perimeter distance and look for valleys and peaks and take the larger of them. Look at function FindNumberOfVertices(). It's meant to work with a labeled image where you pass in the labeled image and the blob number you want it to inspect. If you have questions, then start a new question and attach your image and segmentation code.

poonam 2016-2-11

'MinPeakProminence' is not supported in 'matlab13a'. is there any other way to get minimum and maximum distance in graph.

This new thread is started for this error.

https://in.mathworks.com/matlabcentral/answers/267617-problem-with-minpeakvariance-in-findpeak

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Answer 1

Image Analyst 2012-12-3

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So you said that you already got the boundary/edge - presumably from using bwboundaries() - and you already got the centroid - presumably from using regionprops, so what help do you need? You just need to run around the boundary and for each boundary point, calculate the angle using simple trigonometry and the distance using the Pythagorean theorem or hypot(). It's just a simple for loop. Write back if you still need help figuring it out.

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Diah 2012-12-3

编辑：Diah 2012-12-3

在 MATLAB Online 中打开

Yes, It like so simple, but I can't solve it. Hiks... Here are the code I have.

clear  %clear workspace & variable editor
clc   %clear command window
close all;
img1=imread('3.jpeg')
iw=img1
img2=iw(:,:,1)>10 | iw(:,:,2)>10 | iw(:,:,3)>10;
ir=bwmorph(img2,'erode',2);
ir2=bwmorph(ir,'dilate',2);
ir3=imfill(ir2,'holes')
img3=bwperim(ir3,8)
cc = bwconncomp(ir3);
stats = regionprops(cc, 'all');
luas=stats.Area
perimeter=stats.Perimeter
% look for centroid
B = bwboundaries(img3); 
boundary=B{1};
[m n]=size(boundary);
gy=(sum(boundary(:,1)))/m
gx=(sum(boundary(:,2)))/m
for i=1:m
    xbar=(boundary(i,1)-gx)^2
    ybar=(boundary(i,2)-gy)^2
    boundary(i,3)=sqrt(xbar+ybar)
end
co=1
[maxccond indx]=max(boundary(i,3))
for i=1:m
a=boundary(indx,3)/boundary(i,3)
theta(i,1)=acosd(a)
if theta(i,1)==0 | theta(i,1)==10 | theta(i,1)==20 | theta(i,1)==30 | theta(i,1)==40 | theta(i,1)==50 | theta(i,1)==60 | theta(i,1)==70 |theta(i,1)==80 | theta==90 | theta==100 | theta==110 |theta==120 | theta==130 | theta==140 | theta==150 |theta==160 | theta==170 | theta==180 | theta==190 |theta==200 | theta==210 | theta==220 | theta==230 |theta==240 | theta==250 | theta==260 | theta==270 |theta==280 | theta==290 | theta==300 | theta==310 |theta==320 | theta==330 | theta==340 | theta==350 |theta==360
    ccd(co,1)=theta(i,1)
    ccd(co,2)=boundary(i,2)
    ccd(co,3)=boundary(i,1)
end
end

At ccd, I must have the degrees every 10 degrees. But the code only show 1 row, at 0 degrees. Then I check theta, at row 513, it values 48.762026742399100 + 0.000000000000000i. It happens at all row after 513. I have 941 rows.

This is an image that I used: http://imgur.com/BsOZW

Image Analyst 2012-12-4

在 MATLAB Online 中打开

No, that really won't get the job done. Try it this way instead. It's well commented so I think you can follow along:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
imtool close all;  % Close all imtool figures.
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
format longg;
format compact;
fontSize = 20;
% Read in the color image.
folder = 'C:\Users\Diah\Documents';
baseFileName = 'flower.jpg';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, 'file')
  % Didn't find it there.  Check the search path for it.
  fullFileName = baseFileName; % No path this time.
  if ~exist(fullFileName, 'file')
    % Still didn't find it.  Alert user.
    errorMessage = sprintf('Error: %s does not exist.', fullFileName);
    uiwait(warndlg(errorMessage));
    return;
  end
end
rgbImage = imread(fullFileName);
% Get the dimensions of the image.  numberOfColorBands should be = 3.
[rows columns numberOfColorBands] = size(rgbImage);
% Display the original color image.
subplot(2, 2, 1);
imshow(rgbImage, []);
axis on;
title('Original Color Image', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Get the binary image by thresholding the red channel.
binaryImage = rgbImage(:,:,1) > 34;
subplot(2,2,2);
imshow(binaryImage, []);
axis on;
title('Binary Image', 'FontSize', fontSize);
[labeledImage numberOfBlobs] = bwlabel(binaryImage, 8);     % Label each blob so we can make measurements of it
% Get the centroid.
blobMeasurements = regionprops(labeledImage, 'Centroid');
xCenter = blobMeasurements(1).Centroid(1);
yCenter = blobMeasurements(1).Centroid(2);
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
% Plot the borders of all the coins on the original grayscale image using the coordinates returned by bwboundaries.
subplot(2,2,3);
imshow(rgbImage);
axis on;
title('Outlines, from bwboundaries()', 'FontSize', fontSize);
hold on;
% Plot the centroid.
plot(xCenter, yCenter, 'r+', 'MarkerSize', 20, 'LineWidth', 3);
boundaries = bwboundaries(binaryImage);
numberOfBoundaries = size(boundaries);
for k = 1 : numberOfBoundaries
  thisBoundary = boundaries{k};
  plot(thisBoundary(:,2), thisBoundary(:,1), 'r', 'LineWidth', 3);
  % for each boundary, calculate the angle an distance as a function of boundary point.
  numberOfBoundaryPoints = length(thisBoundary);
  angles = zeros(1, numberOfBoundaryPoints);
  distances = zeros(1, numberOfBoundaryPoints);
  for p = 1 : numberOfBoundaryPoints
    xb = thisBoundary(p,2);
    yb = thisBoundary(p,1);
    angles(p) = atand((yb-yCenter) / (xb-xCenter));
    distances(p) = sqrt((xb-xCenter)^2+(yb-yCenter)^2);
  end
  % Plot them
  subplot(4, 2, 6);
  plot(angles, 'LineWidth', 3);
  caption = sprintf('Angles for blob #%d', k);
  title(caption, 'FontSize', fontSize);
  grid on;
  subplot(4, 2, 8);
  plot(distances, 'LineWidth', 3);
  grid on;
  caption = sprintf('Distances for blob #%d', k);
  title(caption, 'FontSize', fontSize);
  % Prompt user for the next blob, if there is one.
  if numberOfBoundaries > 1
    promptMessage = sprintf('Do you want to Continue processing,\nor Cancel processing?');
    titleBarCaption = 'Continue?';
    button = questdlg(promptMessage, titleBarCaption, 'Continue', 'Cancel', 'Continue');
    if strcmpi(button, 'Cancel')
      break;
    end
  end
end
hold off;

Diah 2012-12-4

在 MATLAB Online 中打开

Yes walter, the calculation of the centroid can use regionprops. But I want to try the formula that i read on paper.

To handle the result of acosd(), how about use round(acosd())? To handle multiple boundary pixels at the same angle, may be i will use the shortest distance from the center.

Sorry, my code above have a little mistake at [maxccond indx]=max(boundary(i,3)). It should be [maxccond indx]=max(boundary(:,3)). So, the true code that i have run is:

clear  %clear workspace & variable editor
clc   %clear command window
close all;
img1=imread('DataBunga\1\1\3.jpeg')
iw=img1
img2=iw(:,:,1)>10 | iw(:,:,2)>10 | iw(:,:,3)>10;
ir=bwmorph(img2,'erode',2);
ir2=bwmorph(ir,'dilate',2);
ir3=imfill(ir2,'holes')
img3=bwperim(ir3,8)
cc = bwconncomp(ir3);
stats = regionprops(cc, 'all');
luas=stats.Area
perimeter=stats.Perimeter
% look for centroid
B = bwboundaries(img3); 
boundary=B{1};
[m n]=size(boundary);
gy=(sum(boundary(:,1)))/m
gx=(sum(boundary(:,2)))/m
for i=1:m
    xbar=(boundary(i,1)-gx)^2
    ybar=(boundary(i,2)-gy)^2
    boundary(i,3)=sqrt(xbar+ybar)
end
co=1
[maxccond indx]=max(boundary(:,3))
for i=1:m
a=boundary(indx,3)/boundary(i,3)
theta(i,1)=acosd(a)
if theta(i,1)==0 | theta(i,1)==10 | theta(i,1)==20 | theta(i,1)==30 | theta(i,1)==40 | theta(i,1)==50 | theta(i,1)==60 | theta(i,1)==70 |theta(i,1)==80 | theta==90 | theta==100 | theta==110 |theta==120 | theta==130 | theta==140 | theta==150 |theta==160 | theta==170 | theta==180 | theta==190 |theta==200 | theta==210 | theta==220 | theta==230 |theta==240 | theta==250 | theta==260 | theta==270 |theta==280 | theta==290 | theta==300 | theta==310 |theta==320 | theta==330 | theta==340 | theta==350 |theta==360
    ccd(co,1)=theta(i,1)
    ccd(co,2)=boundary(i,2)
    ccd(co,3)=boundary(i,1)
end
end

All the values of tabel theta give the number 0+.... I can't see the 60 degrees, 90 degrees, 270 degrees, etc. Only row 209 that have 0 degrees, that means the starting poin (fartest distance). How I get the other degrees or how i can get the edge coordinate at every 10 degrees?

Image Analyst 2012-12-4

编辑：Image Analyst 2012-12-4

在 MATLAB Online 中打开

Diah, I don't know what paper told you that you could get the centroid using this method:

gy=(sum(boundary(:,1)))/m
gx=(sum(boundary(:,2)))/m

where you say the centroid is the average of the x and the average of the y values, but it's wrong. Walter and I have both told you to use regionprops. To prove it, run this code and look where your code puts the centroid and compare it with where regionprops correctly locates it.

binaryImage = false(400, 400);
binaryImage(50:200, 100:300) = true;
binaryImage(50:350, 301:302) = true;
imshow(binaryImage);
% Get centroid via bwboundaries and incorrect method.
boundaries = bwboundaries(binaryImage);
boundary = boundaries{1};
[m n]=size(boundary)
gy=(sum(boundary(:,1)))/m
gx=(sum(boundary(:,2)))/m
hold on;
plot(gx, gy, 'r+');
text(gx+10, gy, 'average method', 'Color', [1 0 0]);
% Get centroid correctly via regionprops
measurements = regionprops(binaryImage, 'Centroid');
centroid = [measurements.Centroid];
xCenter = centroid(1)
yCenter = centroid(2)
plot(xCenter, yCenter, 'b+');
text(xCenter+10, yCenter, 'regionprops method', 'Color', [0 0 1]);

Diah 2012-12-6

Walter, class(boundary) is double

Mr. Image Analyst, about gx and gy, I have read "An Interactive Flower Image Recognition System" by TH Hsu from http://link.springer.com/article/10.1007%2Fs11042-010-0490-6?LI=true.

To get the shape features, we first define the centroid (gx, gy) of the flower region as the flower center, which is computed as follows:

gx=1/N(sum of Xi from i=1 until i=N), gy=1/N(sum of Yi from i=1 until i=N)

where N is the number of pixels on the flower boundary, Xi and Yi are respectively the X and Y coordinates of the i-th boundary pixel (TH Hsu, 2010).

If I try regionprops and average sum to my data, it is not significantly different. Ok, thank you Mr. Image Analyst and Walter for the advice using Regionprops. It gives me more knowledge.

“You need to look at where each point is located. For example a vertical line could be +90 or -90, but if you know which point it your origin, it's simple. It's just simple 10th grade trigonometry. If you still can't figure it out, let us know”. --> I still can’t figure it. :-( I'm sorry, I just a beginner in matlab. :-(

Image Analyst 2012-12-7

在 MATLAB Online 中打开

Diah, you can run my demo and see for yourself. Whose algorithm do you believe gave a centroid closer to what you would expect? I think you'll find that it's our regionprops method and not their method.

You don't need to know MATLAB to come up with a series of criteria about where points are located to figure out what quadrant they are in and what number you need to add to get it into the 0-359.9999999 range. Nonetheless, run this demo and see what happens:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
format longg;
format compact;
fontSize = 20;
angles = 0 : 10 : 360;
x = cosd(angles);
y = sind(angles);
plot(x, y, 'ro-');
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
grid on;
xCenter = 0;
yCenter = 0;
for k = 1 : length(angles)
  % Draw a line from the center to the edge of the circle.
  line([xCenter x(k)], [yCenter y(k)], 'LineWidth', 3);
  % Calculate the angle of that line
  angle = atand(y(k)/x(k));
  % Convert to 0-360
  if y(k) >= yCenter && x(k) >= xCenter
    quadrant = 1;
  elseif y(k) >= yCenter && x(k) <= xCenter
    angle = 180 + angle;
    quadrant = 2;
  elseif y(k) <= yCenter && x(k) < xCenter
    angle = 180 + angle;
    quadrant = 3;
  elseif y(k) <= yCenter && x(k) >= xCenter
    angle = 360 + angle;
    quadrant = 4;
  end
  promptMessage = sprintf('This latest line, going from (0,0) to (%.8f, %.8f),\nis at angle %f, and in quadrant %d.', ...
    x(k), y(k), angle, quadrant);
  fprintf('%s\n', promptMessage);
  titleBarCaption = 'Continue?';
  button = questdlg(promptMessage, titleBarCaption, 'Continue', 'Cancel', 'Continue');
  if strcmp(button, 'Cancel')
    break;
  end
end

Ruksana siddique 2019-3-7

demo 1.png

hello mr. walter, i also want to draw the angle but in the image at 30 degree. more i want to crop each angle part. i am showing the image like this

Muhammad Andi Yusran 2019-12-9

在 MATLAB Online 中打开

hello Mr. Image Analyst, how can i represent this code to the image, im just beginner, thanks before :).

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
format longg;
format compact;
fontSize = 20;
angles = 0 : 10 : 360;
x = cosd(angles);
y = sind(angles);
plot(x, y, 'ro-');
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
grid on;
xCenter = 0;
yCenter = 0;
for k = 1 : length(angles)
  % Draw a line from the center to the edge of the circle.
  line([xCenter x(k)], [yCenter y(k)], 'LineWidth', 3);
  % Calculate the angle of that line
  angle = atand(y(k)/x(k));
  % Convert to 0-360
  if y(k) >= yCenter && x(k) >= xCenter
    quadrant = 1;
  elseif y(k) >= yCenter && x(k) <= xCenter
    angle = 180 + angle;
    quadrant = 2;
  elseif y(k) <= yCenter && x(k) < xCenter
    angle = 180 + angle;
    quadrant = 3;
  elseif y(k) <= yCenter && x(k) >= xCenter
    angle = 360 + angle;
    quadrant = 4;
  end
  promptMessage = sprintf('This latest line, going from (0,0) to (%.8f, %.8f),\nis at angle %f, and in quadrant %d.', ...
    x(k), y(k), angle, quadrant);
  fprintf('%s\n', promptMessage);
  titleBarCaption = 'Continue?';
  button = questdlg(promptMessage, titleBarCaption, 'Continue', 'Cancel', 'Continue');
  if strcmp(button, 'Cancel')
    break;
  end
end

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Answer 2

Kay CH 2015-6-12

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Thank you very much. It's very useful for me too.

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