enter image description hereI checked the earlier question on "How do I recover the signal from an ECG image"


Currently I am trying to write a program to extract the x and y co-ordinates of an ECG signal embedded within the background grid . I was able to use the image from the earlier post and used some inbuilt functions to crop the image and convert the image to grayscale and then to binary.

This is how I did it:

% Cropping the image and saving the co-ordinates 
% select the boundries in the pop up window , hit on crop image to save the
% cropped image x,y co-ordinates
img1=rgb2gray(img);  % Converting the cropped image to grayscale
Bw=im2bw(img1,0.8);  % converting the grayscale image to binary image 
imshow(Bw) ![cropped and binary image of the actual ecg_test.jpg][4]

Now I want to clean the image and then extract the co-ordinates of this image. How do I implement the MATLAB part of:

img = ColorConvert[
    Import["https://i.sstatic.net/500Kg.jpg"], {{0, -20}, {0, 0}}],

 Transpose[Function[With[{m = Min[#]},
     Map[Function[{v}, If[v == m, 1, 0]], #]]] /@ 
   Transpose[ImageData[img, "Real"]]
  • $\begingroup$ what is the application $\endgroup$ Commented Nov 15, 2012 at 8:20
  • $\begingroup$ The project i am trying to do is digitize the scanned documents of ECG charts , I know there are softwares which does this, but i am trying to implement using simple matlab code and try to understand the algorithm on how we can extract the co-ordinates corresponding to the signal in the image $\endgroup$
    – user3278
    Commented Nov 15, 2012 at 19:05
  • $\begingroup$ The way you phrased it, it is more a programming question along the lines of "how do I translate this code in language X in to the language Y". Some other stackexchange sites might be more suitable for programming problems. If you, on the other hand, have the problem understanding the concept, what is being done here, then you are in the right place. But, if that's the case, it would be very good if you edited your question and make it more along the lines of asking for explanation instead of translation to a different programming language. $\endgroup$
    – penelope
    Commented Nov 16, 2012 at 9:31

1 Answer 1


I believe this is an optimisation problem. E.g. what is the path from the left hand side of the image to the right hand of the image that is 'most black'.

As such 'rectangular shortest path' method would be ideal. You need to resample the image something like 10x to ensure that the method gets all the peaks properly, but in the end you will get something like this:

enter image description here


Here is a paper on circular shortest paths, but it is basically the same thing:


Update 2:

MATLAB script:

IM = imread('500Kg.jpg');
IM = rgb2gray(IM);
IM = double(IM);

% Over sample 8 times in horizontal direction
% Amount to oversample is determined by maximum slope, e.g. if max slope is
% 16, oversample by 16.
IM2 = interp2(IM,linspace(1,size(IM,2),size(IM,2)*8),linspace(1,size(IM,1),size(IM,1))');

% Calculate shortest path using 255 - image intensity as energy function
[spath,senergy,slastindex,im] = shortestPath(255-IM2',255,0,1,100);

% Down sample path to account for original oversampling
p_x = 1:size(IM2,2);
p_x = p_x ./ 8;
p_y = spath(:,2);

% Plot on top of original image
imagesc(IM); colormap gray; hold on; plot(p_x,p_y,'r'); hold off;

Shortest path function:

function [ p, e, l, im ] = shortestPath( m, opt, a, ax, bx  )
%SHORTESTPATH Summary of this function goes here
%   Detailed explanation goes here

%Get size of matrix
[sy, sx] = size(m);
im = zeros(sy,sx);

%Convert to value different from optimal
value = abs(opt - m);

%Preallocate index matrix and energy matrix
lastindex = zeros(sy, sx);
energy = zeros(sy, sx);

%Initialise first rows
energy(1,:) = value(1,:);
lastindex(1,:) = 1:sx;

%Initialise temp matrix
t = zeros(1,3);

%Loop through remaining rows
for row = 2:sy
    for col = ax:bx
        %Get the last energies and current values
        if col == ax 
            t = [inf energy(row-1,col:(col+1))];
            cv = [inf value(row,col:(col+1))];
        elseif col == bx
            t = [energy(row-1,(col-1):col) inf];
            cv = [value(row,(col-1):col) inf];
            t = energy(row-1,(col-1):(col+1));
            cv = value(row,(col-1):(col+1));

        %Add energy for moving
        t = t + [a 0 a];

        %Add energy from difference from optimum
        t = t + cv;

        %Find minimum
        [v,i] = min(t);

        %Save new values
        energy(row,col) = v;
        lastindex(row,col) = col + i - 2;


[v,li] = min(energy(sy,ax:bx));
li = ax + li - 1;
p(sy,:) = [sy li];
im(sy,li) = 1;
for row = (sy-1):-1:1
    i = lastindex(row,li);
    p(row,:) = [row,i];
    im(row,i) = 1;
    li = i;

e = energy;
l = lastindex;

  • $\begingroup$ Can you kindly explain the interpolation technique with an example, I took the most black pixel value and i was able to clean the image .I want to understand how to extract the co-ordinates corresponding to the interpolated red line in the image above, so that i can save them in a text doc .Also correct me if iam wrong , from the image above i can see that you didn't remove the background and you where able to interpolate the signal ,can you kindly explain how you did this $\endgroup$
    – user3278
    Commented Nov 16, 2012 at 3:22
  • $\begingroup$ @user3278 please see the pdf I have linked, it explains shortest path more clearly than I can do :) The result of shortest path is a set of coordinates like you desire. I have just plotted a line through these coordinates on top of the original picture. You might also need to think about detecting the grid lines so you can match the pixel coordinates to the actual scale of the graph. $\endgroup$ Commented Nov 16, 2012 at 13:14
  • $\begingroup$ ,Thanks for sharing the paper, I will study and implement the algo for my project. $\endgroup$
    – user3278
    Commented Nov 16, 2012 at 16:53
  • $\begingroup$ If you get stuck let me know. I do have the code but wheres the fun in that :P $\endgroup$ Commented Nov 16, 2012 at 23:08
  • $\begingroup$ ,I am trying to understand the paper ,I will try before i could ask some further help ,as mentioned i am a signal processing engineer but a beginner at image processing , I am completely new to the algorithm you showed me , let me try and then i will get back to you $\endgroup$
    – user3278
    Commented Nov 28, 2012 at 7:18

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