All possible combinations of 2 vectors.

2012 11 22
4 个回答
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1 个投票

Hi everyone.
I have one vector and one number. For example [1 3 5] and 0.
How do I generate all possible combinations? Like this:
0 3 5
1 0 5
1 3 0
0 0 5
0 3 0
1 0 0
0 0 0

2 个评论
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Matt Fig
Matt Fig 2012-11-22
Why is the last row all zeros? It looks like the rule is: take at least one element from each vector, with repetition allowed only for the shorter vector. But then the last row breaks this. So what is the rule?
Artyom
Artyom 2012-11-22
The rule is:
1) we have an n - dimensional vector.
2) replace one number with zero and find all combinations
3) replace two numbers with zero and find all combinations
4) ...
5) replace n-1 number with zero and find all combinations
6) replace n number with zero

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Matt Fig
Matt Fig 2012-11-22
编辑：Matt Fig 2012-11-23

0 个投票

Here is a solution:
function H = mycomb(V)
% Help
L = length(V);
H = cell(1,L);
for ii = 1:L-1
C = nchoosek(1:L,L-ii);
R = cumsum(ones(size(C)));
M = max(R(:,1));
H{ii} = zeros(M,L);
H{ii}(R+(C-1)*M) = V(C);
end
H{L} = zeros(1,L);
H = vertcat(H{:});
Now try it out from the command line:
>> mycomb([4 5 6])
ans =
4 5 0
4 0 6
0 5 6
4 0 0
0 5 0
0 0 6
0 0 0
>> mycomb([4 5 6 7])
ans =
4 5 6 0
4 5 0 7
4 0 6 7
0 5 6 7
4 5 0 0
4 0 6 0
4 0 0 7
0 5 6 0
0 5 0 7
0 0 6 7
4 0 0 0
0 5 0 0
0 0 6 0
0 0 0 7
0 0 0 0

更多回答（3 个）

Andrei Bobrov
Andrei Bobrov 2012-11-22
编辑：Andrei Bobrov 2012-11-22

1 个投票

variant
t = [1 3 5];
ii = perms([t, zeros(size(t))]);
out = unique(sort(t(:,1:numel(t)),2),'rows');
or
t = [1 3 5];
out = [];
n = numel(t);
for jj = 1:n
k = nchoosek(t,n - jj);
out = [out;[zeros(size(k,1),jj),k]];
end
or
k = ones(1,numel(t)) * 2.^(numel(t)-1:-1:0)';
out = bsxfun(@times,t,dec2bin(0:k - 1,numel(t))-'0');
Azzi Abdelmalek
Azzi Abdelmalek 2012-11-22
编辑：Azzi Abdelmalek 2012-11-23

0 个投票

save this function
function y=arrangement(v,n)
m=length(v);
y=zeros(m^n,n);
for k = 1:n
y(:,k) = repmat(reshape(repmat(v,m^(n-k),1),m*m^(n-k),1),m^(k-1),1);
end
then type
x=arrangement([1 3 5 0],3)
out=x(~all(x,2),:)
If you don't need repetition add
s=arrayfun(@(t) sort(out(t,:)),(1:size(out,1))','un',0)
out1=unique(cell2mat(s),'rows')
Matt J
Matt J 2012-11-23
编辑：Matt J 2012-11-23

0 个投票

t=[1 3 5];
n=length(t);
result = bsxfun(@times, [1,3,5], dec2bin(2^n-1:-1:0)-'0')

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