most frequent word in cell array

Hi, I have a cell array "P" of size 2000 by 20. Each cell value is either "Yes" or "No". How can I make a new cell array "vote" of size 2000 by 1 that each cell contains the most frequent word of each row in P?

Answer 1

[Walter Roberson]

tf = ismember(lower(P), 'yes');
votes = sum(tf, 2);

[dpb]

Or, you could mix the two... :)

>> vote=categorical(sum(tf,2)>size(tf,2)/2,[true false],{'Yes','No'})
vote = 
   Yes 
   No 
>>

although still need to handle ties if possible.

[Walter Roberson]

Sorry, I wasn't paying attention. What my code calculates is the number of 'yes' in each row. The number of 'no' is the width of the row minus the number of 'yes'

[dpb]

I was just throwing in the categorical variable into the mix in the end on top of your solution for the total by row using string matching thereby "mixing the two" of the cellstr variable to the categorical variables I had suggested totally (which did the sums via countcats) by then using the categorical to display the name in English of the winner...

Wasn't imply anything at all was wrong, just adding the final step and that primarily to "show off" categorical to the OP as worth looking at.

[Walter Roberson]

Right, but I had overlooked that the question asked about the most common entry -- which can be found by testing the count against width/2

Answer 2

[dpb]

Good place to use categorical variables instead of the cellstr...

Example:

>> yn={'yes' 'no' 'Yes';'no' 'No', 'NO'};   % minimal dataset including capitaliztion differences
>> ync=categorical(lower(yn));              % convert to categorical and normalize spelling
>> cnts=countcats(ync,2)                    % count responses on 2nd dimension
cnts =
   1     2
   3     0
>> vote=cnts(:,2)>cnts(:,1);                  % see which is greater (Y>N --> True)
>> vote=categorical(vote,[true false],{'Yes','No'})  % convert to categorical to display
vote = 
   Yes 
   No 
>> yn      % original table to compare -- looks like right choice.
yn = 
  'yes'    'no'    'Yes'
  'no'     'No'    'NO' 
>>

NB: The above doesn't have the extra logic to check for tie--in case that is possible will need to test for == as well and add the third category of TIE as possible output.

ADDENDUM

If TIE is possible, look at computing difference between counts and then the SIGN function will generate the tri-state variable needed.

Answer 3

[Sarah Palfreyman]

Try tokenizing with Text Analytics Toolbox and you can easily get a histogram count.