how to shift trapped zeros to the bottom keeping leading zeros in given matrix fixed?
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Hi all, I have a problem while finding the probablity transition matrix. The code I have written to find the transition matrix calculate the transition probability matrices of each columns (in the given matrix below) but it excludes the transition between two states if zeros are trapped like
1
0
2 (shown below in bold) So, I wanted shift the trapped zeros to last.
I encountered trapped zeros in the matrix for instance, a matrix like
A = [0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
0 1 0 1 0 0 0 1 2 1 1 2 2;
2 1 2 1 0 0 0 0 1 1 1 1 1;
1 0 0 1 1 1 1 1 2 2 2 1 1]
Now, I want to change this matrix into new matrix like
new_A =[0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
2 1 2 1 1 1 1 1 2 1 1 2 2;
1 1 0 1 0 0 0 1 1 1 1 1 1;
0 0 0 1 0 0 0 0 2 2 2 1 1]
shifting all trapped zeros to the bottom of each column. Any help will be greatly appreciated.
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Arif Hoq
2022-3-21
if your matrix A almost same dimension( 5 X 13) everytime, then
A = [0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
0 1 0 1 0 0 0 1 2 1 1 2 2;
2 1 2 1 0 0 0 0 1 1 1 1 1;
1 0 0 1 1 1 1 1 2 2 2 1 1];
B=A(:,1);
C=B(2:end);
C([2 end])=C([end 2]);
D=[B(1);C];
output=[D A(:,2:end)]
另请参阅
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