how to get the most repeated element of a cell array?

Question

DuckDuck 2013-4-26

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编辑： Peter Saxon 2021-1-23

i have an cell array like this

 {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';}

is there a way to get the label of this array as rj?

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Matt J 2013-4-26

maybe i did ask the question wrong, but if i have more {''} it will give me that name. instead i want {'rj'}. is there a workaround to counting?

Yes, you'll need to clarify the question. Why should {''} be ignored? Are there any other strings that should be ignored?

the cyclist 2013-4-26

In other words, you need to provide a more precise definition of the rule that defines the label.

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Answer 1

Cedric 2013-4-26

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编辑：Cedric 2013-4-26

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the cyclist >> I knew I had overlooked something easier. :-)

Well, look at how I did overlook something easier ;-D :

 C = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'ab';'ab'} ;
 setMatch  = @(s,c) struct('string', s, 'count', c) ;
 match     = setMatch('', 0) ;
 hashtable = java.util.Hashtable() ;
 for k = 1 : length(C)
     if isempty(C{k}), continue ;  end
     if hashtable.containsKey(C{k})
         count = hashtable.get(C{k}) ;
         if count >= match.count,  match = setMatch(C{k}, count+1) ;  end
         hashtable.put(C{k}, count+1) ; 
     else
         if match.count == 0,  match = setMatch(C{k}, 1) ;  end
         hashtable.put(C{k}, 1) ; 
     end
 end

Running this leads to;

 >> match
 match = 
    string: 'rj'
     count: 3

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DuckDuck 2013-4-26

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i did it but i'm geting an error when trying to save results from columns with more than one string!!

for k=1:length(tsf)
cellData = labelw(:,tsf(k));
indexToEmpty = cellfun(@isempty,cellData);
cellData(indexToEmpty) = [];
[uniqueCellData(:,k),~,whichCell] = unique(cellData);
end;

Cedric 2013-4-26

编辑：Cedric 2013-4-26

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You don't need to save the content of temporary variables at each iteration of the loop. You just need to save results, and you should have something like (where the ".." have to be adapted to your case):

 maxCountElements = cell(size(..), 1) ;
 for k = 1 : ..
    cellData = ..
    indexToEmpty = cellfun(@isempty,cellData);
    cellData(indexToEmpty) = [];
    uniqueCellData = unique(cellData);
    [uniqueCellData,~,whichCell] = unique(cellData);
    cellCount = hist(whichCell,unique(whichCell));
    [~,indexToMaxCellCount] = max(cellCount);
    maxCountElements{k} = uniqueCellData(indexToMaxCellCount) ;
 end

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Answer 2

the cyclist 2013-4-26

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编辑：the cyclist 2013-4-26

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Quite convoluted, but I think this works:

cellData = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';'rj';}
indexToEmpty = cellfun(@isempty,cellData);
cellData(indexToEmpty) = {''};
uniqueCellData = unique(cellData);
[~,whichCell] = ismember(cellData,unique(uniqueCellData))
cellCount = hist(whichCell,unique(whichCell));
[~,indexToMaxCellCount] = max(cellCount);
maxCountElement = uniqueCellData(indexToMaxCellCount)

The essence of the algorithm is using the hist() function to count up the frequency. Unfortunately, that function only works on numeric arrays, so I had to use the ismember() command to map the cell array values to numeric values.

A further complication was the existence of the empty cell elements. I replaced them with empty strings. You'll need to be careful if your original array has empty strings.

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Matt J 2013-4-26

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No need for ismember,

[uniqueCellData,~,whichCell] = unique(cellData);

the cyclist 2013-4-26

I knew I had overlooked something easier. :-)

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Answer 3

Peter Saxon 2021-1-23

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编辑：Peter Saxon 2021-1-23

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Found a neat solution with categories, just posting this here so when I forget how to do this and google it again I'll see it...

C = {[];[];[];[];[];[];[];[];'rj';'rj';'rj';'ab';'ab'} ;
catC=categorical(C);
catNames=categories(catC);
[~,ix] = max(countcats(catC));
disp(catName{ix}) % rj