How to find pixel in one square block?
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Hello,
I have an Image called checkerboard, now i know the size of one square block is 34 x 34 mm. how can i find that in that one square block 34mm has how many pixel?
is there any formula to find or function?
Thank you in advance.
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采纳的回答
Image Analyst
2020-8-16
Try this. You'll find meanArea = 102.3 pixels.
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;
%--------------------------------------------------------------------------------------------------------
% READ IN IMAGE
folder = pwd;
baseFileName = 'checkerboard.jpg';
% Get the full filename, with path prepended.
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
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, []);
title('Original Grayscale Image', 'FontSize', fontSize, 'Interpreter', 'None');
impixelinfo;
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;
% Display histogram
subplot(2, 2, 2);
imhist(grayImage);
grid on;
title('Histogram of original gray image', 'FontSize', fontSize);
%--------------------------------------------------------------------------------------------------------
% SEGMENTATION OF IMAGE
binaryImage = grayImage < 40;
% Display the image.
subplot(2, 2, 3);
imshow(binaryImage, []);
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
impixelinfo;
% Measure areas
labeledImage = bwlabel(binaryImage, 4);
props = regionprops(labeledImage, 'Area', 'BoundingBox');
allAreas = [props.Area]
bb = vertcat(props.BoundingBox);
widths = bb(:, 3);
heights = bb(:, 4);
aspectRatios = widths ./ heights
keepers = find(allAreas' > 80 & aspectRatios > 0.8 & aspectRatios < 1.3);
% Get new image with just the squares.
binaryImage = ismember(labeledImage, keepers);
% Display the image.
subplot(2, 2, 4);
imshow(binaryImage, []);
title('New Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
impixelinfo;
% Measure areas of only hte "good" blobs.
labeledImage = bwlabel(binaryImage, 4);
props = regionprops(labeledImage, 'Area', 'BoundingBox');
allAreas = [props.Area]
meanArea = mean(allAreas)
13 个评论
Image Analyst
2020-8-19
Just invert it and multiply by 34, but it really shouldn't even be necessary since we're not ever going to use that information, remember?
% Pixels = mm * pixels/mm
mmPerPixel = sqrt(areaInMm / areaInPixels);
pixelsPerMm = 1 / mmPerPixel; % or sqrt(areaInPixels / areaInMm)
pixelDistance34 = 34 * pixelsPerMm % Number of pixels across a 34 mm span.
更多回答(1 个)
Walter Roberson
2020-8-16
Use bwconncomp with connection 4. Pass the result to regionprops asking for BoundingBox . You will get a struct array returned. You can put them together into one array using vertcat(). Now you can use median() on the 3rd and 4th columns to find out what the median box size is in pixels. Then knowing that represents 34 mm of physical space, you can calculate the image resolution.
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