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How to switch two values in a matrix?

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Changoleon
Changoleon on 16 Nov 2016
Commented: Changoleon on 17 Nov 2016
Hi. I have a matrix :
x = [ 6.5 7 23 24 25]
x =
6.5000 7.0000 23.0000 24.0000 25.0000
However, the numbers inside matrix x are not in the correct spots. I need to move some of them. In order to find out which ones to move, I have to look at another matrix which is called c and c equals to:
c = [ 1 2 3 4 5 ;
3 2 4 1 5]
c =
1 2 3 4 5
3 2 4 1 5
I want to write a code which can detect which columns in matrix c do not have the same numbers in its two rows. For example in matrix c the answer is 1st, 3rd, and 4th columns. Following is what I have done.
y = find(c(1,:)~=c(2,:));
Now this line of code helps me to find where in numbers do not match.
Now I want a code which can apply the findings of y to swap the values in matrix x.
In other words, the out put of the code that I am looking for should give me
J = blahblah(x);
J = [ 24 7 6.5 23 25];
This is very important and I have no clue how to do it.
Any help would be appreciated.

Accepted Answer

Roger Stafford
Roger Stafford on 16 Nov 2016
If I understand the method in your example correctly, then using your definition of x and c, the following should accomplish the necessary transformation using matlab's 'circshift' function:
y = find(c(:,1)~=c(2,:));
J = x;
J(y) = J(circshift(y,1));
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More Answers (1)

Walter Roberson
Walter Roberson on 16 Nov 2016
If you have two indices, I, J, at which you need to swap, then
YourMatrix([I, J]) = YourMatrix([J, I]);
will do the swap.
If you indices are expressed through a row vector of length 2, y, then
YourMatrix(y) = YourMatrix(fliplr(y));
Use flipud if it is a row vector.
This will not work as well with length greater than 2: if your vector were an odd length then one of the elements would be set to itself.
  2 Comments
Walter Roberson
Walter Roberson on 16 Nov 2016
The question becomes how to match the source and destination.
The general syntax
YourMatrix(ListOfDestiantions) = Yourmatrix(ListOfCorrespodingSources);
reads everything from Yourmatrix(ListOfCorrespodingSources) before doing any assignments at YourMatrix(ListOfDestiantions), so it is safer than using
for K = 1 : length(ListOfCorrespodingSources)
YourMatrix(ListOfDestinations(K)) = YourMatrix(ListOfCorrespodingSources(K));
end
The "for" version of it messes up on swapping values.

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