Repeative selecting unique values of matrix

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Lukasz
Lukasz 2014-5-13
回答： Andrei Bobrov 2014-5-13
Helo All!
I am new on this forum so please be understanding. My problem is follows:
I have matrix M [3x9] containing three 9-element vectors (rows) of values 1..9 in a different order. For example:
M(1,1:9) = [2 3 8 1 4 7 6 5 9]
M(2,1:9) = [1 3 8 9 4 6 5 2 7]
M(3,1:9) = [2 8 1 4 6 3 9 5 7]
I want to take 9 elements from M, by 3 elements from each row with no duplicates, starting from begin of each row. In this example will be: 2 -> 1 -> 8 (because 2 was before) -> 3 -> 9 (because 3 and 8 were before) -> 4 (because 8 and 1 was before) etc...
As a result I want to obtain R matrix [3x3] of values R = [2 3 7; 1 9 6; 8 4 5]
I have started like this:
P = zeros(3,9); % buffer containing already selected elements of M
for X=1:9
for Y=1:3
if ismember(M(Y,X),P)==0 % if the current element has not been already selected
P(Y,X)=M(Y,X); % copy this element from matrix M to buffer P
else
*** What should be here? ***
end
end
end
Thanks in advance!
Lukas

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采纳的回答

Jos (10584)
Jos (10584) 2014-5-13
Here is a working algorithm
M = [2 3 8 1 4 7 6 5 9 ;
1 3 8 9 4 6 5 2 7 ;
2 8 1 4 6 3 9 5 7]
P = zeros(n,m,1) ;
Mcopy = M ;
for rankX = 1:3,
for userX = 1:3,
tmp = Mcopy(userX,:) ; % select current user
tmp = tmp(tmp>0) ; % find the remaining values
P(userX, rankX) = tmp(1) ; % select the first (most desired)
Mcopy(Mcopy==tmp(1)) = 0 ; % remove them from M
end
end
P
  1 个评论
Lukasz
Lukasz 2014-5-13
@ Jos (10584), this is exactly what I was looking for, it works! Thank you very much!
@ W. Owen Brimijoin - Thanks also to you. I think I messed things up when explaining and it was hard to understand me.
Regards

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更多回答（4 个）

W. Owen Brimijoin
W. Owen Brimijoin 2014-5-13
You don't really need the three vectors to begin with, you can make a matrix containing a random permutation (randperm.m) of the numbers 1:9 as follows:
R = reshape(randperm(9),3,3);
Unless I am misunderstanding what you're trying to achieve?
Lukasz
Lukasz 2014-5-13
W. Owen Brimijoin thanks for your answer.
Selecting values from mentioned matrix M cannot be random. Each row of M contains elements arranged in order from most desirable to least. Imagine we have 3 users, each user should get 3 elements but also, each user wants to receive the best (most desirable) values of M. Moreover, values selected by user #1 cannot be chosen by user #2 or #3.
W. Owen Brimijoin
W. Owen Brimijoin 2014-5-13
Ok, now I understand what you're trying to do. The first three values will always be the first three three from user 1, then the first different ones from user 2, and so on.
R = M(1,1:3);
r = M(2,setdiff(M(2,:),M(1,1:3)));
R(2,:) = r(1:3);
R(3,:) = M(2,setdiff(M(2,:),R));
Better?
  1 个评论
Lukasz
Lukasz 2014-5-13
I think you are very close but let me explain one more thing:
I would like users to choose values in fair manner like below:
1. User #1 selects first value from row #1 that has not been chosen before (in first step, it will be value M(1,1).
2. User #2 selects first value from row #2, that has not been chosen before by user #1.
3. User #3 selects first value from row #3 that has not been chosen before by users #2 and #1.
4. Back to step 1 to select next value for user #1.
I want to users select values one-by-one.
Thanks a lot and sorry for the confusion.

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Andrei Bobrov
Andrei Bobrov 2014-5-13
M = [2 3 8 1 4 7 6 5 9
1 3 8 9 4 6 5 2 7
2 8 1 4 6 3 9 5 7];
ii = ndgrid(1:size(M,1),1:size(M,2));
[a,b] = unique(M(:),'stable');
i0 = ii(b);
out = zeros(3);
for jj = 1:3
out(jj,:) = a(find(i0==jj,3));
end

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