Find maximum array values avoiding mat2cell.

Hello Matlab community.

I was wondering if someone could give me a hand with a code problem I've been having for a while.

So I have an array A =1xn such as:

(NaN, NaN, NaN, NaN, 1, 6, 8, 32, -5, NaN,NaN,NaN,NaN,NaN,NaN, 65, 2 ,16 80, 35, 26, 12, NaN, ..., n).

As you can tell, both the NaN and numeric lenghts do not follow any pattern. The main idea is trying to get the maximum value of each numeric string and being able to get this sort of array as an output:

(NaN, NaN, NaN, NaN, 0, 0, 0, 32, 0, NaN,NaN,NaN,NaN,NaN,NaN, 0, 0 ,0 80, 0, 0, 0, NaN, ..., n).

I'd like to do it without using the mat2cell function since later on I'll have to use the code for Simulink and this kind of function is not allowed by the software.

My first idea was using the common max function:

[value, position] = max(A);

Obviously this will only give me the maximum number of the whole array.

It would be great if someone could help me out. Thank's in advanced!

[Rik]

Can't you just call any function with a function block? I haven't worked with Simulink in a very long time, so I don't know which function would or wouldn't be allowed.

[Santos García Rosado]

No. There are a couple of matlab functions that simulink's function block doesn't support.

Answer 1

[Bruno Luong]

Just a for-loop (is it allowed in Simulink?)

A=[NaN, NaN, NaN, NaN, 1, 6, 8, 32, -5, NaN,NaN,NaN,NaN,NaN,NaN, 65, 2 ,16 80, 80, 35, 26, 12, NaN];

imax = [];
Amax = -Inf;
for i=1:length(A)
    if isnan(A(i))
        imax = [];
        Amax = -Inf;
    else
        if A(i) > Amax 
            A(imax) = 0;
            imax = i;
            Amax = A(imax);
        else
            A(i) = 0;
        end
    end
end

[Santos García Rosado]

Thank's Bruno! I've just checked it and it works perfectly. And yes, it is absolutely compatible with Simulink. Thank's again.

[Santos García Rosado]

Hi Bruno, sorry to bother you again.

I´ve been trying to upgrade your code so that I could not only get the maximum value for each string, but the second, third and fourth maximums aswell.

I've managed to get a correct output by making the max value zero and reusing your code again for every maximumm needed. However this seems inefficient at first sight. How do you think we could rewrite your code to make this possible?

Thank you,

Santos

[Bruno Luong]

No this method cannot not be generalized to keep k-largest maximum in one scan.

[Santos García Rosado]

Ok, thanks anyways!

[Jan]

See my answer to solve this different problem.

Answer 2

[Jan]

For the expanded question in the comment to Bruno's solution you ask for the n largest elements:

A = [NaN, NaN, NaN, NaN, 1, 6, 8, 32, -5, NaN,NaN,NaN,NaN,NaN,NaN, 65, 2 ,16 80, 35, 26, 12, NaN]
n = 3;
BlockMaxK(A, n)

% NaN, NaN, NaN, NaN, 0, 6, 8, 32, 0, NaN, NaN, NaN, NaN, NaN, NaN, 65, 0, 0, 80, 35, 0, 0, NaN

function B = BlockMaxK(A, n)
m   = [true, isnan(A), true];
ini = strfind(m, [true, false]);
fin = strfind(m, [false, true]) - 1;

B = nan(size(A));
for k = 1:numel(ini)
   ik       = ini(k);
   fk       = fin(k);
   [v, ind] = maxk(A(ik:fk), n);
   B(ik:fk) = 0;
   B(ik - 1 + ind) = v;
end
end

[Santos García Rosado]

Thank's Jan! This works great, now the code is much more efficient.

Answer 3

[Bruno Luong]

This is a method for multiple maxima by block.

I post this for illustration of algorithmic mainly, because the real performance is rather poor, but it should be good theoretically. It uses the so called red-black tree structure, something that MATLAB really miss for decades, and I lost all my fate to see it available one day.

Such method is kind of sequential, meaning it just scan once the data and more adatped for problems where the data are avaiblable sequantially in time (thus simullink framework). Unfortunately the RedBlack library is not as fast as if it was implemented natively in C/C++.

This method can also be seen as an extension of my answer for the original question of single maximum by block.

Here I use a MATLAB implementation by Brain Moore.

A=[NaN, NaN, NaN, NaN, 1, 6, 8, 32, -5, NaN,NaN,NaN,NaN,NaN,NaN, 65, 2 ,16 80, 80, 35, 26, 12, NaN]
k = 3;

% FEX file By Brian Moore
% https://www.mathworks.com/matlabcentral/fileexchange/45123-data-structures
T = RedBlackTree();
for i=1:length(A)
    if isnan(A(i))
        T.Clear();
    else
        Insert = T.Count < k;
        if ~Insert
            Min = T.Minimum();
            Insert = A(i) > Min.key;
            if Insert
                A(Min.value) = 0;
                T.Delete(Min);
            else
                A(i) = 0;
            end
        end
        if Insert
            T.Insert(A(i),i);
        end
    end
end

For a better solution, see Jan's answer.

[Santos García Rosado]

Thank you Bruno. I'll take a good look at it.

Answer 4

[Image Analyst]

If you have the Image Processing Toolbox you can use regionprops() to ask for the MaxIntensity:

A=[NaN, NaN, NaN, NaN, 1, 6, 8, 32, -5, NaN,NaN,NaN,NaN,NaN,NaN, 65, 2 ,16 80, 80, 35, 26, 12, NaN];
[labeledGroups, numGroups] = bwlabel(~isnan(A))
props = regionprops(labeledGroups, A, 'MaxIntensity');
maxIntensities = [props.MaxIntensity]
output = zeros(size(A));
output(isnan(A)) = nan;
for group = 1 : numGroups
	thisGroup = ismember(labeledGroups, group);
	% Find out where in this group the max intensity occurs.
	indexes = (A .* thisGroup) == maxIntensities(group);
	output(indexes) = maxIntensities(group);
end
output % Show in command window.

You'll get

numGroups =

2

maxIntensities =

32 80

output =

Columns 1 through 19

NaN NaN NaN NaN 0 0 0 32 0 NaN NaN NaN NaN NaN NaN 0 0 0 80

Columns 20 through 24

80 0 0 0 NaN

[Santos García Rosado]

Unfortunately I do not have acces to Image Processing Toolbox at the moment. In case I ever do, I'll try using the code you propose. Thank you anyways.