How do I create an Array of random binary?

Essentially, I have ten matrices in my array of 60x96x10 and for each matrix I need to have random binary numbers, but the indices of the 1's cannot repeat through the array and each of the 60x96 needs to have a one only once.

5 个评论

I don't understand. If you have 60 by 96 matrixes, and 10 slices of those, then in any given vertical column through the slices in the "z" direction, the "1" can appear at a slice number of 1 through 10. So you can only have 10 vertical columns that have unique slice values before you'll need to repeat a slice value. So what does "cannot repeat" mean to you?
It means that when you slice in the "z" direction, anywhere you do it, 1 will appear once and only once.
The question is not clear: Do you have 10 ones finally, or 60*96?
sorry - yes i have 60*96 ones
tash, if you have 10 slices and only one "1" in each slice, then you will have only 10 1's, not 60x96. If you do have 60x96 ones, then you MUST have some z slices that have more than one 1 in them. If you didn't then you'd have only 10 1's, not 60x96 of them.

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回答(2 个)

Perhaps you want 10 unique numbers in the range of 1:60*96 as indices?
Index = randperm(1:60*96, 10);
Result = zeros(60, 96, 10);
Ind = sub2ind([60*96, 10], Index, 1:10);
Result(Ind) = 1;

1 个评论

thank you for this response but i would like 60*96 ones finally, so how would i tweak this response?

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Jon
Jon 2015-8-3
编辑:Jon 2015-8-3
M = zeros(60,96,10);
for j = 1:size(M,3)
rand_row_idx = randi([1 size(M,1)]);
rand_col_idx = randi([1 size(M,2)]);
M(rand_row_idx,rand_col_idx,j) = 1;
end
Edit: just saw that you want 60*96 ones finally; this is for 10 ones. For 5760 ones, you can use
M = zeros(60,96,10);
for j = 1:size(M,1)
for jj = 1:size(M,2)
rand_z = randi([1 size(M,3)]);
M(j,jj,rand_z) = 1;
end
end

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