Correspondence label and coordinates' points - matlab

Question

How to obtain the coordinates of the first and the last appearances (under column-major ordering) of each label present in a matrix?
Example of a label matrix (where labels are 1 to 4):
L = [
1 1 1 1 0 0 0 0
0 0 0 0 2 2 0 0
0 0 0 0 0 0 2 0
0 0 0 0 0 0 0 0
0 0 0 0 0 3 0 0
0 0 0 0 0 0 3 3
0 0 0 4 0 0 0 0
4 4 4 0 0 0 0 0
];
For the above example L, I would like to obtain a matrix of coordinates like:
M = [
1 1 1
1 4 1
2 5 2
3 7 2
5 6 3
6 8 3
8 1 4
7 4 4 ];
Where the 1st column of M contains horizontal coordinates, the 2nd contains vertical coordinates, and the 3rd column contains the label. There should be 2 rows for each label.

Answer 1

With for-loop you can do it like that:
M=zeros(2*max(L(:)),3);
for k=1:max(L(:))
[r,c]=find(L==k);
s=sortrows([r c],2);
M(k*2-1:k*2,:)=[s(1,:) k; s(end,:) k];
end
M =
1 1 1
1 4 1
2 5 2
3 7 2
5 6 3
6 8 3
8 1 4
7 4 4
Maybe somehow with regionprops options you can do it without the loop...

Answer 2

I just had to try it with accumarray:
R = size(L, 1);
[rowIndex, colIndex, values] = find(L); % Find nonzero values
index = (colIndex-1).*R+rowIndex; % Create a linear index
labels = unique(values); % Find unique values
nLabels = numel(labels);
minmax = zeros(2, nLabels);
minmax(1, :) = accumarray(values, index, [nLabels 1], #min); % Collect minima
minmax(2, :) = accumarray(values, index, [nLabels 1], #max); % Collect maxima
temp = ceil(minmax(:)/R);
M = [minmax(:)-R.*(temp-1) temp repelem(labels, 2, 1)]; % Convert index to subscripts
M =
1 1 1
1 4 1
2 5 2
3 7 2
5 6 3
6 8 3
8 1 4
7 4 4
Here's what I got for timing with Dev-iL's script and Adiel's newest code (Note that the number of labels can't go above 127 due to how Adiel's code uses the uint8 values as indices):
| Adiel | Dev-iL | gnovice
-----------------------+---------+---------+---------
20 labels, 1000x1000 | 0.0753 | 0.0991 | 0.0889
20 labels, 10000x10000 | 12.0010 | 10.2207 | 8.7034
120 labels, 1000x1000 | 0.1924 | 0.3439 | 0.1387
So, for moderate numbers of labels and (relatively) smaller sizes, Adiel's looping solution looks like it does best, with my solution lying between his and Dev-iL's. For larger sizes or greater numbers of labels, my solution starts to take the lead.

Answer 3

If you're looking for a vectorized solution, you can do this:
nTags = max(L(:));
whois = bsxfun(#eq,L,reshape(1:nTags,1,1,[]));
% whois = L == reshape(1:nTags,1,1,[]); % >=R2016b syntax.
[X,Y,Z] = ind2sub(size(whois), find(whois));
tmp = find(diff([0; Z; nTags+1])); tmp = reshape([tmp(1:end-1) tmp(2:end)-1].',[],1);
M = [X(tmp), Y(tmp), repelem(1:nTags,2).'];
Or with extreme variable reuse:
nTags = max(L(:));
Z = bsxfun(#eq,L,reshape(1:nTags,1,1,[]));
[X,Y,Z] = ind2sub(size(Z), find(Z));
Z = find(diff([0; Z; nTags+1]));
Z = reshape([Z(1:end-1) Z(2:end)-1].',[],1);
M = [X(Z), Y(Z), repelem(1:nTags,2).'];
Here's my benchmarking code:
function varargout = b42973322(isGPU,nLabels,lMat)
if nargin < 3
lMat = 1000;
end
if nargin < 2
nLabels = 20; % if nLabels > intmax('uint8'), Change the type of L to some other uint.
end
if nargin < 1
isGPU = false;
end
%% Create L:
if isGPU
L = sort(gpuArray.randi(nLabels,lMat,lMat,'uint8'),2);
else
L = sort(randi(nLabels,lMat,lMat,'uint8'),2);
end
%% Equality test:
M{3} = DeviL2(L);
M{2} = DeviL1(L);
M{1} = Adiel(L);
assert(isequal(M{1},M{2},M{3}));
%% Timing:
% t(3) = timeit(#()DeviL2(L)); % This is always slower, so it's irrelevant.
t(2) = timeit(#()DeviL1(L));
t(1) = timeit(#()Adiel(L));
%% Output / Print
if nargout == 0
disp(t);
else
varargout{1} = t;
end
end
function M = Adiel(L)
M=[];
for k=1:max(L(:))
[r,c]=find(L==k);
s=sortrows([r c],2);
M=[M;s(1,:) k; s(end,:) k];
end
end
function M = DeviL1(L)
nTags = max(L(:));
whois = L == reshape(1:nTags,1,1,[]); % >=R2016b syntax.
[X,Y,Z] = ind2sub(size(whois), find(whois));
tmp = find(diff([0; Z; nTags+1])); tmp = reshape([tmp(1:end-1) tmp(2:end)-1].',[],1);
M = [X(tmp), Y(tmp), repelem(1:nTags,2).'];
end
function M = DeviL2(L)
nTags = max(L(:));
Z = L == reshape(1:nTags,1,1,[]);
[X,Y,Z] = ind2sub(size(Z), find(Z));
Z = find(diff([0; Z; nTags+1]));
Z = reshape([Z(1:end-1) Z(2:end)-1].',[],1);
M = [X(Z), Y(Z), repelem(1:nTags,2).'];
end

Answer 4

You can retrive the uniqe values (your labels) of the matrix with unique.
Having them retrived you can use find to get their indices.
Put together your matrix with it.