How to automatically obtain shape coordinates

2022 1 24
3 个回答
更新时间：2022 1 25
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I have an image (attached) that I want to crop. But to crop them I always need to manually take the center "coordinates" (index) of the center of each black box in the red circle. I need to automate it but I don't know where to start.

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回答（3 个）

Matt J
Matt J 2022-1-24
编辑：Matt J 2022-1-24
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0 个投票

Ran in:
Perhaps as follows
load Image
B=medfilt2(A,[5,5])<60;
B=bwareafilt(B,5) & ~bwareafilt(B,1);
T=regionprops('table',B,'Centroid'); %square centroids
LT=min(T.Centroid); %%left top corner
SZ=max(T.Centroid)-LT+1; %size fo box
A=imcrop(A,[LT,SZ]); %ignore projective warping
imshow(A,[])

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Jacob Ebilane
Jacob Ebilane 2022-1-24
Kind of close to what I need, but I need 4 points because I have to straighten it using a program I found. I could use LT to be set as my first point, but I'd need the location of the other 3 boxes.
Matt J
Matt J 2022-1-25
You have the 4 points in T.Centroid.

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yanqi liu
yanqi liu 2022-1-25
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Ran in:
clc; clear all; close all;
img = imread('https://ww2.mathworks.cn/matlabcentral/answers/uploaded_files/871715/uno.png');
if ndims(img) == 3
img = rgb2gray(img);
end
bw = imbinarize(img,'adaptive','ForegroundPolarity','dark','Sensitivity',0.4);
bw2 = ~bw;
bw2 = imopen(bw2, strel('square', 5));
bw3 = imclose(bw2, strel('line', size(bw2,1), 90));
bw4 = imclose(bw2, strel('line', size(bw2,2), 0));
% find left and right
[L,num] = bwlabel(bw3);
stats = regionprops(L);
rects = cat(1, stats.BoundingBox);
ind1 = find(rects(:,4)>size(bw2,1)*0.8);
[~,ind2] = min(rects(ind1,1));
[~,ind3] = max(rects(ind1,1));
bw3 = L==ind1(ind2) | L == ind1(ind3);
% find top and bottom
[L,num] = bwlabel(bw4);
stats = regionprops(L);
rects = cat(1, stats.BoundingBox);
ind1 = find(rects(:,3)>size(bw2,2)*0.8);
[~,ind2] = min(rects(ind1,2));
[~,ind3] = max(rects(ind1,2));
bw4 = L==ind1(ind2) | L == ind1(ind3);
% make square
bw5 = logical(bw3 + bw4);
bw5 = imfill(bw5, 'holes');
[r,c] = find(bw5);
rect = [min(c) min(r) max(c)-min(c) max(r)-min(r)];
% get 4 square
figure; imshow(img);
hold on; rectangle('position', rect, 'EdgeColor', 'g', 'LineWidth', 2)
Image Analyst
Image Analyst 2022-1-25

0 个投票

Here is yet another way:
grayImage = imread('uno.png');
subplot(2, 2, 1);
imshow(grayImage, []);
title('Original Image.')
if ndims(grayImage) == 3
grayImage = rgb2gray(grayImage);
end
topHatImage = imbothat(grayImage, true(51));
subplot(2, 2, 2);
imshow(topHatImage, [])
title('Top Hat Filtered Image.')
impixelinfo;
mask = topHatImage > 60; %~imbinarize(grayImage,'adaptive','ForegroundPolarity','dark','Sensitivity',0.4);
mask = imfill(mask, 'holes');
props = regionprops(mask, 'Area')
allAreas = sort([props.Area])
mask = bwareafilt(mask,[400, 7000]);
mask = bwconvhull(mask);
subplot(2, 2, 3);
imshow(mask, []);
title('Mask.')
props = regionprops(mask, 'BoundingBox')
croppedImage = imcrop(grayImage, props.BoundingBox);
subplot(2, 2, 4);
imshow(croppedImage, []);
title('Cropped Image.')
It could be made faster if you started with a good image, like one from a scanner instead of a poorly lit paper and a mobile phone camera.

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