how to get the most repeated element of a cell array?
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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?
3 Commenti
  Matt J
      
      
 il 26 Apr 2013
				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
      
      
 il 26 Apr 2013
				In other words, you need to provide a more precise definition of the rule that defines the label.
Risposta accettata
  Cedric
      
      
 il 26 Apr 2013
        
      Modificato: Cedric
      
      
 il 26 Apr 2013
  
      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
5 Commenti
  Cedric
      
      
 il 26 Apr 2013
				
      Modificato: Cedric
      
      
 il 26 Apr 2013
  
			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
Più risposte (2)
  the cyclist
      
      
 il 26 Apr 2013
        
      Modificato: the cyclist
      
      
 il 26 Apr 2013
  
      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.
3 Commenti
  Matt J
      
      
 il 26 Apr 2013
				No need for ismember,
[uniqueCellData,~,whichCell] = unique(cellData);
  Peter Saxon
 il 23 Gen 2021
        
      Modificato: Peter Saxon
 il 23 Gen 2021
  
      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
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