Repeative selecting unique values of matrix
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Lukasz
le 13 Mai 2014
Réponse apportée : Andrei Bobrov
le 13 Mai 2014
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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Jos (10584)
le 13 Mai 2014
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
Plus de réponses (4)
W. Owen Brimijoin
le 13 Mai 2014
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?
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W. Owen Brimijoin
le 13 Mai 2014
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?
Andrei Bobrov
le 13 Mai 2014
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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