构建在Matlab的多阶马尔可夫链转移矩阵(Constructing a multi-order M

6个状态中的第一级转变矩阵可以非常优雅构造为如下

 x = [1 6 1 6 4 4 4 3 1 2 2 3 4 5 4 5 2 6 2 6 2 6]; % the Markov chain
 tm = full(sparse(x(1:end-1),x(2:end),1)) % the transition matrix.

因此，这里是我的问题，你如何构建一个二阶转移矩阵优雅？ 我想出了一个解决方案如下

 [si sj] = ndgrid(1:6);
 s2 = [si(:) sj(:)]; % combinations for 2 contiguous states
 tm2 = zeros([numel(si),6]); % initialize transition matrix
 for i = 3:numel(x) % construct transition matrix
   tm2(strmatch(num2str(x(i-2:i-1)),num2str(s2)),x(i))=...
   tm2(strmatch(num2str(x(i-2:i-1)),num2str(s2)),x(i))+1;
 end

是否有一个/双内胆，无环的选择？

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编辑：我试图对荷银与 “X =轮（5 *兰特（[1,1000]）+ 1）;” 我的比较溶液

 % ted teng's solution
 Elapsed time is 2.225573 seconds.
 % Amro's solution
 Elapsed time is 0.042369 seconds. 

一有什么区别！ 仅供参考， grp2idx可在网上。

Answer 1:

尝试以下方法：

%# sequence of states
x = [1 6 1 6 4 4 4 3 1 2 2 3 4 5 4 5 2 6 2 6 2 6];
N = max(x);

%# extract contiguous sequences of 2 items from the above
bigrams = cellstr(num2str( [x(1:end-2);x(2:end-1)]' ));

%# all possible combinations of two symbols
[X,Y] = ndgrid(1:N,1:N);
xy = cellstr(num2str([X(:),Y(:)]));

%# map bigrams to numbers starting from 1
[g,gn] = grp2idx([xy;bigrams]);
s1 = g(N*N+1:end);

%# items following the bigrams
s2 = x(3:end);

%# transition matrix
tm = full( sparse(s1,s2,1,N*N,N) );
spy(tm)

文章来源: Constructing a multi-order Markov chain transition matrix in Matlab