Horizontal Projection of the Histogram

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Assad
Assad 2012-6-13
评论： Walter Roberson 2017-10-19
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 个评论
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sayar chit
sayar chit 2017-10-14
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 2017-10-19
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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回答（2 个）

Walter Roberson
Walter Roberson 2012-6-15
hist(sum(double(BW),2))
  18 个评论
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Walter Roberson
Walter Roberson 2012-6-20
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 2012-6-25
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 2012-6-25
I don't know if you have to use that algorithm, but there are plenty to choose from. How about this one:
http://maven.smith.edu/~nhowe/research/code/
  2 个评论
Assad
Assad 2012-6-30
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 2012-6-30
编辑：Walter Roberson 2012-6-30
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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