## Splitting up a matrix into X sizes/ based on specified values in a matrix

mark

### mark (view profile)

님이 질문을 제출함. 28 Dec 2013
최근 활동 mark

### mark (view profile)

님이 댓글을 추가함. 28 Dec 2013
Wayne King

### Wayne King (view profile)

님이 답변을 채택함.
If I have a matrix of size 2x100 of time vs position values (or whatever). I we say time goes from 0s to 2s then I would like to write a script that says:
" Find where X(1,:) = 1 and split it, and X(2,:), such that I have two new matrices A and B which correspond to the values from 0s to 1s and 1.1s to 2s" I tried using the find feature but that didn't work. Any ideas?
Essentially, if I have 10s of run time then I'd like to be able to separate those 10s into 10 separate matrices.
Cheers, Mark.
EDIT:
I should add that the time scale may not be equal as the time sampling is automated based on when an event occurs. This is to reduce the computational time substantially and maintain a high enough resolution at an event. So, the important aspect is to be able to specify the points on the matrix that I'd like to separate out!

Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 28 Dec 2013
It's better to post your data or code as a text than an image

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## 답변 수: 2

### Wayne King (view profile)

on 28 Dec 2013
Edited by Wayne King

### Wayne King (view profile)

on 28 Dec 2013

Are the time instants just integers? If not, that may be why find() does not work depending on how you are using it.
X = ones(2,100);
X(1,:) = 0:99;
idx1 = find(X(1,:)<=50,1,'last');
idx2 = find(X(1,:)>50,1,'first');
A = X(:,1:idx1);
B = X(:,idx2:end);

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### Baalzamon (view profile)

on 28 Dec 2013

Logical indexing is one way. Both Doug and Loren have a mini tutorial on it. To get say all elements of X(1,:), that are say 0-50, replace the ':' with a logical argument. e.g. A = X(1, X(1,:) > 50 ); B = X(1, X(1,:) < 51); for both rows A = X(:, X(1,:) > 50); B = X(:, X(1,:) < 51);
The 'X(1,:) > 50' bit is looking up all elements in the first row of X that is GT 50, and then using that to create the new array.
As you can guess it is possible to use multiple logical arguments with '&&' etc.

mark

### mark (view profile)

on 28 Dec 2013
Thanks guys, the logical stuff works.
if true
X1=0:0.5:100;
X2=0:0.25:50;
X=[X1' X2'];
Ax = find(X(:,1)<=50);
A1=[X1(Ax)];
A2=[X2(Ax)];
A=[A1' A2'];
Bx = find(X(:,1)>50);
B1=[X1(Bx)];
B2=[X2(Bx)];
B=[B1' B2'];
end

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