How to find max of each iteration in cell array?
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Hello!
i have A = 20x2 cell , in each cell i have 3 values of 3 iterations, i want to find the max of each iteration in 20 cell in order to get just two vectors
A1 [max1 max2 max3] A2[max1,max2,max3] after this i will find the max between A1 and A2
i don't know if cell2mat can support this knowing that i will use a large number of cell (100x100)
thanks in advance
4 Comments
Matt J
on 14 Jan 2023
Why use cells at all, if their contentsare all the same size?
Jan
on 14 Jan 2023
100 x 100 is not considered as "large" in times of giga bytes.
It is not clear, what "A1 [max1 max2 max3] A2[max1,max2,max3] " means. You want to find a maximum value in 20 cells, but you have 40.
Concatenating the cells and 1 or 2 calls of the max function should solve the problem:
B = cat(3, A{:}); % Maybe, or:
B = cat(4, cat(3, A{:, 1}), cat(3, A{:, 2}))
Safia
on 14 Jan 2023
Accepted Answer
Star Strider
on 14 Jan 2023
Edited: Star Strider
on 15 Jan 2023
It would help to have your cell array.
Try something like this —
M = randn(3, 20, 2);
C = squeeze(mat2cell(M,3,ones(20,1),[1 1]))
Cmax = cellfun(@max, C)
Check_1 = C{1,1} % Check Results
Check_1_Max = max(Check_1)
Check_2 = C{1,2} % Check Results
Check_2_Max = max(Check_2)
EDIT — (15 Jan 2023 at 19:05)
LD = load(websave('safia%20answer','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1264850/safia%20answer.mat'));
Result_10_element = LD.Result_10_element
Result_10_elementMax = cellfun(@max, Result_10_element) % Desired Result: Cell Maxima
MaxColResult_10_elementMax = max(Result_10_elementMax) % 'Global' Column Maxima
MaxRowResult_10_elementMax = max(Result_10_elementMax,[],2) % 'Global' Row Maxima
One of these should be the desired final result (I hope).
.
16 Comments
Safia
on 15 Jan 2023
That calculation is straightforward —
LD = load(websave('safia%20answer','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1264850/safia%20answer.mat'));
Result_10_element = LD.Result_10_element
Result_10_elementMax = mat2cell([max(cell2mat(Result_10_element))], 1, [3 3])
This approach converts the cell arrays to a matrix, takes the maximum of each column (creating a (1x6) vector) then converts that into a (1x2) cell array.
.
Safia
on 15 Jan 2023
Star Strider
on 15 Jan 2023
As always, my pleasure!
Safia
on 16 Jan 2023
Star Strider
on 16 Jan 2023
It would help to have a representative sample of the cell array.
Try something like this:
Result_10_elementMax = mat2cell([max(cell2mat(Result_10_element))], 1, [51 51])
That will create a (1x2) cell array with 51 elements in each of the two cells.
If you want them apportioned differently, for example a (1x34) cell array of 3 elements each, try this:
Result_10_elementMax = mat2cell([max(cell2mat(Result_10_element))], 1, [repmat(3,1,34)])
The elements of the second argument must sum to 102 in this instance, so any vector that accomplishes that will work.
.
Safia
on 16 Jan 2023
Image Analyst
on 16 Jan 2023
I thought this was all solved. Anyway, @Star Strider he did attach a .mat file in one of his replies to my answer. I'm attaching it again here.
Safia
on 16 Jan 2023
@Safia — It must be a factor of 102, so there are only a limited number of combinations that will work.
Factors_Of_102 = factor(102)
Size__1 = 102/4
Since that is not an integer, 4 will not work, however other options will —
Size_2 = 102/6
The sizes must be multiples of the ‘Factors_Of_102’ to work here. They have to add to 102, so for example:
ColSizes = [ones(1,25)*4 2]
SumColSizes = sum(ColSizes)
ColSizes = [ones(1,51)*2]
SumColSizes = sum(ColSizes)
will both work, however in the first of these, not all the cells will have the same number of columns. The last cell will have only 2 columns. In the second, all cells will have 2 columns per cell.
Any vector that sums to 102 will work.
@Image Analyst — Thank you. That’s where I got the original file I used earlier.
.
Safia
on 17 Jan 2023
Safia
on 17 Jan 2023
Star Strider
on 17 Jan 2023
Edited: Star Strider
on 17 Jan 2023
I understand the concept. What I do not understand is your cell array, or what you want to do with it because I do not have it to work with.
I cannot respond to the 3D array approach until I understand the problem you are having with the cell array you have.
Image Analyst
on 17 Jan 2023
Safia
on 17 Jan 2023
Star Strider
on 17 Jan 2023
As always, my pleasure! (Even though I had nothing to do with coding your 3D array!)
More Answers (2)
Matt J
on 14 Jan 2023
The more natural thing would be to have A be a 20x2xN array to keep track of the N iterations. Then you could simply do,
max(A,[],[1,2])
8 Comments
Safia
on 14 Jan 2023
Matt J
on 14 Jan 2023
It will work, but not with A as a cell array. Explore my suggestion that you recreate A instead as a numeric array.
Safia
on 14 Jan 2023
Image Analyst
on 14 Jan 2023
@Safia you keep forgetting to attach your data in a .mat file, probably because you did not read the Community Guidelines like the posting form advised you to do.
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
We'll check back later for your attached file.
@Safia I don't know what process you are using to produce A, but when rewritten to my suggestion, it should look something like this:
numIterations=3;
A=nan(20,2,numIterations); %Preallocate
for i=1:numIterations
A(:,:,i)=rand(20,2);
end
maxima = max(A,[],[1,2]);
whos A maxima
Safia
on 14 Jan 2023
Image Analyst
on 15 Jan 2023
@Safia, did you expand the comments to show the hidden ones? Did you overlook my question where I asked you directly for you to attach your A in a .mat file?
save('safia answers.mat', 'A');
Are you just refusing to do so? Frankly I'm surprised for the volunteers to keep trying to help you when you are not making it easy for them to help you. Why are you not coooperating so we can finally get your problem solved? Do none of the two solutions posted so far work, or you don't understand them? Maybe I'll post a less cryptic, simple for loop based solution.
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
Image Analyst
on 15 Jan 2023
Edited: Image Analyst
on 15 Jan 2023
Is this what you want? A simple for loop based solution that's easy to understand?
% Create cell array, A because Safia won't upload it.
for row = 1 : 51 % or 20 or 100 or whatever you want. Not sure which since you've said all 3 things.
A{row, 1} = randi(9, 3, 1); % Each cell has 3 "iterations" as he calls them.
A{row, 2} = randi(9, 3, 1);
end
% Now that we have "A", we can begin.
% Find max in each column of A for all rows of A
for row = 1 : size(A, 1)
A1(row) = max(A{row, 1}); % Get max for this row in column 1.
A2(row) = max(A{row, 2}); % Get max for this row in column 2.
end
If not, say why.
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
If you've read this part of the FAQ, I think you could do it yourself:
2 Comments
Image Analyst
on 16 Jan 2023
No problem. Now you'll know for next time how to prepare your answers so that people can quickly answer them. Anyway, you accepted Star's answer so it sounds like he figured it out and solved it for you. Thanks for accepting his answer.
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