Generate random integers with a condition

Hi all,
I have a matrix A 84x1 (attached). I need to replace ZEROS by randi(3). However, I need to ensure that the number of 1s, 2s or 3s does not exceed 33.
Hope the problem is clear.
Thank you

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Jan
Jan 2018년 3월 7일
편집: Jan 2018년 3월 7일
The number of 1s and 2s and 3s together, or for each value? Do you want to replace all zeros by the same random value, or should each zero replaced by an individual random value?
N/A
N/A 2018년 3월 7일
Thanks for your response. Number of 1s, 2s and 3s for each values, i.e. I cannot have more than 33 1s or 33 2s or 33 3s. All zeros are to be replaced by an individual random value
Thanks
Jan
Jan 2018년 3월 7일
I've adjusted the code. See [EDITED]

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Jan
Jan 2018년 3월 7일
편집: Jan 2018년 3월 7일

0 개 추천

% [EDITED] after the comment:
A = randi([0,5], 84, 1);
z = (A == 0);
nz = sum(z);
ready = false;
while ~ready
A(z) = randi(3, nz, 1);
ready = sum(A==1) < 33 && sum(A==2) < 33 && sum(A==3) < 33;
end
To be sure, add a maximum number of loops. Better stop with an error than run an infinite loop:
count = 0;
while ~ready && count < 1e6
...
count = count + 1;
end
if count == 1e6
error('Cannot find matching values.')
end
A constructive method might be better also: Create a pool with 32 1s, 2s and 3s, and select the elements randomly out of this pool:
pool = repelem(1:3, 32);
index = randperm(numel(pool), nz);
value = pool(index);
Now the vector value must contain less than 33 ones, twos or threes.

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N/A
N/A 2018년 3월 7일
Thanks for your insights. Creating a pool and selecting the elements randomly out of this pool is a good point
N/A
N/A 2018년 3월 7일
Works perfectly! Cheers!

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