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

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

Cedric
Cedric le 26 Avr 2013
Is the content either empty or equal to the same string? What happens if there are more empty cells than cells containing strings?
Matt J
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?
In other words, you need to provide a more precise definition of the rule that defines the label.

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 Réponse acceptée

Cedric
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
Cedric le 26 Avr 2013
Modifié(e) : Cedric le 26 Avr 2013
Effess: I voted for the answer given by The Cyclist, because it is better. You can update it a little if you want to get rid of the empty cells, as follows: replace
cellData(indexToEmpty) = {''};
with
cellData(indexToEmpty) = [];
supposed i have a matrix with these columns, how do i get another matrix with the results match for each column in separate column of match
Cedric
Cedric le 26 Avr 2013
Modifié(e) : Cedric le 26 Avr 2013
If C is a "2D" cell array, yo can get column e.g. 4 as follows:
C_col = C(:,4) ;
The simplest way to apply either solution (The Cyclist or mine) to all columns is to loop over columns of your cell array, and to save the result in another cell array e.g.
matches = cell(1, size(C, 2)) ;
for c = 1 : size(C, 2)
C_col = C(:,c) ;
% .. whichever method, applied to C_col ..
matches{c} = .. the solution, e.g. match.string
end
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
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

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Plus de réponses (2)

the cyclist
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

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?
No need for ismember,
[uniqueCellData,~,whichCell] = unique(cellData);
I knew I had overlooked something easier. :-)

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Peter Saxon
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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