How do I find absolute max value in a vector without losing the sign when the value is negative in MatLab R2011b?

I have many vectors that contain force and joint moment data and I need to find the max values without losing the sign. For example, if the values were [2 6 -7] I'd need to get -7, if the vector was [-2 5 6] I'd need to get 6. For certain reasons, I have to work with MatLab R2011b.

Answer 1

[Stephen23]

Method one: use indexing to select from the original vector:

>> V = [2,6,-7];
>> [~,X] = max(abs(V));
>> V(X)
ans = -7

>> V = [-2,5,6];
>> [~,X] = max(abs(V));
>> V(X)
ans =  6

Method two: use sign to multiply with the output value:

>> V = [2,6,-7];
>> [A,X] = max(abs(V));
>> A.*sign(V(X))
ans = -7

>> V = [-2,5,6];
>> [A,X] = max(abs(V));
>> A.*sign(V(X))
ans =  6

[Richard Gerbino]

I was wondering if there is a way to do this for a matrix of column vectors. My solution is below. I guess it's pretty straightforward, but can the indexing be made simpler?

   myData = [2 6 -7; -5 2 1]';
   [~, idx] = max(abs(myData));
   plotData = [myData(idx(1),1) myData(idx(2),2)]

[Stephen23]

It would be nice if max et al would return linear indices. Your indexing is probably about the simplest, but because you hard-code the columns it is not very generalized. Here is a generalized solution:

M = [2,6,-7;-5,2,1;0,-3,9].'

M = 3×3

2 -5 0 6 2 -3 -7 1 9

[~,R] = max(abs(M),[],1);
X = sub2ind(size(M),R,1:size(M,2));
V = M(X)

V = 1×3

-7 -5 9

[Steven Lord]

The max and min functions can return linear indices as of release R2019a.

M = [2,6,-7;-5,2,1;0,-3,9].'

M = 3×3

2 -5 0 6 2 -3 -7 1 9

[~, ind] = max(abs(M), [], 1, 'linear')

ind = 1×3

3 4 9

M(ind)

ans = 1×3

-7 -5 9