Find maximum array values avoiding mat2cell.

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Santos García Rosado
评论： Santos García Rosado 2021-1-4
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!
  2 个评论
Rik
Rik 2020-12-28
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
Santos García Rosado 2021-1-4
No. There are a couple of matlab functions that simulink's function block doesn't support.

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采纳的回答

Bruno Luong
Bruno Luong 2020-12-28
编辑：Bruno Luong 2020-12-28
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
  5 个评论
显示 3更早的评论
Jan
Jan 2020-12-30
See my answer to solve this different problem.

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更多回答（3 个）

Jan
Jan 2020-12-30
编辑：Jan 2020-12-30
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
Bruno Luong
Bruno Luong 2020-12-30
编辑：Bruno Luong 2020-12-30
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.
Image Analyst
Image Analyst 2020-12-28
编辑：Image Analyst 2020-12-28
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
  1 个评论
Santos García Rosado
Santos García Rosado 2020-12-29
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.

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