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 commentaires
Matt J
le 26 Avr 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
le 26 Avr 2013
In other words, you need to provide a more precise definition of the rule that defines the label.
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Cedric
le 26 Avr 2013
Modifié(e) : Cedric
le 26 Avr 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 commentaires
Cedric
le 26 Avr 2013
Modifié(e) : Cedric
le 26 Avr 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
Plus de réponses (2)
the cyclist
le 26 Avr 2013
Modifié(e) : the cyclist
le 26 Avr 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 commentaires
Matt J
le 26 Avr 2013
No need for ismember,
[uniqueCellData,~,whichCell] = unique(cellData);
Peter Saxon
le 23 Jan 2021
Modifié(e) : Peter Saxon
le 23 Jan 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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