How can i manipulate a matrix with this characteristics?

2014 4 9
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更新时间：2014 4 10
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Hi, i have a matrix wtih repeated elements like:
[1 4 1 4 2 3 3 5 5 6 6 5]
i want break that matrix. And then in each other matrix, running in a cycle, have only the index referent to each number (1, 2, 3, ...), something like:
1 2 3 4 5 6
[1 4] [2 3] [2 3 [1 4] [3 5 [5 6
3 5] 5 6 6 5]
6 5]
I tried creating a 3 dimension matrix (i,j,k) and each 3rd dimension having above matrix. That hypothesis gives me the repeated solutions inside. I tried to get them out but I might not be able to.
I hope someone can help me out.
PS: Some question that may have, just do it pls.
Regards, Vitor Ribeiro.

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Joseph Cheng
Joseph Cheng 2014-4-9
编辑：Joseph Cheng 2014-4-9

0 个投票

You can use cells for this. I do not see the pattern you have for [1 4 1 4 2 3 3 5 5 6 6 5] to the broken up matrix. but to set the broken up matrix with index references, cells should work.
MATRIX{1} = [1 4];
MATRIX{2} = [2 3];
MATRIX{3} = [2 3;3 5];
etc.
Which should be simple enough to accomplish in a for loop with whatever initial matrix breakup pattern is.

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Vitor Ribeiro
Vitor Ribeiro 2014-4-10
I have implemented this code, following your ideia.
z=1;
for a=1:size(bus,1)
z=1;
for i=1:size(branch_,1)
for j=1:2
if branch_(i,j)==bus(a)
brx{a,z}=branch_(i,1:2);
z=z+1;
end
end
end
end
Now i need to ensure that in the matrix brx i don't have repeated contents in each row.
[1,4] [1,4] [] [] [] []
[8,2] [8,2] [] [] [] []
[3,6] [3,6] [] [] [] []
[1,4] [1,4] [4,5] [9,4] [4,5] [9,4]
[4,5] [4,5] [5,6] [5,6] [] []
[3,6] [5,6] [5,6] [3,6] [6,7] [6,7]
[6,7] [6,7] [7,8] [7,8] [] []
[8,2] [7,8] [7,8] [8,2] [8,9] [8,9]
[9,4] [8,9] [8,9] [9,4] [] []
Considering that the real aspect of inside matrix brx is represented above. I tried to do a comparison with a conditional structure if. Just like:
if brx{i,j}==brx{i,j+1}
brx{i,j+1}={};
end
But in fact, a comparison between this 2 elements results in a [1 1] matrix, cause it compares each one of the 2 element of inside cells of the matrix brx.
If you know a way to eliminate repeated cells of brx matrix I appreciate you answer me.
I tried below code, and obviously did not result. I have no experience working with cells. Hope someone can help me out.
for i=1:size(brx,1)
for j=1:size(brx,2)-1
for k=1:size(brx,1)
if brx{i,j}==brx{i,k}
brx{i,j+1}={};
end
end
end
end
Best reagards, Vitor Ribeiro.

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Vitor Ribeiro
Vitor Ribeiro 2014-4-10

0 个投票

I have implemented this code, following your ideia.
z=1;
for a=1:size(bus,1)
z=1;
for i=1:size(branch_,1)
for j=1:2
if branch_(i,j)==bus(a)
brx{a,z}=branch_(i,1:2);
z=z+1;
end
end
end
end
Now i need to ensure that in the matrix brx i don't have repeated contents in each row.
[1,4] [1,4] [] [] [] []
[8,2] [8,2] [] [] [] []
[3,6] [3,6] [] [] [] []
[1,4] [1,4] [4,5] [9,4] [4,5] [9,4]
[4,5] [4,5] [5,6] [5,6] [] []
[3,6] [5,6] [5,6] [3,6] [6,7] [6,7]
[6,7] [6,7] [7,8] [7,8] [] []
[8,2] [7,8] [7,8] [8,2] [8,9] [8,9]
[9,4] [8,9] [8,9] [9,4] [] []
Considering that the real aspect of inside matrix brx is represented above. I tried to do a comparison with a conditional structure if. Just like:
if brx{i,j}==brx{i,j+1}
brx{i,j+1}={};
end
But in fact, a comparison between this 2 elements results in a [1 1] matrix, cause it compares each one of the 2 element of inside cells of the matrix brx.
If you know a way to eliminate repeated cells of brx matrix I appreciate you answer me.
I tried below code, and obviously did not result. I have no experience working with cells. Hope someone can help me out.
for i=1:size(brx,1)
for j=1:size(brx,2)-1
for k=1:size(brx,1)
if brx{i,j}==brx{i,k}
brx{i,j+1}={};
end
end
end
end
Best reagards, Vitor Ribeiro.

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