How do I create a vector combinations in pairs
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How do I create a vector combinations in pairs ,with no repeated elements ,for example :
A(1,2,3,4)
1,2
1,3
1,4
2,1
2,3
2,4
3,1
3,2
3,4
4,1
4,2
4,3
I tried to use some commands like : perms And combnk , thanks in advanced
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Akzeptierte Antwort
Azzi Abdelmalek
am 18 Sep. 2013
Bearbeitet: Azzi Abdelmalek
am 18 Sep. 2013
a=fliplr(fullfact([4 4]))
a(~diff(a')',:)=[]
or
[ii,jj]=ndgrid(1:4,1:4);
a=[jj(:) ii(:)];
a(~(a(:,1)-a(:,2)),:)=[]
2 Kommentare
Azzi Abdelmalek
am 19 Sep. 2013
n=4
m=3 % number of combinations
a=fliplr(fullfact(ones(1,m)*n));
b=sort(a,2);
idx=any(~diff(b')',2);
a(idx,:)=[]
Weitere Antworten (5)
Roger Stafford
am 19 Sep. 2013
What you are asking for in this comment are known as the partial permutations. I don't know if matlab has such a routine but you can use 'nchoosek' and 'perms' to create one. Let A be a row vector of n elements and let the number of these to be selected in each permutation be called r.
c = nchoosek(A,r)';
ncr = size(c,2);
p = perms([1:r]);
pr = size(p,1);
p = reshape(p',1,[]);
B = zeros(ncr*pr,r);
for k = 1:ncr
B((k-1)*pr+1:k*pr,:) = reshape(c(p,k),r,[])';
end
B will be the desired list of partial permutations.
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Roger Stafford
am 19 Sep. 2013
Here is a more compact way of using 'nchoosek' and 'perms'.
c = nchoosek(A,r)';
B = reshape(c(perms(1:r)',:),r,[])';
where A, r, and B are as before.
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Andrei Bobrov
am 19 Sep. 2013
Bearbeitet: Andrei Bobrov
am 19 Sep. 2013
d = fullfact([4 4]);
out = d(diff(d,[],2)~=0,:);
and
A = [8 2 9 6 1];
n = 3;
ix = fullfact(ones(1,n)*numel(A));
out = A(ix(all(diff(sort(ix,2),[],2),2),:));
and using the ideas by Roger Stafford (they very nice)
c = nchoosek(A,r)';
p = perms([1:r]);
s = size(c);
c(reshape(bsxfun(@plus,p',reshape((0:s(2)-1)*s(1),1,1,[])),s(1),[])');
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Jos (10584)
am 19 Sep. 2013
The simplest way is to create all N*N combinations and weed out the N ones that have the same value.
N = 4 ;
A = 1:N ;
[b2,b1] = ndgrid(A) ; % generalization
q = b1~=b2 ;
% q = ~eye(N) ; % they are all on the diagonal
B = [b1(q) b2(q)]
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