Detect the shape in MATLAB
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Hello, I hope you are doing well.
I have the image of 1000x1000.I have attached the image below I want to detect the number of lines in the image and pixel value of each line. How can i do it in MATLAB
as you can see in image, there is four lines, is Hough transform work on this or any image processing.
For example line is at 720, then it show pixel value 720
2 个评论
Jan
2022-3-15
Are the lines parallel to the X axis in all cases? Then there are much cheaper methods than a Hough transformation.
Med Future
2022-3-16
采纳的回答
Image Analyst
2022-3-15
% Make the image binary, if it's not already.
binaryImage = grayImage > 128;
% Count the lines.
[LabeledImage, numLines] = bwlabel(binaryImage);
18 个评论
Med Future
2022-3-16
Image Analyst
2022-3-16
You can get the intensity for each blob (line) individually using regionprops():
props = regionprops(labeledImage, grayImage, 'MeanIntensity');
meanIntensities = [props.MeanIntensity]
The "pixel value for the image" is the image variable itself, but you haven't really explained what that means. You can get the pixel value anywhere just by giving the row and column, like
grayLevel = grayImage(row, col);
where you assign some numerical value to row and col.
If you want the MEAN pixel value for the whole image, use mean2()
meanValue = mean2(grayImage);
Med Future
2022-3-16
Image Analyst
2022-3-16
OK, here is a full demo. Tell me if this is what you want:
% Demo by Image Analyst
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 = 22;
markerSize = 40;
%--------------------------------------------------------------------------------------------------------
% READ IN IMAGE
folder = pwd;
baseFileName = 'image_2732.png';
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
% The file doesn't exist -- didn't find it there in that folder.
% Check the entire search path (other folders) for the file by stripping off the folder.
fullFileNameOnSearchPath = baseFileName; % No path this time.
if ~exist(fullFileNameOnSearchPath, 'file')
% Still didn't find it. Alert user.
errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
rgbImage = 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(rgbImage)
if numberOfColorChannels > 1
% It's not really gray scale like we expected - it's color.
fprintf('It is not really gray scale like we expected - it is color\n');
% Extract the blue channel.
grayImage = rgbImage(:, :, 3);
else
grayImage = rgbImage;
end
%--------------------------------------------------------------------------------------------------------
% Display the image.
subplot(2, 1, 1);
imshow(grayImage);
impixelinfo;
axis('on', 'image');
title('Original Gray Scale Image', 'FontSize', fontSize, 'Interpreter', 'None');
% Maximize window.
g = gcf;
g.WindowState = 'maximized';
drawnow;
%--------------------------------------------------------------------------------------------------------
% Binarize the image to get a mask.
mask = grayImage >= 128;
% Display mask image.
subplot(2, 1, 2);
imshow(mask);
hold on;
impixelinfo;
axis('on', 'image');
drawnow;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
% Get a labeled image and count of the number of distinct regions.
[labeledImage, numRegions] = bwlabel(mask);
caption = sprintf('Binary Image with %d lines', numRegions);
title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
% Get all the blob properties. Can only pass in originalImage in version R2008a and later.
props = regionprops(labeledImage, 'Area', 'Centroid');
allLengths = [props.Area]
centroids = vertcat(props.Centroid);
% Put a thick red line wherever there is a line. This is because on the figure the lines are so thin they're not visible.
hold on;
for k = 1 : numRegions
x1 = centroids(k, 1) - allLengths(k)/2;
if x1 < 1
x1 = 1;
elseif x1 > columns
x1 = columns;
end
y1 = centroids(k, 2);
x2 = centroids(k, 1) + allLengths(k)/2;
if x2 < 1
x2 = 1;
elseif x2 > columns
x2 = columns;
end
y2 = y1; % Assumes perfectly horizontal lines.
plot([x1,x2], [y1,y2], 'r-', 'LineWidth', 4);
xt = props(k).Centroid(1);
yt = props(k).Centroid(2);
str = sprintf('Length = %d at line %d', allLengths(k), y1);
text(xt, yt, str, 'Color', 'r', 'FontWeight', 'bold', 'HorizontalAlignment', 'Center', 'VerticalAlignment','bottom', 'FontSize',15);
end
hold off;
% Tell the user
message = sprintf('Done!\n');
uiwait(helpdlg(message))
Med Future
2022-3-19
Image Analyst
2022-3-19
@Med Future you just need to use unique(). This will throw out duplicate line numbers. Add these lines to the end of my code (also attached as testGrayImage.m):
% Get the row number the white lines are on
whiteLineRows1 = centroids(:, 2)
% Some of the lines are separated but are on the same line.
% The poster wants any on the same line to be considered as they are on the same line.
% To do that use unique.
whiteLineRows2 = unique(whiteLineRows1)
You'll see
whiteLineRows1 =
577
686
392
577
whiteLineRows2 =
392
577
686
I think simple thresholding is simpler than using a hough transform.
By the way, the code also works for your second image.
Med Future
2022-3-19
Image Analyst
2022-3-19
Yes, probably. There could be qualifications depending on exactly what you want but you could use
[r, c] = find(binaryImage);
topLine = min(r);
If not, then start a new question. Like maybe the line is actually thicker than a single pixel wide curve, and you want the top line and bottom line for each and every column in the image, or some variation of that.
Med Future
2022-3-19
Image Analyst
2022-3-19
Yes it works with that image. It will find hundreds of separate lines because your lines are broken up into many, many separate segments, each with their own endpoints. However extracting the y values and using unique() will get you just the line numbers that they're on even though they're all broken up.
Here is what you get:
whiteLineRows2 =
104
108
224
407
867
Code is attached.
Med Future
2022-3-19
Med Future
2022-3-19
Image Analyst
2022-3-19
Then don't plot individual ones in the loop and call yline() after the loop. Solution attached.
% Put a thick red line wherever there is a line. This is because on the figure the lines are so thin they're not visible.
hold on;
for k = 1 : numRegions
x1 = centroids(k, 1) - allLengths(k)/2;
if x1 < 1
x1 = 1;
elseif x1 > columns
x1 = columns;
end
y1 = centroids(k, 2);
x2 = centroids(k, 1) + allLengths(k)/2;
if x2 < 1
x2 = 1;
elseif x2 > columns
x2 = columns;
end
y2 = y1; % Assumes perfectly horizontal lines.
% plot([x1,x2], [y1,y2], 'r-', 'LineWidth', 4);
% xt = props(k).Centroid(1);
% yt = props(k).Centroid(2);
% str = sprintf('Length = %d at line %d', allLengths(k), y1);
% text(xt, yt, str, 'Color', 'r', 'FontWeight', 'bold', 'HorizontalAlignment', 'Center', 'VerticalAlignment','bottom', 'FontSize',15);
end
hold off;
% Get the row number the white lines are on
whiteLineRows1 = centroids(:, 2)
% Some of the lines are separated but are on the same line.
% The poster wants any on the same line to be considered as they are on the same line.
% To do that use unique.
whiteLineRows2 = unique(whiteLineRows1)
hold on;
yline(whiteLineRows2, 'LineWidth', 2, 'Color', 'r');
Med Future
2022-3-20
Med Future
2022-3-21
Image Analyst
2022-3-21
I don't know what that means. It plots a line all the way across - that is how yline() works. You said that line segments on the same line should be considered one line even if they have a gap. If you don't want the line drawn all the way across, you can use plot() or line() to plot line segments.
Med Future
2022-3-25
Image Analyst
2022-3-25
I've given it to you both ways. First I found line segments, with the starting and ending points, then you said you wanted "only unique lines" so if two blobs were on the same line/row, I used unique() to get only unique lines. So if two blobs were on the same row, I'd draw only one line across the whole row. So now you have it both ways. Does that not solve the problem?
更多回答（1 个）
img = imread('https://ww2.mathworks.cn/matlabcentral/answers/uploaded_files/928244/image_2732.png');
bw = im2bw(img);
% 霍夫分析
[H,T,R] = hough(bw);
P = houghpeaks(H,5,'threshold',ceil(0.3*max(H(:))));
lines = houghlines(bw,T,R,P,'FillGap',5,'MinLength',7);
max_len = 0;
line_r = [];
figure; imshow(img, []); hold on;
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','cyan');
p = [];
xy = round(xy);
for j = min(xy(:,1)) : max(xy(:,1))
p = [p img(xy(1,2), j)];
end
text(xy(1,1),xy(1,2)-15, sprintf('pixel value = %.2f', mean(p)), 'color', 'r');
end
9 个评论
Med Future
2022-3-16
Med Future
2022-3-16
编辑：Med Future
2022-3-16
Med Future
2022-3-16
Image Analyst
2022-3-16
Attach the actual image that you read in, not a screen capture of the final output.
yanqi liu
2022-3-17
yes，sir，may be upload your real image，or modify parameter，such as
P = houghpeaks(H,3,'threshold',ceil(0.1*max(H(:))));
lines = houghlines(bw,T,R,P,'FillGap',5,'MinLength',3);
Med Future
2022-3-19
Image Analyst
2022-3-19
It should. It will find hundreds of separate lines because your lines are broken up into many, many separate segments, each with their own endpoints. However extracting the y values and using unique() will get you just the line numbers that they're on even though they're all broken up.
Med Future
2022-3-19
Image Analyst
2022-3-19
Sorry - I put it in the wrong place. it was meant to be a reply to your comment to me
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