Combine and sorting array values

Hello!

I got a result matrix A

A = [3 9 4]

And another matrix B

  [ 3  8
    4  8
   15  5
B= 16  1
   16  2
   16  7
   17  6
   17  2 ]

First i want to combine A with all values of the second column of B that have a unique value in the first column -> Matrix AB

AB = [3 9 4 8 5]

As a final result i want the combinations of AB with all values of the second column of B that have not a unique value in the first column ->Result

Result= [3 9 4 8 5 1 6   3 9 4 8 5 1 2
         3 9 4 8 5 2 6   3 9 4 8 5 2 2
         3 9 4 8 5 7 6   3 9 4 8 5 7 2]

How can i do that?

Many thanks in advance!

Answer 1

[DGM]

Maybe something like:

A = [3 9 4];
B = [3  8
    4  8
    15  5
    16  1
    16  2
    16  7
    17  6
    17  2];

U = unique(B(:,1));
c = histcounts(B(:,1),U);
col1unique = ismember(B(:,1),U(c==1));

AB = [A unique(B(col1unique,2),'stable').']

AB = 1×5

3 9 4 8 5

combpool = unique(B(~col1unique,2));
combs = nchoosek(combpool,2);
result = [repmat(AB,[size(combs,1) 1]) combs]

result = 6×7

3 9 4 8 5 1 2 3 9 4 8 5 1 6 3 9 4 8 5 1 7 3 9 4 8 5 2 6 3 9 4 8 5 2 7 3 9 4 8 5 6 7

If you want the result formatted as shown, then:

result = reshape(result.',size(result,2)*2,[]).'

result = 3×14

3 9 4 8 5 1 2 3 9 4 8 5 1 6 3 9 4 8 5 1 7 3 9 4 8 5 2 6 3 9 4 8 5 2 7 3 9 4 8 5 6 7

[L. Edwin M.]

Thanks! Works for the example, but if B is

B = [3  8
     4  8
    15  5
    16  1
    16  2
    16  7]

then i get AB without the "5"

AB = 1     2     3     8

I don't clearly get it why..

[L. Edwin M.]

And the result combinations are also wrong..

It's like the values of 16 are in one group and the values of 17 are in another group.

Now i want all combinations of group 16 with group 17..

Would be great if you can help me!! Thanks a lot!

[DGM]

Auugh. I should've seen that coming. I wrote that using the old histc() and just substituted the newer histcounts() and ignored how it behaves differently on the ends. I'm just going to stick with histc() instead of trying to set up bins for histcounts().

A = [3 9 4];
B = [3  8
    4  8
    15  5
    16  1
    16  2
    16  7];

U = unique(B(:,1));
c = histc(B(:,1),U); %#ok<HISTC>
col1unique = ismember(B(:,1),U(c==1));
AB = [A unique(B(col1unique,2),'stable').']

AB = 1×5

3 9 4 8 5

Everything else should be the same

combpool = unique(B(~col1unique,2));
combs = nchoosek(combpool,2);
result = [repmat(AB,[size(combs,1) 1]) combs]

result = 3×7

3 9 4 8 5 1 2 3 9 4 8 5 1 7 3 9 4 8 5 2 7

EDIT: I went ahead and wrangled a histcounts() solution anyway

A = [3 9 4];
B = [3  8
    4  8
    15  5
    16  1
    16  2
    16  7];

[c ed] = histcounts(B(:,1),'binmethod','integers');
col1unique = ismember(B(:,1),ed(c==1)+0.5);
AB = [A unique(B(col1unique,2),'stable').']

AB = 1×5

3 9 4 8 5

[L. Edwin M.]

Perfect thanks!

Still didn't get the last step yet..thought about to store all values of the second column which have

the same value in the first column in one cell and then get all combinations of them.

C = B(~col1unique, :) 
C =

    16     1
    16     2
    16     7
    17     6
    17     2
    

Combinations wanted

CC = [1 6; 1 2; 2 6; 2 2; 7 6; 7 2] 

So at the end i can combine AB with CC..

Any idea?

[DGM]

Oh I misunderstood what kind of combinations you were after. I was just doing combinations of all remaining (unique) elements of col2.

This does combinations of the groups within col2 associated with the nonunique blocks in col1

A = [3 9 4];
B = [3  8
    4  8
    15  5
    16  1
    16  2
    16  7
    17  6
    17  2];

[c ed] = histcounts(B(:,1),'binmethod','integers');
col1unique = ismember(B(:,1),ed(c==1)+0.5);
AB = [A unique(B(col1unique,2),'stable').']

AB = 1×5

3 9 4 8 5

% split col2 of B into groups associated with nonunique elements of col1
Bnonu = B(~col1unique,:);
g = findgroups(Bnonu(:,1));
ngroups = max(g);
elements = cell(1,ngroups);
for k = 1:ngroups
    elements{k} = Bnonu(g==k,2);
end

% find combinations of elements between groups
combinations = cell(1,ngroups); %set up the varargout result
[combinations{:}] = ndgrid(elements{:});
combinations = cellfun(@(x) x(:), combinations,'uniform',false);
result = [combinations{:}];
result = [repmat(AB,[size(result,1) 1]) result]

result = 6×7

3 9 4 8 5 1 6 3 9 4 8 5 2 6 3 9 4 8 5 7 6 3 9 4 8 5 1 2 3 9 4 8 5 2 2 3 9 4 8 5 7 2

based on this:

https://www.mathworks.com/matlabcentral/answers/98191-how-can-i-obtain-all-possible-combinations-of-given-vectors-in-matlab#answer_252633

[L. Edwin M.]

Many thanks!!!!