Transition Matrix that compares two elements to the next two elements.

Hi all, I am trying to modify this code to create a transition matrix that compares two elements to the next two elements. As of now the code successfully compares two elements to the next one element. I have been trying to modify it, but have not had any luck. Any help would be greatly appreciated. Thanks.

close all
clc

dataset = [1 2 1 1 1 2 2 2 1 1 1 2 2];

precision = 1;
markovChain = (round(dataset/precision)).*precision;
%number of states
Nstates = max(markovChain);

      %get Norder-contiguous sequences of the markov chain
      ngrams = [];
      for i = 0:1
          ngrams = [ngrams, circshift(markovChain,[0 -1*(i)])'];
      end
      ngrams = cellstr(num2str( ngrams));
      ngrams = ngrams(1:end-2);

      %create  all combinations of Norder-contiguous sequences
      [x{1:2}] = ndgrid(1:Nstates);

      %format x to cell
      evalStr = ['xCell = cellstr(num2str(['];
      for i = 1:2
          evalStr = [evalStr 'x{' num2str(i) '}(:) '];            
      end
      evalStr = [evalStr ']));'];
      eval(evalStr);

      %map ngrams to numbers
      [gn,~,g]=unique([xCell;ngrams]);
      s1 = g(Nstates^2+1:end);

      %states following the ngrams
      s2 = markovChain(3:end);

      %get transition matrix
      tm = full(sparse(s1,s2,1,Nstates^2,Nstates^2) );

transitionMatrix = bsxfun(@rdivide, tm, sum(tm,2));

[Azzi Abdelmalek]

Can you post the expected result. There may be a better solution then modifying your code

[Ellie]

The expected matrix would be a N^2 x N^2 where N would be the number of states.

I did it by hand and the expected result should be [0 1 0 0; 0.5 0 0 0.5; 1 0 0 0; 0.5 0 0.5 0] where the element in the firs row first col would be the transition from the two elements [1 1] to the next two elements [1 1].

[Azzi Abdelmalek]

where the element in the firs row firs col would [1 1]. What does that mean? Can you explain what do you mean by comparison

[Ellie]

So in the set of data, the subset [1 1] will transition to [1 1] zero times, to [1 2] twice, to [2 1] zero times, and to [2 2] twice. Then those numbers are divided by the total number of times a change occurs from [1 1] which would be 4. I calculated the expected result incorrectly before, it should be [0 0.5 0 0.5; 0.5 0 0 0.5; 1 0 0 0; 0.5 0 0.5 0]

[Azzi Abdelmalek]

For the first column of the result it's clear, what about the second one?

[Ellie]

Well that is the firs row. The second row would be the transitions from [1 2] to [1 1], to [1 2], to [2 1] and to [2 2]. The third row would be the transitions from [2 1] to the above. and The last would be the transitions form [2 2].

[Walter Roberson]

You should have just answered in http://uk.mathworks.com/matlabcentral/answers/231381-error-using-accumarray-third-input-sz-must-be-a-full-row-vector-with-one-element-for-each-column-of

[Walter Roberson]

This is not good code :(

[Walter Roberson]

I have given a solution in http://uk.mathworks.com/matlabcentral/answers/231381-error-using-accumarray-third-input-sz-must-be-a-full-row-vector-with-one-element-for-each-column-of to the form of the question presented there, which does not use those messy eval().

Answer 1

[Azzi Abdelmalek]

a=[1 2 1 1 1 2 2 2 1 1 1 2 2];
b=[0 1 0 0; 0.5 0 0 0.5; 1 0 0 0; 0.5 0 0.5 0];
state=[1 1;1 2;2 1;2 2];
q=zeros(4);
for k=1:4
  for p=1:4
        q(p,k)=numel(strfind(a,[state(k,:)  state(p,:)]));
  end
  q(:,k)=q(:,k)/sum(q(:,k));
end
q

The result

q =

         0    0.5000    1.0000    0.5000
    0.5000         0         0         0
         0         0         0    0.5000
    0.5000    0.5000         0         0