I want to get chain code but it is showing error. Please help me.

Question

MONIKA SAINI 2015-2-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/175650-i-want-to-get-chain-code-but-it-is-showing-error-please-help-me

评论： Image Analyst 2015-2-8

I=imread('C:\Users\abcd\Desktop\mini.jpg');
bw=im2bw(I)
bw=~bw;
BW2 = bwperim(bw,8);
image=BW2;
[x y]=find(image==1);
b=[x y];
[cc]=chaincode(b)

This is showing eror: ??? Undefined function or method 'chaincode' for input arguments of type. Please tell me where I am making mistake. 'double'.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2015-2-6

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/175650-i-want-to-get-chain-code-but-it-is-showing-error-please-help-me#answer_167129

Why do you think there is a function called chaincode()? I think you want bwboundaries() instead.

10 个评论
显示 8更早的评论隐藏 8更早的评论

Image Analyst 2015-2-7

编辑：Image Analyst 2015-2-7

在 MATLAB Online 中打开

Just threshold it do get a binary image, then use bwboundaries. Then put into a loop and build the chain code. Here's a start:

% Code to create a chain code from the boundaries of an binary image.
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 25;
% Create sample image.
z=peaks(500);
subplot(1,2,1);
imshow(z, []);
axis on;
title('Original Grayscale Image', 'FontSize', fontSize);
% Threshold it to form a binary image.
binaryImage = z > 2;
subplot(1,2,2);
imshow(binaryImage);
axis on;
title('Binary Image', 'FontSize', fontSize);
subplot(1,2,1);
% 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') 
% Find the outer perimeter of the two blobs.
boundaries = bwboundaries(binaryImage);
for k = 1 : length(boundaries)
  % Get the k'th boundary.
    thisBoundary = boundaries{k};
  % Get the x and y coordinates.
    x = thisBoundary(:, 2);
    y = thisBoundary(:, 1);
  % Plot this boundary over the original image.
  hold on;
  plot(x, y, 'r-', 'LineWidth', 2);
  % Create the chain code.
    for p = 2 : length(y)
    % See where the current point is relative to the prior point.
        if x(p) < x(p-1) && y(p) < y(p-1)
            % Moved to the upper left
            thisChainCode(p) = 1;
        elseif x(p) == x(p-1) && y(p) < y(p-1)
            % Moved straight up
            thisChainCode(p) = 2;
        elseif x(p) > x(p-1) && y(p) < y(p-1)
            % Moved to the upper right
            thisChainCode(p) = 3;
        elseif x(p) > x(p-1) && y(p) == y(p-1)
            % Moved right
            thisChainCode(p) = 4;
        elseif x(p) > x(p-1) && y(p) > y(p-1)
            % Moved right and down.
            thisChainCode(p) = 5;
        elseif x(p) == x(p-1) && y(p) > y(p-1)
            % Moved down
            thisChainCode(p) = 6;
        elseif x(p) <= x(p-1) && y(p) > y(p-1)
            % Moved down and left
            thisChainCode(p) = 7;
        elseif x(p) <= x(p-1) && y(p) == y(p-1)
            % Moved left
            thisChainCode(p) = 8;
        end
    end
    % Save the chain code that we built up for this boundary
  % into the cell array that will contain all chain codes.
    chainCodes{k} = thisChainCode
end
% Display the chain codes to the command window.
celldisp(chainCodes);

As you can see it's pretty easy. I just did that off the top of my head and is what you were probably thinking of but just didn't know how to code up. It may need a few tweaks and initializations though. If you can't finish it, let me know.

As you probably know since you're a Ph.D. in the image processing field, all or most of the handwriting published papers are listed here in Vision Bib: http://www.visionbib.com/bibliography/contentschar.html#OCR,%20Document%20Analysis%20and%20Character%20Recognition%20Systems in Section 23.4.6 and subsections.