How to generate a random matrix with conditions?

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Firas Al-Kharabsheh
评论： Firas Al-Kharabsheh 2016-4-9
How to generate a matrix ( n x m ) where i need this number represent a number of ones in each row like this
A = [ 2 3 1
3 1 1
4 1 0
2 2 0 ]
I need to generate a matrix like this
X = [ 1 1 0 1 1 1 0 1
1 1 1 0 1 0 0 1
1 1 1 1 0 0 1 0
0 1 1 0 0 0 1 1 ]
Where every numbers in matrix A must be at least zero between its numbers of one
  1 个评论
Steven Lord
Steven Lord 2016-4-8
It sounds like the poster wants something like run-length decoding but where only the length of the runs of 1's are given and it's assumed there are 0's between those runs. But you're right, the poster needs to clarify the rules for how many 0's should be between the runs. It's not just one 0 between each run, as seen in rows 2, 3, and 4.

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回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2016-4-8
编辑：Azzi Abdelmalek 2016-4-8
It's almost what you are asking for!
A = [ 2 3 1
3 1 1
4 1 0
2 2 0 ]
[n,m]=size(A);
ii=max(sum(A,2));
no=ii+m-1;
out=zeros(n,no);
for k=1:n
r=A(k,:);
nz=no-sum(r);
q=num2cell(zeros(1,m-1));
idx=randi(m-1,1,nz-m+1);
for kk=1:numel(idx)
q{idx(kk)}(end+1)=0;
end
aa=[];
for pp=1:m-1
aa=[aa ones(1,r(pp)) q{pp}];
end
aa=[aa ones(1,r(end))];
out(k,:)=aa;
end
out
Andrei Bobrov
Andrei Bobrov 2016-4-8
[m,n] = size(A);
X = zeros(m,sum(A,2) + n - 1);
a = arrayfun(@(x)ones(1,x),A,'un',0);
Xc(:,1:2:2*n-1) = a ;
Xc(:,2:2:end) = {0};
for ii = 1:m
b = [Xc{ii,:}];
X(ii,1:numel(b)) = b;
end
Azzi Abdelmalek
Azzi Abdelmalek 2016-4-8
编辑：Azzi Abdelmalek 2016-4-8
I think this is What you want:
A = [ 2 3 1
3 1 1
4 1 0
2 2 0 ]
[n,m]=size(A);
ii=max(sum(A,2));
no=ii+m-1;
out=zeros(n,no);
for k=1:n
r=A(k,:);
nz=no-sum(r);
q=[ cell(1) num2cell(zeros(1,m-1)) cell(1)];
idx=randi(m+1,1,nz-m+1);
for kk=1:numel(idx)
q{idx(kk)}(end+1)=0;
end
aa=q{1};
for pp=1:m
aa=[aa ones(1,r(pp)) q{pp+1}];
end
out(k,:)=aa;
end
out
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