How to compare a matrix rows such as:

How to compare a matrix rows such as:

A=[

7 8 0 5 6 0 1 2 3 4

7 8 0 5 6 0 1 2 3 5

7 8 0 5 6 0 1 2 3 6

7 8 0 5 6 0 1 2 3 7

7 8 0 5 6 0 1 2 3 8

7 8 0 5 6 0 1 2 4 5

7 8 0 5 6 0 1 2 4 6

]

I want to get the results

B=[7 8 0 5 6 0 1 2 3 4] Each row 1-8 each appear once

Who can help me to discuss this issue with me?

[Stephen23]

This does not really explain anything. If there is a method to this, then you need to explain it clearly for us.

[Stephen23]

And now you are deleting your comments that explain the problem? Why? Because you have deleted some comments, my following comment will not make much sense now, but maybe it will help us to solve your problem anyway.

[Stephen23]

This is still a poorly defined problem, although I am sure that it is quite easy to solve. It seems that you wish to exclude any row with repeated elements (your first and second points), but what is the rule for excluding A(4,:) and A(5,:)?

• you compare A(4,[1,2]) with A(2,[3,4]), but why row 2? Why not row 1 or any other row? How do I know that I should be comparing to row 2?
• you give two examples that show that order and position is unimportant ( A(4,[1,2]) vs A(2,[3,4]) and A(5,[1,2]) vs A(2[4,3]). Is the position or number of compared elements important? Can I compare one element? Or three? Do they need to be adjacent?
• In your examples you have different columns with zeros exclusively:

[7,8,0,5,6,0,1,2,3,4] % your original question
[1,0,2,0,3,4,5,6,7,8] % your comment to my first question.

Which of these is correct? Are the zeros significant?

[Wang]

Numbers 1-8 divided into three groups

For example : [3 4 0 1 2 0 5 6 7 8]

First block is 3,4

Second block is 1,2

Third block is 5,6,7,8

Block combination can not be repeated

For example : %A(1,1:2) and A(2,3:4) and A(4,3:4) is repeated

The final result for the B matrix

How can I to do this?

A=

[

3 4 0 1 2 0 5 6 7 8

1 2 0 3 4 0 5 6 7 8

1 2 0 1 2 0 5 6 7 8

2 1 0 4 3 0 5 6 7 8

]; % four row

B=

[

1 2 0 3 4 0 5 6 7 8

];

[Stephen23]

Given the matrix A from your comment above:

A = [3 4 0 1 2 0 5 6 7 8;...
     1 2 0 3 4 0 5 6 7 8;...
     1 2 0 1 2 0 5 6 7 8;...
     2 1 0 4 3 0 5 6 7 8];

We can try your proposed algorithm:

• row 1: repeated blocks on same row? no -> continue.
• row 1: repeated blocks from previous rows? no previous blocks -> continue.
• row 2: repeated blocks on same row? no -> continue.
• row 2: repeated blocks from previous rows? [1,2] is also in row 1 -> remove row 2.

And yet you give row 2 as the only remaining row in your solution.

So we have a contradiction between your explanation and the desired output that you give. You say that you do not want rows to contain blocks that are repeats of blocks from previous rows (which row 2 has) and yet you give row 2 as the only remaining line in your proposed output.

Also all rows repeat the block [5,6,7,8], so really row 1 can be the only remaining row, not row 2. Or, if we restrict ourselves to comparing only the first two blocks on each row, then as shown row 2 anyway repeats a block from row 1, so must be excluded.

Your explanation is not adequate, or is inconsistent. Please explain it more carefully.

Answer 1

[Stephen23]

Try this:

>> [A(1,1:8),unique(A(:,9)).']
ans =
     7     8     0     5     6     0     1     2     3     4

This takes the first eight elements of the first row, and concatenates these with the unique elements of the ninth column, which matches your example output. It seems that your example does not reference the tenth column: is this intentional?

[Ced]

Can we assume that the numbers are sorted as in your example?

Answer 2

[Honglei Chen]

B = sort(A,2);
A(ismember(B(:,3:end),1:8,'rows'),:)

Answer 3

[Stephen23]

You explanations are not very clear, but it appears that you want an operation that returns each unique row of your input matrix A. This can be done very easily using the unique function. Read its documentation carefully and you will find the second argument quite useful for your code:

>> A = [1,2,3;4,5,6;1,2,3;1,2,3;7,8,9;4,5,6] % six rows
A =
     1     2     3
     4     5     6
     1     2     3
     1     2     3
     7     8     9
     4     5     6
>> B = unique(A,'rows') % finds all three unique rows
B =
     1     2     3
     4     5     6
     7     8     9

Answer 4

[Ced]

As an alternative, I came up with this code which seems to do the same thing.

[ ~, ind_sort ] = sort(A(:,1));
B = A(ind_sort,:);
B([false; (sum(B(1:end-1,:)==B(2:end,:),2)==size(A,2)) ],:) = [];

From a short test, it's faster than unique by a factor of at least 2. If your matrix is already sorted, you can leave out the first two lines and simply set B = A;