Generation of a matrix based on a defined order of numbers

Hi all,

I need your help regarding the generation of a set of numbers.

Suppose, I have two vectors:

v1 = [2 3 4 5 6 7 8 9]

v2 = [3 4 5 6 7]

From v1, I would like to draw five numbers such that these five numbers include only (and any) two values of v1 (always only two numbers in the sequence), i.e.,

2 2 9 9 9

3 3 7 7 7

2 9 9 9 9

.........etc

Then I would like to draw five number from v2, however, the five numbers drawn from v1 are based on only one number that is (always one number in the sequence),

3 3 3 3 3

4 4 4 4 4

........

In the end, the sequences would include 10 numbers where every sequence of the five numbers drawn from v1 are combined with every other sequence of numbers drawn from v2. That is,

2 2 9 9 9 3 3 3 3 3

2 2 9 9 9 4 4 4 4 4

2 2 9 9 9 5 5 5 5 5

. ............

I would like to find out all the combinations that are possible, i.e., starting from

2 2 2 2 3 3 3 3 3 3

2 2 2 2 3 4 4 4 4 4

2 2 2 2 3 5 5 5 5 5

2 2 2 2 3 6 6 6 6 6

2 2 2 2 3 7 7 7 7 7

.........

.......

8 9 9 9 9 7 7 7 7 7

Many thanks.

Answer 1

[Guillaume]

One way to do it:

v1 = [2 3 4 5 6 7 8 9];
v2 = [3 4 5 6 7];

pick1 = nchoosek(v1, 2)';
idx1 = hankel([1 1 1 1 2], [2 2 2 2]) + permute(2*(0:size(pick1, 2)-1), [1 3 2]);
pick1 = reshape(pick1(idx1), 5, [])
pick2 = repmat(v2', 1, 5);
[pick2, pick1] = ndgrid(num2cell(pick2, 2), num2cell(pick1', 2));
result = cell2mat([pick1(:), pick2(:)])

[Fayyaz]

Many thanks.

Answer 2

[Image Analyst]

I believe this works:

v1 = [2 3 4 5 6 7 8 9]
v2 = [3 4 5 6 7]
% Get first vector of 5 numbers from v1.
twoIndexes = randperm(numel(v1), 2)
twoNumbers = sort(v1(twoIndexes))
index = randi(4)
firstVec = [twoNumbers(1) * ones(1, index), twoNumbers(2) * ones(1, 5 - index)]
% Get second vector of 5 numbers from v2.
secondIndex = randperm(numel(v2), 1)
secondVec = v2(secondIndex) * ones(1, 5)
% Stitch together to get the final vector.
finalVec = [firstVec, secondVec]

[Fayyaz]

Many thanks for this. The code works, however, I believe I didn't explain my question clearly. I would like to find out all the combinations that are possible, i.e., starting from

(2 2 2 2 3) (3 3 3 3 3)

.........

.......

(8 9 9 9 9) (7 7 7 7 7)

[Image Analyst]

A set of for loops can do it. Can't you figure it out?

What's the use case for this? Why do you want/need to do this unusual thing?

[Fayyaz]

Thanks for the hint. I need every combination so that at the end I get 10 combinations which minimise the determinant of the asymptotic variance-covariance matrix for a utility function

Answer 3

[Andrei Bobrov]

v1 = [2 3 4 5 6 7 8 9];
v2 = [3 4 5 6 7];
n = numel(v2);

q = nchoosek(v1,2)';
m = reshape(permute(...
               q(2 - tril(ones(n-1,n)) + reshape(0:size(q,2)-1,1,1,[])),[2,1,3]),n,[])';
v = v2(:)*ones(1,5);
n2 = size(m,1);
out = [m(kron((1:n2)',ones(n,1)),:), repmat(v,n2,1)];

Added

m = reshape(permute(q( bsxfun(@plus,2 - tril(ones(n-1,n)),...
               reshape(0:size(q,2)-1,1,1,[])) ),[2,1,3]),n,[])'; % for MATLAB <= R2016a

[Fayyaz]

Dear Andrei Bobrov, Many thanks for this. It works perfectly :)