Horizontal Projection of the Histogram

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Assad
Assad on 13 Jun 2012
Commented: Walter Roberson on 19 Oct 2017
i am using matlab for the ligature recognition and i want to get the horizontal Histogram projection of the Image so that i can segment the Lines can any one guide me regarding this
here is my code
close all;
clear all
clc;
imtool close all; % Close all figure windows created by imtool.
workspace; % Make sure the workspace panel is showing.
I = imread('C:\Users\ENZEE\Desktop\MUS\shah.jpg');
Im=im2double(I);%use to double the precision of the Image
Ir=imresize(Im,[100 150]);
%figure,imshow(Ir), title('Extracted Image');
text(size(I,2),size(I,1)+15, ...
'Image courtesy', ...
'FontSize',7,'HorizontalAlignment','right');
text(size(Ir,2),size(Ir,1)+25, ....
'Trying.........', ...
'FontSize',7,'HorizontalAlignment','right');
I = imnoise(Ir,'salt & pepper',0.02);
K = medfilt2(Ir);
%imshow(Ir), title('Noise Image'), figure, imshow(K), title('Filtered Image');
%otsu's method
[level EM]= graythresh(K);
%display(level)
BW = im2bw(K,level);
%figure, imshow(BW),title('Converted Image');
[imx,imy]=size(BW);
%imhist(K), title('Histogram')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%Segmentation%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
numofpixels=size(I,1)*size(I,2);% size(image array,dimensions)
L=bwlabel(BW);%L = bwlabel(BW, n) returns a matrix L, of the same size as BW,
%containing labels for the connected objects in BW.
mn=min(L);
%disp('~~~~~~~~~~~~~~~~~~~~~~~~~~MIN~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~')
%disp(mn)
%disp('~~~~~~~~~~~~~~~~~~~~~~~~~FINDED~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~')
fd=find(mn<1);
%disp(fd)
%disp('~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~')
calc=length(fd)/numofpixels;
disp(length(fd))
disp(numofpixels)
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%[r c]=size(BW);
%contour = bwtraceboundary(BW, [r,c], 'W', 8, 50,...
% 'counterclockwise');
%STATS = regionprops(BW,'FilledArea');
% seD = strel('arbitrary',numofpixels);
% BWfinal = imerode(BW,seD);
% BWfinal = imerode(BW,seD);
% %figure, imshow(BWfinal), title('segmented image');
% BWoutline = bwperim(BWfinal);
% Segout = Ir;
% Segout(BWoutline) = 255;
%figure, imshow(Segout), title('outlined original image');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%NEW TRY%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure
subplot(2,2,1)
hist(BW)
subplot(2,2,2)
barh(~BW,2)
% [pixelCount grayLevels] = imhist(BW);
%
%
%
% bar(pixelCount); title('Histogram of original image');
% xlim([0 grayLevels(end)]); % Scale x axis manually.
%x1=size(I,1);
%y1=size(I,2)
%disp(STATS)
%crp= imcontour(BW,150);
%figure,imshow(crp),title('Manually Image')
%crp=imcrop(BW);
%imcontour(crp,2);
%figure,imshow(crp),title('Manually Croped Image')
%a=bwperim(BW);
%imshow(a)
%bwdist()
%disp(calc)
%a=imcontour(I,2);
%imshow(a)
%CR=imcrop(BW,'rect');
%figure,imshow(CR)
%t=wpdec2(BW,2,'db1');
%plot(t)
% figure;
% hi=hist(BW);
% hi1=hi(1:2:256);
% horz=1:2:256;
% bar(horz,hi);
% axis([0 255 0 1400]);
% set(gca, 'xtick', 0:50:255);
% set(gca, 'ytick', 0:2000:15000);
% xlabel('Gray level' );
% ylabel('No of pixels' );
% title('Histogram before opening the image');
}
  5 Comments
sayar chit
sayar chit on 14 Oct 2017
Your answer isn't quite clear so if you have clean suggestion for lines,words and character segmentation, give me my account (chitsanlwin.maths.mm@gmailcom). I wish you good luck and health for your whole life
Walter Roberson
Walter Roberson on 19 Oct 2017
sayar chit, who were you addressing that to? Whose answer is not quite clear? Which answer? What are some of the parts you are finding confusing?

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Answers (2)

Walter Roberson
Walter Roberson on 15 Jun 2012
hist(sum(double(BW),2))
  18 Comments
Walter Roberson
Walter Roberson on 20 Jun 2012
Hmmm, I would assume that the foreground and background would have been reversed somewhere along the way, or else the regionprops() and labeling are going to produce pretty useless information. I guess that wasn't a good assumption.
The assignment to H and the thresholding against 50 (or whatever works) done on that, could go anywhere after the bwopenarea() and before the "for a".
I notice that the "for a" code has not been repaired even though I gave specific replacement code :(
Assad
Assad on 25 Jun 2012
what is BWopenarea() ?
are you talking about bwareaopen() i was thinking if i can segment the lines on the be half of thresolds\histograms whats the techniques of doing that

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Image Analyst
Image Analyst on 25 Jun 2012
I don't know if you have to use that algorithm, but there are plenty to choose from. How about this one:
  2 Comments
Assad
Assad on 30 Jun 2012
can you suggest me how to crop on the basis of Histogram values such as crop all those who are less than in Histogram
Walter Roberson
Walter Roberson on 30 Jun 2012
Edited: Walter Roberson on 30 Jun 2012
If H is your row histogram,
first = find(H >= 10, 1, 'first');
last = find(H >= 10, 1, 'last');
cropped_image = YourImage(first:last, :);

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