ten*_*eng 8 matlab transition probability matrix markov-chains
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.
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所以这是我的问题,你如何优雅地构建二阶转移矩阵?我想出的解决方案如下
[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
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是否有单/双线,无环替代?
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编辑:我尝试将我的解决方案与Amro的比较与"x = round(5*rand([1,1000])+ 1);"
% ted teng's solution
Elapsed time is 2.225573 seconds.
% Amro's solution
Elapsed time is 0.042369 seconds.
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有什么区别!仅供参考,grp2idx可在线获取.
请尝试以下方法:
%# 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)
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