# How to return the row vector containing the highest values from a matrix?

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Lewis on 22 Nov 2016
Commented: Lewis on 22 Nov 2016
Hi, I can't seem to find a solution to this but it's a pretty simple problem. Say I have a matrix:
M = [10 9 11 7 8;
10 9 8 7 6;
9 6 7 8 5]
What I want to do is to somehow extract the row vector which holds the largest sequence. I have tried the max() function but it just doesn't retain the individual rows. It is important that it returns a specific row vector from M.
The rows won't always necessarily be sequential like this, so I cannot just use the row that sums to the largest value either.
Thanks

James Tursa on 22 Nov 2016
Please define "largest sequence" for us. What is the exact criteria you want for this? What would be returned for your example above?
Lewis on 22 Nov 2016
Ah, the example isn't so good actually, but the largest sequence in that example would be
[10 9 11 7 8]
A better example might be:
M = [13 9 8 7 6;
13 12 11 9 8;
13 11 5 3 2]
Where I would want to return the following vector:
[13 12 11 9 8]
because if we were to sort all elements of each vector, then this one would contain the highest, the second highest, the third highest, etc.

Image Analyst on 22 Nov 2016
Try this:
M = [13 9 8 7 6;
13 12 11 9 8;
13 11 5 3 2]
% Find the maxima in each column independently.
columnMaxima = max(M, [], 1)
% Now sort in decending order.
columnMaxima = sort(columnMaxima, 'Descend')

Lewis on 22 Nov 2016
If possible then just taking the first row should be fine... I'd quite like to keep an index of which row from M it came from.
The same issue also occurs in the following example:
M = [9 6 7 5 8; 6 7 5 8 4];
columnMaxima = max(M, [], 1);
columnMaxima = sort(columnMaxima, 'Descend')
columnMaxima = [9 7 7 8 8]
Image Analyst on 22 Nov 2016
The second, optional output of max supplies the index from which the max value came from. Use it.
Lewis on 22 Nov 2016
Thank you for the help - seems to be working perfectly now.

dpb on 22 Nov 2016
Sort first by row (already done here in these examples), then by column, retaining the sort index. Then (I think, altho didn't prove concluseively) the row will be minimum of the sum of those indices (fewest permutations to arrange).
>> [~,ix]=sort(M,1);
>> [~,ix]=min(sum(ix,2))
>> M(ix,:)
ans =
13 12 11 9 8
>>

#### 1 Comment

Lewis on 22 Nov 2016
Thanks for the suggestion