Calculation of pixel shift

Question

Swati Jain 2017-7-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/350867-calculation-of-pixel-shift

评论： Walter Roberson 2017-7-31

采纳的回答： Image Analyst

Binary.jpg

在 MATLAB Online 中打开

Hi,

I have a slated edge image and I want to know the value of required pixel shift to make it vertical. I know the pixel value at first and end point of the edge. But want to scan the image line by line and know the values of shifted pixels for each line. But I am getting an error: *Matrix dimensions must agree for the following line: dif = abs(bw(i:j)- bw(i:j+1))

close all;
clear all;
clc;
I=imread('C:\Users\swati\OneDrive\Documents\MATLAB\Examples\Cropped Image.jpg');
figure;
imshow(I); title('Original Image');
impixelinfo;
m=167;n=200;
%Averaging The Pixels
h=ones(10,10)/100; %Create a normalized pixel; averaging filter.
I_avg=imfilter(I,h); %apply filter to image 
figure(); imshow(I_avg);
title(I_avg);
impixelinfo;
%Scanning of Image
v=I_avg(1,:); %scanning the I_avg Intensity
figure;plot(v);
%converting GrayScale image to Binary Image
for x=1:200
    for y=1:m
        if I_avg(x,y)>mean(I_avg(:))
            bw(x,y)=1;
        elseif I_avg(x,y)<mean(I_avg(:))
            bw(x,y)=0;
        end
    end
end
figure; imshow(bw);
impixelinfo;
%finding first left most pixel when intensity drops to zero in first row
fisrt_value=zeros([1,m]);
fist_value=bw(1,:)
%finding first left most pixel when intensity drops to zero in last row
second_value=zeros([1,m]);
last_value=bw(200,:)
p2x=1;
p2y=56;
p1x=200;
p1y=73;
m=(p2y-p1y)/(p2x-p1x) %find the slope
theta=atand(m)
shifted_pixel=p1y-p2y  % number of shifted pixels
 j=55;
for i = 1:20:200
 dif = abs(bw(i:j)- bw(i:j+1))
 fprintf('Row Number=%0.0f',i);
 pos=find(dif==1)
 shift=p1y-pos 
 j=j+1;
 i= i+1;
end

Could anyone please suggest me what is wrong in my code or is there any other way to do that?

Thanks, Swati

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 2017-7-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/350867-calculation-of-pixel-shift#answer_276127

在 MATLAB Online 中打开

Swati, get rid of all the programming you did for this and replace it with this:

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 short g;
format compact;
fontSize = 25;
%===============================================================================
% Get the name of the image the user wants to use.
baseFileName = 'binary.jpg';
% Get the full filename, with path prepended.
folder = []; % Determine where demo folder is (works with all versions).
fullFileName = fullfile(folder, baseFileName);
%===============================================================================
% Read in a demo image.
grayImage = imread(fullFileName);
% Get the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage)
if numberOfColorChannels > 1
  % It's not really gray scale like we expected - it's color.
  % Use weighted sum of ALL channels to create a gray scale image.
  grayImage = rgb2gray(grayImage);
  % ALTERNATE METHOD: Convert it to gray scale by taking only the green channel,
  % which in a typical snapshot will be the least noisy channel.
  % grayImage = grayImage(:, :, 2); % Take green channel.
end
% Display the image.
subplot(2, 2, 1);
imshow(grayImage, []);
axis on;
axis image;
caption = sprintf('Original Gray Scale Image');
title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
drawnow;
hp = impixelinfo();
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0, 1, 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
% set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')
drawnow;
% Create binary image from grayscale image.
binaryImage = grayImage > 128;
% Display the image.
subplot(2, 1, 1);
imshow(binaryImage, []);
axis on;
axis image;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
% Find edge between row 10 and 190
for row = 10:190
  x(row-9) = find(binaryImage(row, :), 1, 'last');
  y(row-9) = row;
end
coefficients = polyfit(x, y, 1);
angle = atand(coefficients(1))
% Rotate the image
rotatedImage = imrotate(binaryImage, angle-90);
% Display the image.
subplot(2, 1, 2);
imshow(rotatedImage, []);
axis on;
axis image;
title('Rotated Image', 'FontSize', fontSize, 'Interpreter', 'None');

2 个评论
显示无隐藏无

Swati Jain 2017-7-31

编辑：Swati Jain 2017-7-31

在 MATLAB Online 中打开

Could you please explain this portion of the code:

for row = 10:190
    x(row-9) = find(binaryImage(row, :), 1, 'last');
    y(row-9) = row;
  end

What is meaning of

x(row-9)

and

find(binaryImage(row, :), 1, 'last')

here, I think it is searching for 1 in binary image. Am I correct? What does 'last' mean? And why did you use row values from 10 to 190?

Walter Roberson 2017-7-31

在 MATLAB Online 中打开

For clarity you could instead write,

rowvals = 10:190;
% Find edge between row 10 and 190
for rowidx = 1 : length(rowvals)
      row = rowvals(rowidx);
  x(rowidx) = find(binaryImage(row, :) ~= 0, 1, 'last');
  y(rowidx) = row;
end

The syntax for find is

find( expression, number_of_matches_to_use, start_from_beginning_or_end )

so find(expression, 1, 'last') means to look for 1 value starting from the end (last position) and keep proceeding forward until the corresponding expression value is non-zero. In other words, the location of the last non-zero value.

The 1 does not mean to search for 1's: it is the number of values to take.

Another way of writing result = find(expression, N, 'last') is

temp = find(expression);
M = length(temp);
if M > N
  M = N;
end
result = temp(end-M+1:end);

That is, if there are fewer than N available values then it will use what is there, but if there are more than N available values then it will use the last N of them.

请先登录，再进行评论。

Answer 2

Walter Roberson 2017-7-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/350867-calculation-of-pixel-shift#answer_276112

在 MATLAB Online 中打开

In

abs(bw(i:j)- bw(i:j+1))

the left side, bw(i:j) has j-i+1 elements. The right side, bw(i:j+1) has j+1-i+1 = j-i+2 elements.

I suggest

abs(diff(bw(i:j+1)))

9 个评论
显示 7更早的评论隐藏 7更早的评论

Walter Roberson 2017-7-30

在 MATLAB Online 中打开

You have

j=55;
for i = 1:20:200
 dif = abs(bw(i:j)- bw(i:j+1))
 fprintf('Row Number=%0.0f',i);
 pos=find(dif==1)
 shift=p1y-pos 
 j=j+1;
 i= i+1;
end

so in the first iteration, i = 1, j = 55. 1:55 is valid subscripts so you could potentially get some results from the dif. Then, the j=j+1 increments j to 56. The i = i+1 changes i to 2. Then i gets overwritten with 21 by the next iteration of "for i = 1:20:200" . When you have a "for" loop, any changes you make to the loop control variable will be discarded as soon as the next iteration starts. Think of it has having a hidden "real" loop control variable.

So second iteration, i = 21, j = 56. 21:56 is valid subscripts so you could potentially get some results from the dif. j gets incremented to 57. i gets changed to 22, but that doesn't matter, as explained above.

Third iteration, i = 41, j = 57, 41:57 is valid, you might get something in dif. j becomes 58, i becomes 42 but that doesn't matter.

Fourth iteration, i = 61, j = 58, 61:58 is the empty range, so you cannot get any results from the loop. j becomes 59, i becomes 62 but that doesn't matter.

All subsequent iterations of i also lead to empty vectors because i > j.

Image Analyst 2017-7-31

Do we really need to keep discussing this without an image to even look at? What's the big secret? Show us the image.

Swati Jain 2017-7-31

Binary.jpg

Sorry, Please find the attached Image.

请先登录，再进行评论。

Answer 3

Image Analyst 2017-7-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/350867-calculation-of-pixel-shift#answer_276113

When you say this: abs(bw(i:j)- bw(i:j+1)), then the first vector is (j-i+1) elements long. The second vector, since you're going one element further along the array is (j+1-i+1) = (j-i+2). The second vector is one element longer so you cannot subtract them element-by-element. Not sure what you're going due to the disappointing lack of comments.

If you know the endpoints of the line, you can get the deltay and deltax and use atan2d() and imrotate() to straighten the edge (align it perfectly vertical or horizontal).