Grouping identical matrices in cell array
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Consider two cell arrays (here of size 6x1), where each entry contains a matrix, e.g.
a = { [1 2; 3 4] [2 2 3; 2 2 3] [1 2; 3 4] [1 2; 3 4] [2 3 2; 3 4 5] [2 2 3; 2 2 3] }';
b = { [5 6; 7 8] [2 2 3; 2 2 3] [9 9; 9 9] [5 6; 7 8] [2 3 2; 3 4 5] [2 2 3; 2 2 3] }';
I want to find an array, which assigns a group to each of the six entries, i.e.
groups = [1 2 3 1 4 2];
A group is defined by identical a{i} and b{i} entries (or up to a tolerance). I came up with the following brute-force code
n = length(a);
groups = zeros(n,1);
counter = 0;
for i = 1:n
if groups(i)~=0
continue;
end
counter = counter + 1;
groups(i) = counter;
for j = i+1:n
if groups(j)~=0
continue;
end
if isequal(a{i},a{j}) && isequal(b{i},b{j})
groups(j) = counter;
end
end
end
which is quite inefficient due to the for-loops. Is there a smarter way of finding these groups? Thanks :)
1 commentaire
Stephen23
le 31 Mai 2021
"which is quite inefficient due to the for-loops"
I doubt that the loops themselves are consuming much time. Have you run the profiler?
Réponse acceptée
Jan
le 31 Mai 2021
Modifié(e) : Jan
le 31 Mai 2021
Start with a simplified version of your code:
n = numel(a);
groups = zeros(n, 1);
counter = 0;
for i = 1:n
if groups(i) == 0
counter = counter + 1;
groups(i) = counter;
for j = i+1:n
if groups(j) == 0 && isequal(a{i}, a{j}) && isequal(b{i}, b{j})
groups(j) = counter;
end
end
end
end
Now let's assume the cell arrays a and b are huge, e.g. 1e6 elements. Then comparing 1e6 with 1e6-1 elements takes a lot of time. It might be cheaper to create a hash at first:
Hash = cell(1, n);
for k = 1:n
Hash{k} = GetMD5({a{k}, b{k}}, 'Array', 'bass64');
end
[~, ~, groups] = unique(Hash, 'stable');
% With 1e6 elements per cell, R2018b, Win10:
% Elapsed time is 21.341034 seconds. % Original
% Elapsed time is 21.286879 seconds. % Cleaned
% Elapsed time is 6.252804 seconds. % Hashing
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