Reputation: 25
I want to produce a cell array as below in matlab:
P= {100;010;000;000;001}
{100;000;010;000;001}
{100;000;000;010;001}
{000;100;010;000;001}
{000;100;000;010;001}
{000;000;100;010;001}
where P= {5x3 cell} {5x3 cell} {5x3 cell} {5x3 cell} {5x3 cell} {5x3 cell};
from the 5x3 matrix, the pattern should be only one '1' in each column and the third column is '1' at final row in every cell.
i.e : P{1}= 100;010;000;000;001
how should I do that?
From what I had made up till now:
PP=zeros(5,3);
P={1,6};
P={PP,PP,PP,PP,PP,PP};
how can I put the ones in the cell?
thank you so much
Upvotes: 2
Views: 214
Reputation: 19880
Try this:
x = unique(perms([0 0 2 4]),'rows'); %# we'll convert 2 to 10 and 4 to 100 later
xi = randperm(size(x,1)); %# permute rows randomly
n = 6;
xi = xi(1:n); %# k random rows
x = x(xi,:);
x = [ x ones(n,1) ]'; %# add 1s and transpose
%# convert to strings representing binary numbers, then to cell array
out = reshape(cellstr(dec2bin(x)), size(x,1), n);
%# split the cell array by columns
out = mat2cell(out, size(out,1), ones(1,n));
If you actually need numbers (0, 10, 100), not strings, the code will be a little shorter:
x = unique(perms([0 0 10 100]),'rows');
xi = randperm(size(x,1));
n = 6;
xi = xi(1:n);
x = x(xi,:);
x = [x ones(n,1)]';
out = mat2cell(x,size(x,1),ones(1,n));
UPDATE:
According your comment, here is the new code:
n = 6;
P = repmat({zeros(5,3)},1,n); %# output array preallocation
x = unique(perms([0 0 1 2]),'rows');
xi = randperm(size(x,1));
xi = xi(1:n);
x = x(xi,:);
c = [x ones(n,1)+2]'; %# column index
r = repmat((1:5)',1,n); %# row index
for k=1:n
P{k}(sub2ind( [5 3], r(c(:,k)>0,k), c(c(:,k)>0,k) )) = 1;
end
If you need the array of logical matrices just substitute zeros
with false
in the 2nd line:
P = repmat({false(5,3)},1,n);
Upvotes: 1