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 Kommentare

Cedric
Cedric am 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
Matt J am 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?
In other words, you need to provide a more precise definition of the rule that defines the label.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Cedric
Cedric am 26 Apr. 2013
Bearbeitet: Cedric am 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 Kommentare

Cedric
Cedric am 26 Apr. 2013
Bearbeitet: Cedric am 26 Apr. 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 am 26 Apr. 2013
Bearbeitet: Cedric am 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
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 am 26 Apr. 2013
Bearbeitet: Cedric am 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

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

the cyclist
the cyclist am 26 Apr. 2013
Bearbeitet: the cyclist am 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 Kommentare

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. :-)

Melden Sie sich an, um zu kommentieren.

Peter Saxon
Peter Saxon am 23 Jan. 2021
Bearbeitet: Peter Saxon am 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

Produkte

Gefragt:

am 26 Apr. 2013

Bearbeitet:

am 23 Jan. 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by