Getting all the combinations of 4 vectors?

Problem: I want to get the combinations of 4 vectors so that my output would look something like this:

(all 4 vectors are the same vec=[1:9])

1 1 1 1

1 1 1 2

1 1 1 3

...

1 1 1 9

1 1 2 1

1 1 2 2

1 1 2 3 etc etc

and it would just keep counting up until it reaches 9 9 9 9. Normally, I would use for loops in other languages but I thought I could use the combo feature in matlab.

My Solution: My goal was to test my idea with smaller numbers to see if I could get the combos of 2 vectors from [1:3] and only choosing 2.

Here is what I tried,

vec1=[1:2];

vec2=[1:2];

combos=combnk([vec1 vec2],2)

Output was:

     1     2
     2     2
     2     1
     1     2
     1     1
     1     2

The problem is, it is double counting the combo: 1 2. Am i using the function wrong? I would appreciate any help.

Answer 1

[Konstantinos Sofos]

Hi,

The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired n-column matrix:

As an example:

    vectors = { [1 2], [3 6 9], [10 20] }; %input data: cell array of vectors

    n = numel(vectors); % number of vectors
    combs = cell(1,n); % pre-define to generate comma-separated list
    [combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); % the reverse order in these two
    % comma-separated lists is needed to produce the rows of the result matrix 
    combs = cat(n+1, combs{:}); %concat the n n-dim arrays along dimension n+1
    combs = reshape(combs,[],n); %reshape to obtain desired matrix

Regards,

[Lliam]

Thanks this is exactly what I was looking for, as it allows me to easily modify the vectors.

However, now i need to figure out how: combs, cat, and ngrid actually work lol, Thanks!

Answer 2

[Roger Stafford]

This is the wrong function for your problem. Your problem has 9^4 = 6561 rows of values, which does not correspond to anything generated by 'combnk'. I would suggest Matt Fig's COMBINATOR function in the File Exchange using "permutations with replacement". (A "permutation" with replacement is something of a misuse of the term 'permutation' since it allows such vectors as 1 1 1 1, but that is what you need.)

[Star Strider]

COMBINATOR

[Lliam]

Thanks

Answer 3

[Dean Ranmar]

What about allcomb?

Answer 4

[John D'Errico]

A = dec2base(1111:9999,10) - '0';
A(1:20,:)
ans =
     1     1     1     1
     1     1     1     2
     1     1     1     3
     1     1     1     4
     1     1     1     5
     1     1     1     6
     1     1     1     7
     1     1     1     8
     1     1     1     9
     1     1     2     0
     1     1     2     1
     1     1     2     2
     1     1     2     3
     1     1     2     4
     1     1     2     5
     1     1     2     6
     1     1     2     7
     1     1     2     8
     1     1     2     9
     1     1     3     0
...

A(end,:)
ans =
     9     9     9     9

[Roger Stafford]

The base should be 9 instead of 10 with 1 added to each digit afterward.

A = dec2base(0:9^4-1,9,4)-'/';

[John D'Errico]

Yes. Asleep.

[Lliam]

Thank you