Asked by effess
on 26 Apr 2013

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?

Answer by Cedric Wannaz
on 26 Apr 2013

Edited by Cedric Wannaz
on 26 Apr 2013

Accepted answer

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

Show 2 older comments

Cedric Wannaz
on 26 Apr 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

effess
on 26 Apr 2013

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 Wannaz
on 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

Answer by the cyclist
on 26 Apr 2013

Edited by 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.

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## 3 Comments

## Cedric Wannaz (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/73681#comment_145694

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 (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/73681#comment_145698

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 (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/73681#comment_145699

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