Identify the minimum number of rows in a matrix meeting a condition

3 次查看（过去 30 天）
Amirhossein Moosavi
编辑： Matt J 2020-7-8
Hi everyone,
Please assume that I have a matrix as follows:
A = [0 1 0 0
1 1 0 0
0 0 1 0
0 0 0 1
1 0 1 1]
I would like to identify and selcet the minimum number of rows that have value of one in all columns (altoghether). For example, rows 2 and 5 are the best choice for matrix A. Do you have any suggestion?
Regards,
Amir
  2 个评论
Matt J
Matt J 2020-7-8
编辑：Matt J 2020-7-8
What about this case?
A = [1 1 0 0
0 0 0 1
0 0 1 1]
You could argue that A([1,3],:) is the minimal selection because only those two rows contain a cluster of two [1,1] whereas all three rows contain the single-element cluster [1], possibly as sub-clusters of the [1,1] sequences.
On the other hand, if you make the rule that a cluster of length n cannot belong to a longer cluster of length m>n and can therefore be neighbored only by zeros, then A(2,:) is the minimal selection because now it is the only row qualifying as having a single-element cluster [1].

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2020-7-8
编辑：Matt J 2020-7-8
This solution uses the Optimization Toolbox, although I am still only half certain what row-selection rule you are looking for (see my question here).
[m,n]=size(A);
irp=1:m;
rp=randperm(m); %row permutation
irp(rp)=irp; %inverse permutation
Ap=A(rp,:);
x=optimvar('x',[1,m],'LowerBound',0,'UpperBound',1,'Type','integer');
prob=optimproblem('Objective',sum(x));
prob.Constraints=(x*Ap>=ones(1,n));
sol=solve(prob);
selection=find(round(sol.x(irp)));
  2 个评论
Amirhossein Moosavi
Amirhossein Moosavi 2020-7-8
Yes, it seems to be a solution. However, there are two problems:
1) It is not fast.
2) It is baised toward one combination when multiple combinations meet the condition. For example, see the below matrix:
A = [0 1 0 0
0 0 1 0
0 0 1 0
0 0 1 0
1 0 0 1]
Matt J
Matt J 2020-7-8
编辑：Matt J 2020-7-8
2) Just randomly pre-permute the rows of A (see my edited version above).
1) I find that my version is faster than yours for larger problems even when adding in the above permutation. For example, with A = randi([0,1],40,100), I find
Elapsed time is 0.193646 seconds.%Matt J
Elapsed time is 0.367822 seconds.%Amir

请先登录，再进行评论。

更多回答（2 个）

madhan ravi
madhan ravi 2020-7-8
编辑：madhan ravi 2020-7-8
According to what I understood, see if this does what you want to do:
[a, b] = unique(A, 'rows', 'stable');
s = sum(a, 2);
rows = b(s > 1)
  1 个评论
Amirhossein Moosavi
Amirhossein Moosavi 2020-7-8
Thanks! Unfortunately, this is not a solution for my quesiton. You would follow my above examples (1 and 2) for further clarification. If you still need more explanation, please let me know.

请先登录，再进行评论。

Amirhossein Moosavi
This is a code that addresses my question, but I am not sure whether it is the fastest way or not.
EG = cell(1, 100); % A place to store eligible combinations
cn1 = 1; cn2 = 1; FG = 0;
for i = 1:size(A, 2)
C = nchoosek(1:size(A, 1), cn1); % A function to create all combinations of rows (starting from the minimum number)
for j = 1:size(C, 1)
S = sum(A(C(j, :), :));
if all(S >= 1) && (sum(S) >= size(A, 2)) % The condition that I explained in my question
EG{cn2} = C(j, :);
cn2 = cn2 + 1;
FG = 1;
end
end
if FG == 1 % If at least one combination has been found, break the loop (no need to check combinations with more number of rows)
break;
end
cn1 = cn1 + 1;
end
EG(cn2:end) = [];
selection = EG{randi([1 numel(EG)])}; % Randomly select one of the eligible combinations
Do you have any suggestion to make it faster?
  2 个评论
Matt J
Matt J 2020-7-8
编辑：Matt J 2020-7-8
My method is faster for larger problems. For this matrix,
A = randi([0,1],40,100);
I obtain these timings,
Elapsed time is 0.193646 seconds.%Matt J
Elapsed time is 0.367822 seconds.%Amir

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Elementary Math 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by