- Do you want the sum of A is the same as the sum of B?
- Do you want exactly 3 (for examples) values in A to be already repeated in B?
could anyone help me how to generate two different random numbers of same value.
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I am having two matrices A and B generated randomly with different values
A=rand(4,3)
B=rand(4,3)
Now I want to know is there any other way such that the some of the values generated in A can be made equal to the values generated in B.
댓글 수: 2
Yongjian Feng
2021년 7월 5일
Not very sure what you need here. Since rand is random, it is possible already some of the values in A can be repeated in B.
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Steven Lord
2021년 7월 5일
A = rand(5, 5);
A1 = A; A2 = A; % Make copies for later comparison
B = rand(5, 5);
numberOfMatches = 6;
locationOfMatchesInB = randperm(numel(B), numberOfMatches);
If you want the elements from B to be in exactly the same locations in A:
A1(locationOfMatchesInB) = B(locationOfMatchesInB);
If you want the elements from B to be put in some location, but not necessarily the same locations, in A:
locationOfMatchesInA = randperm(numel(A), numberOfMatches);
A2(locationOfMatchesInA) = B(locationOfMatchesInB);
Now compare:
A
A1
A2
B
Where did A1 flip?
[sort(locationOfMatchesInB).', find(A1 ~= A)]
What changed?
A1(A1 ~= A)
B(A1 ~= A)
댓글 수: 2
Walter Roberson
2021년 7월 6일
A = rand(5, 5);
A1 = A; A2 = A; % Make copies for later comparison
B = rand(5, 5);
locationOfMatchesInB = randperm(numel(B), 6);
추가 답변 (2개)
Amit Bhowmick
2021년 7월 5일
편집: Amit Bhowmick
2021년 7월 5일
Check the values you will alyas get zero that is they are not equal. More preciesly probability of getting same value is very less. Basically the algorithm of generating random number is such you can learn here
https://en.wikipedia.org/wiki/Random_number_generation
rand(4,3)==rand(4,3) or isequal(rand(4,3),rand(4,3))
댓글 수: 0
Walter Roberson
2021년 7월 5일
A = sort(rand(300,300));
imagesc(A)
title('original')
p = 45700/48000;
mask = rand(size(A)) <= p;
B = [A(mask); rand(numel(A)-nnz(mask),1)];
B = reshape(B(randperm(numel(B))), size(A));
imagesc(B)
title('rerandom')
mean(ismember(B(:), A)) * 100
p * 100
So in this case, the target fraction to keep was 95.2083% and the actual fraction kept was 95.2233%
If it was necessary to have exactly the ratio 45700/48000 stay the same then that could be done... though it becomes trickier if you are working with integers.
댓글 수: 3
Walter Roberson
2021년 7월 6일
A = arrayfun( @my_rand, repelem((1:10).',30), 'UniformOutput', false);
B = arrayfun( @my_rand, repelem((1:10).',30), 'UniformOutput', false);
p = 0.9;
nkeep = ceil(p * numel(A)) ;
keepidx = randperm(numel(A), nkeep);
B(keepidx) = A(keepidx);
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