Passing through each element in a matrix just once and randomly

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Carlos 2016-4-25

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评论： John BG 2016-4-27

Hello everyone, I have a very big nxn matrix which I want to visit element-per-element randomly, without repetition, i.e. first the A(4,100) term, then the A(53,20) term, and so on. I'm not saying I want the value of each element to be random, I want the order in which each element is visited to be random, and if possible to go through each element just once, due to the vastness of the matrix (2500x2500). I'm thinking the way to go is in the posible permutations between two 1-to-2500 groups of elements. But I'm not completely sure about that approach. Any help will be highly appreciated, Carlos

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Answer 1

Roger Stafford 2016-4-25

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B = A(randperm(prod(size(A))));

The vector B will have each value of A once and only once in random order.

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Carlos 2016-4-26

Totally did it, thanks a lot.

Jos (10584) 2016-4-26

numel(A) can replace prod(size(A))

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Answer 2

Image Analyst 2016-4-26

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Carlos:

If you want "I want the order in which each element is visited" and "the way to go" then I would get the indexes of that route, path, sequence of indexes, or whatever you want to call it, like this:

linearIndexes = randperm(numel(A)); % All indexes scrambled up.

This gives the indexes to visit. If you want the values at those indexes, then you can use Roger's method, which is essentially

randomValues = A(linearIndexes);

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Answer 3

John BG 2016-4-26

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编辑：John BG 2016-4-26

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With Stafford's answer, there is a chance you do not get certain combinations and repeat some others, if it's an image, certain pixels, while you may repeat other pixels.

The exhaustive way is to randomly choose one by one, but while avoiding repetition, go through all possible combinations.

for that purpose:

P=combinations(1:2500,1:2500);
L=randperm(length(P));   % L is 6250000
for k=L
   A(k,:)
   % process here
end

If you find this answer of any help solving your question,

please click on the thumbs-up vote link, thanks in advance

John

jgb2012@sky.com

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John BG 2016-4-26

编辑：John BG 2016-4-26

With Stafford answer there is a small possibility that with randperm, (not randi) you get same shuffle, because your are not shuffling the numeral referring to all possible combinations, but the bucket stack you pull cards from.

randperm shuffles the stack, more or less randomly, so you may get twice same combination, while missing certain combinations, if not shuffling enough times. And the more times you shuffle, attempting to avoid missing any possible combination, the higher the possibility you repeat a combination.

Yet with a single shuffle of the numerals of all possible permutations, and then reading that single shuffle of pointers, you can neither miss any pixel combination nor hit twice same pixel combination.

Carlos is clear on this point: he wants to read ALL pixels, each once only, randomly.

Why do you think that shuffling the stack you are going to get exactly once each pixel combination, but you are not going to repeat ANY pixel combination at all?

Image Analyst seems to agree, it's the indices of all permutations what you have to shuffle.

You really need to get this right, because if you tell your students that it would work to for instance encrypt/decrypt an image, you are losing/adding noise in a process that should be transparent.

Hope it's clear if not just let me know and I will kindly add more examples.

John

jgb2012@sky.com

Roger Stafford 2016-4-26

To John: You speak of the "small possibility of getting the same shuffle". However, in so doing you seriously misinterpret Carlos' request, in my opinion. He says nothing about a repeated visit with a repeated "shuffle". He wants a single random ordering of the elements in his vector, and if he happens to by chance get the same order as he already had, that is presumably all right - it is one of the possible permutations. (Of course, with 6,250,000-factorial possible permutations the chance of that is mighty small.) Note carefully what the word 'permutation' means, John. It means that each quantity in the original list occurs again in the permuted ordering of that list just once and only once - no repetitions, no omissions.

In any case, this is what Matlab's 'randperm' function does, and I believe it does a very satisfactory job. I have observed the logic behind the 'randperm' operation. For "randperm(N)" it computes N successive random values, each ranging from zero to one using the 'rand' function, and then it sorts these values. It uses the permutation that is involved in that sorting as its output. If you assume that the 'rand' function is doing its job properly, and that 'sort' doesn't lose track of some of its values, this means that the result is a valid permutation and each of the N-factorial possible permutations is equally likely. I believe that is all Carlos is asking. If you doubt that, I suggest you direct your question to him directly.

John BG 2016-4-27