splitting matrix to different row

Ali
2016 5 26
1 个回答
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更新时间：2016 5 26
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0 个投票

I have the following matrix and I want to split it.
A=[0 2 4 ,5 0 4]
it should be like this:
[0 2]
[2 4]
[4 2]
[5 0]
[0 4]
[4 5]
Please write me, If you have answer. Thanks

4 个评论
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Azzi Abdelmalek
Azzi Abdelmalek 2016-5-26
编辑：Azzi Abdelmalek 2016-5-26
A=[0 2 4 ,5 0 4] is a row vector. And you didn't explain how to get the result
Joseph Cheng
Joseph Cheng 2016-5-26
also looking at the little pattern you have should [4 2] be [4 0] as it looks like the [4 5] as a wrap around pairing
Ali
Ali 2016-5-26
编辑：Ali 2016-5-26
Many thanks for the response.. I wrote it wrong.
I have a matrix with 260 row and 3 columns. I would like to produce a matrix with 780 row and 2 columns. for instance:
A=[0 2 4; 5 0 4]
I would like to have something like this:
[0 2]
[2 4]
[4 0]
[5 0]
[0 4]
[4 5]
the cyclist
the cyclist 2016-5-26
I was sent the following via email. I think it is a clearer statement of the request:
# Elements
1 0 2
0 3 2
4 1 6
6 1 2
3 5 7
3 7 2
6 2 7
8 4 6
I would like to have this:
1 0
0 2
2 1
0 3
3 2
2 0
4 1
1 6
6 4
6 1
1 2
2 6
3 5
5 7
7 3
3 7
7 2
2 3
6 2
2 7
7 6
8 4
4 6
6 8

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the cyclist
the cyclist 2016-5-26
编辑：the cyclist 2016-5-26

0 个投票

Trying to piece together all the guesses that these kind volunteers have made in trying to help you. Does this do what you want?
A = [0 2 4; 5 0 4];
At = A';
chunkSize = size(A,2);
shiftedIndex = bsxfun(@plus,mod(1:chunkSize,chunkSize)',[0:chunkSize:numel(At(:))-chunkSize]) + 1;
B = [At(:) At(shiftedIndex(:))]
[ EDIT: I changed this code to correspond to what I wrote in my comment below. Given the new information you provided, I think this is correct.]

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Ali
Ali 2016-5-26
Many thanks the cyclist. I wrote it wrong.
I have a matrix with 260 row and 3 columns. I would like to produce a matrix with 780 row and 2 columns. for instance:
A=[0 2 4; 5 0 4]
I would like to have something like this: [0 2] [2 4] [4 2] [5 0] [0 4] [4 5]
the cyclist
the cyclist 2016-5-26
It is still not easy to interpret what you have, and what you want. You have not defined the rule to get from one to the other.
Please tell us the rule, so we don't have to keep guessing.
the cyclist
the cyclist 2016-5-26
编辑：the cyclist 2016-5-26
Here is my new best guess at what you want:
A = [0 2 4; 5 0 4];
At = A';
chunkSize = size(A,2);
shiftedIndex = bsxfun(@plus,mod(1:chunkSize,chunkSize)',[0:chunkSize:numel(At(:))-chunkSize]) + 1;
B = [At(:) At(shiftedIndex(:))]
Ali
Ali 2016-5-26
编辑：Ali 2016-5-26
i have a text file which I have to convert it in another text file with different format. I read the data from the text file and i want to write it in another way. in file, i read a matrix with 260 row and 3 columns. now i want to write it with 780 rows and 2 columns.
this is a piece of the file :
1 0 2
0 3 2
4 1 6
6 1 2
3 5 7
this is the final format that i need
0: 1 0
1: 0 2
2: 2 1
3: 0 3
4: 3 2
5: 2 0
6: 4 1
7: 1 6
8: 6 4
9: 6 1
10: 1 2
11: 2 6
12: 3 5
13: 5 7
14: 7 3
the cyclist
the cyclist 2016-5-26
编辑：the cyclist 2016-5-26
So, it looks like the rule is what Joseph Cheng guessed. For each
row = [r1 r2 r3]
you want that converted to
M = [r1 r2
r2 r3
r3 r1];
and then for each row, keep appending to the bottom of M.
That is what my code does.
Ali
Ali 2016-5-26
Many thanks for the help.
what about the number in front of each row?
for instance:
1: 0 2
tnx again
Ali
Ali 2016-5-26
Thanks a lot the cyclist!

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