# could anyone help me how to generate two different random numbers of same value.

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jaah navi 2021 年 7 月 5 日
コメント済み: jaah navi 2021 年 7 月 6 日
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.
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jaah navi 2021 年 7 月 5 日
Yes, I know that some of the values in A can be repeated in B.
In my case, the size of A and B is very large i.e, (48000,1). When I generate i can find only minimum number of values remains to be same i.e, fo example 2300 out of 48000. But I want the maximum number of values to be same i.e, 45700/48000.
Could you please hlp me onthis.

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### 採用された回答

Steven Lord 2021 年 7 月 5 日
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
A = 5×5
0.9605 0.1936 0.1763 0.7655 0.5840 0.6432 0.6447 0.2002 0.5192 0.1195 0.8839 0.0211 0.4184 0.5463 0.9373 0.2247 0.7324 0.8192 0.9718 0.4654 0.3539 0.5847 0.8463 0.1017 0.9999
A1
A1 = 5×5
0.3280 0.1936 0.1763 0.7655 0.7112 0.6432 0.6447 0.2002 0.5192 0.7247 0.8839 0.9045 0.3648 0.5463 0.9373 0.2247 0.7324 0.8192 0.9718 0.4654 0.4195 0.5847 0.8463 0.1017 0.9999
A2
A2 = 5×5
0.9605 0.1936 0.1763 0.7655 0.5840 0.7112 0.6447 0.2002 0.5192 0.1195 0.8839 0.7247 0.4184 0.5463 0.9373 0.2247 0.9045 0.3280 0.9718 0.4195 0.3648 0.5847 0.8463 0.1017 0.9999
B
B = 5×5
0.3280 0.3758 0.7772 0.0375 0.7112 0.4655 0.5999 0.1197 0.2314 0.7247 0.4005 0.9045 0.3648 0.0259 0.9559 0.6848 0.5505 0.0491 0.4264 0.4717 0.4195 0.3723 0.9590 0.2997 0.9910
Where did A1 flip?
[sort(locationOfMatchesInB).', find(A1 ~= A)]
ans = 6×2
1 1 5 5 8 8 13 13 21 21 22 22
What changed?
A1(A1 ~= A)
ans = 6×1
0.3280 0.4195 0.9045 0.3648 0.7112 0.7247
B(A1 ~= A)
ans = 6×1
0.3280 0.4195 0.9045 0.3648 0.7112 0.7247
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Walter Roberson 2021 年 7 月 6 日
A = rand(5, 5);
A1 = A; A2 = A; % Make copies for later comparison
B = rand(5, 5);
locationOfMatchesInB = randperm(numel(B), 6);

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### その他の回答 (2 件)

Amit Bhowmick 2021 年 7 月 5 日

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))
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Walter Roberson 2021 年 7 月 5 日
A = sort(rand(300,300));
imagesc(A)
title('original')
p = 45700/48000;
B = reshape(B(randperm(numel(B))), size(A));
imagesc(B)
title('rerandom')
mean(ismember(B(:), A)) * 100
ans = 95.2233
p * 100
ans = 95.2083
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.
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jaah navi 2021 年 7 月 6 日

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