Calculate equation of a curve from an image?

Is there a way to calculate the equation of the curve in an image like the one given below:

Answer 1

[Michael scheinfeild]

if all image is zeros and the line value>0. use

ii=find(image(:));
after you find indexes convert it to x,y

[yy,xx]=ind2sub(size(image),ii);
the use command like polyfit or other fit
n=2 , or n=3 

   p = polyfit(xx,yy,n)
it will give you coeficients !!

Answer 2

[Image Analyst]

Try this code I wrote for you. Simply copy and paste, and change the folder name to wherever you stored your image.

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 a Vrinda's gray scale demo image.
folder = 'C:\Users\Vrinda\Documents';
baseFileName = 'curve1.jpg';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
  % File doesn't exist -- didn't find it there.  Check the search path for it.
  fullFileNameOnSearchPath = baseFileName; % No path this time.
  if ~exist(fullFileNameOnSearchPath, 'file')
    % Still didn't find it.  Alert user.
    errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
    uiwait(warndlg(errorMessage));
    return;
  end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image.  
% numberOfColorBands should be = 1.
[rows, columns, numberOfColorBands] = size(grayImage);
if numberOfColorBands > 1
  % It's not really gray scale like we expected - it's color.
  % Convert it to gray scale by taking only the green channel.
  grayImage = grayImage(:, :, 2); % Take green channel.
end
% Display the original gray scale image.
subplot(2, 2, 1);
imshow(grayImage, []);
title('Original Grayscale Image', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 

% Binarize to get rid of horrible jpeg noise
binaryImage = grayImage < 100;
% Display the original gray scale image.
subplot(2, 2, 2);
imshow(binaryImage, []);
title('Binary Image', 'FontSize', fontSize);
axis on;

% Get the rows (y) and columns (x).
[rows, columns] = find(binaryImage);

% Fit a cubic
coefficients = polyfit(columns, rows, 3); % Gets coefficients of the formula.
% Fit a curve to 500 points in the range that x has.
fittedX = linspace(min(columns), max(columns), 500);
% Now get the y values.
fittedY = polyval(coefficients, fittedX);
% Plot the fitting:
subplot(2,2,3:4);
plot(fittedX, fittedY, 'b-', 'linewidth', 4);
grid on;
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
% Overlay the original points in red.
hold on;
plot(columns, rows, 'r+', 'LineWidth', 2, 'MarkerSize', 10);

[Image Analyst]

Note that the y direction of images and line plots go in different directions. If that is a problem, then let me know what you want to do in more detail.

[Ziv Kassner]

This script is truly helpful! How can I use it if my line is in a shape of a half ellipse? it means that in a few points, X will have 2 Y values, Sincerely, Ziv.

[Image Analyst]

Ziv, fitting an ellipse is not so easy. The FAQ has a link to a paper that shows you how to do it: http://matlab.wikia.com/wiki/FAQ#How_can_I_fit_an_ellipse_or_other_shape_to_a_set_of_XY_data.3F

[nick.x101]

Hi image analyst. I'm doing something similar to Vrinda except in my image I want to iterate through each column and identify the pixel of highest intensity and the plot this curve. Any help would be greatly appreciated.

[Image Analyst]

Nick, I don't think that would get you what you want, which I think might be a line along the centerline of the bright path. Why don't you start your own question? If you do, it would be something like

for col = 1 : columns
    [maxValue(col), row(col)] = max(grayImage(:, col));
end

[nick.x101]

Ok I will try this. I also looked at your Image Segmentation tutorial to try and obtain the line along the centerline but it didn't work too well. It gave the 'blobs' of highest intensity and I tried to find a way around it by changing the property of regionprops but it didn't work to well. Also I will post it as a new question. Thanks Image Analyst.

[jon tan]

Hi image analyst, I am doing something similar but I am not sure how to go about to do this. Basically I have collected raw data in my experiment to plot a displacement time graph and I need to plot the gradient for it which is essentially the velocity time graph as well as its derivative for the acceleration time graph. However, I did not collect sufficient data points to be able to use (y1-y2)/(x1-x2) formula into my points to generate the gradient. Is there anyway that the code you wrote above can be applied to my problem? Thank you

[Image Analyst]

You can generate more points with spline. See attached demos. Start a new question with your data attached if you need more help.

[Albertus Calvin Pratama]

I wondering how if we want to know the Y value by inputting X in command window or script based on the graph?

[Image Analyst]

You just plug the x value into the spline function.

[Albertus Calvin Pratama]

Is it possible to do if i don''t know the spline function and get the Y value if we input X value? or this code can give us the spline function either?

Answer 3

[Panos Tzovaras]

@Image Analyst Hey! Is there a way I can cite you in my PhD thesis where I am using your code?

[Image Analyst]

I imagine it would be similar to how it suggests you cite File Exchange items

Cite As

Image Analyst (2023). Image Segmentation Tutorial (https://www.mathworks.com/matlabcentral/fileexchange/25157-image-segmentation-tutorial), MATLAB Central File Exchange. Retrieved May 12, 2023.

Make changes as appropriate for code on Answers pages.