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 Comments
Cedric
on 26 Apr 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
on 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
on 26 Apr 2013
In other words, you need to provide a more precise definition of the rule that defines the label.
Accepted Answer
More Answers (2)
the cyclist
on 26 Apr 2013
Edited: the cyclist
on 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 Comments
DuckDuck
on 26 Apr 2013
Matt J
on 26 Apr 2013
No need for ismember,
[uniqueCellData,~,whichCell] = unique(cellData);
the cyclist
on 26 Apr 2013
I knew I had overlooked something easier. :-)
Peter Saxon
on 23 Jan 2021
Edited: Peter Saxon
on 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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