Reputation: 471
I have a grid in a binary image (may be rotated). How can I know approximate formula for that grid using MATLAB?
Example image:
(source: sjtu.edu.cn)
Sometimes these black dots are missing, so I need formula or ‘a way’ to estimate possible center of these black dots.
I have tried by using regionprops
, it help me to get center of these exist black dots, but no idea if black dots a missing
clear all
im = imread('print5.jpg');
im = im2bw(im);
[sy,sx] = size(im);
im = imcomplement(im);
im(150:200,100:150) = 0; % let some dots missing!
im = imclearborder(im);
st = regionprops(im, 'Centroid');
imshow(im) hold on;
for j = 1:numel(st)
px = round(st(j).Centroid(1,1));
py = round(st(j).Centroid(1,2));
plot(px,py,'b+')
end
Upvotes: 15
Views: 7983
Reputation: 124563
You could apply Hough transform to detect the grid lines. Once we have those you can infer the grid locations and the rotation angle:
%# load image, and process it
img = imread('print5.jpg');
img = imfilter(img, fspecial('gaussian',7,1));
BW = imcomplement(im2bw(img));
BW = imclearborder(BW);
BW(150:200,100:150) = 0; %# simulate a missing chunk!
%# detect dots centers
st = regionprops(BW, 'Centroid');
c = vertcat(st.Centroid);
%# hough transform, detect peaks, then get lines segments
[H,T,R] = hough(BW);
P = houghpeaks(H, 25);
L = houghlines(BW, T, R, P);
%# show image with overlayed connected components, their centers + detected lines
I = imoverlay(img, BW, [0.9 0.1 0.1]);
imshow(I, 'InitialMag',200, 'Border','tight'), hold on
line(c(:,1), c(:,2), 'LineStyle','none', 'Marker','+', 'Color','b')
for k = 1:length(L)
xy = [L(k).point1; L(k).point2];
plot(xy(:,1), xy(:,2), 'g-', 'LineWidth',2);
end
hold off
(I'm using imoverlay
function from the File Exchange)
The result:
Here is the accumulator matrix with the peaks corresponding to the lines detected highlighted:
Now we can recover the angle of rotation by computing the mean slope of detected lines, filtered to those in one of the two directions (horizontals or verticals):
%# filter lines to extract almost vertical ones
%# Note that theta range is (-90:89), angle = theta + 90
LL = L( abs([L.theta]) < 30 );
%# compute the mean slope of those lines
slopes = vertcat(LL.point2) - vertcat(LL.point1);
slopes = atan2(slopes(:,2),slopes(:,1));
r = mean(slopes);
%# transform image by applying the inverse of the rotation
tform = maketform('affine', [cos(r) sin(r) 0; -sin(r) cos(r) 0; 0 0 1]);
img_align = imtransform(img, fliptform(tform));
imshow(img_align)
Here is the image rotated back so that the grid is aligned with the xy-axes:
Upvotes: 28
Reputation: 26069
here's a way using fft
in 1D over the x and y projection:
First, I'll blur the image a bit to smooth the high freq noise by convolving with a gaussian:
m=double(imread('print5.jpg'));
m=abs(m-max(m(:))); % optional line if you want to look on the black square as "signal"
H=fspecial('gaussian',7,1);
m2=conv2(m,H,'same');
then I'll take the fft of a projection of each axis:
delta=1;
N=size(m,1);
df=1/(N*delta); % the frequency resolution (df=1/max_T)
f_vector= df*((1:N)-1-N/2); % frequency vector
freq_vec=f_vector;
fft_vecx=fftshift(fft(sum(m2)));
fft_vecy=fftshift(fft(sum(m2')));
plot(freq_vec,abs(fft_vecx),freq_vec,abs(fft_vecy))
So we can see both axis yield a peak at 0.07422 which translate to a period of 1/0.07422 pixels or ~ 13.5 pixels.
A better way to get also the angle info is to go 2D, that is:
ml= log( abs( fftshift (fft2(m2)))+1);
imagesc(ml)
colormap(bone)
and then apply tools such as simple geometry or regionprops if you want, you can get the angle and size of the squares. The size of the square is 1/ the size of the big rotated square on the background ( bit fuzzy because I blurred the image so try to do that without that), and the angle is atan(y/x)
. The distance between the squares is 1/ the distance between the strong peaks in the center part to the image center.
so if you threshold ml
properly image say
imagesc(ml>11)
you can access the center peaks for that...
yet another approach will be morphological operation on a binary image, for example I threshold the blurred image and shrink objects to points. It removes pixels so that objects without holes shrink to a point:
BW=m2>100;
BW2 = bwmorph(BW,'shrink',Inf);
figure, imshow(BW2)
Then you practically have a one pixel per lattice site grid! so you can feed it to Amro's solution using Hough transform, or analyze it with fft, or fit a block, etc...
Upvotes: 41