How to switch two values in a matrix?
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Changoleon
am 16 Nov. 2016
Kommentiert: Changoleon
am 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.
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Akzeptierte Antwort
Roger Stafford
am 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));
9 Kommentare
Roger Stafford
am 17 Nov. 2016
@Changoleon. Try this:
y = find(c(:,1)~=c(2,:));
J = x;
J(y) = J(circshift(y,1,2));
The '2' specifies that shifting is along the 2nd dimension (along the row)
Weitere Antworten (1)
Walter Roberson
am 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 Kommentare
Walter Roberson
am 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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