# how do i count unique cell rows?

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Kangkeon Jeon on 7 Dec 2019
Edited: dpb on 8 Dec 2019
I have 10000x2 char cell i want to count how many of each unique rows there is.
There are 9 possible unique rows. How can i code so that it counts how many of those each 9 types are in the cell array?

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Kangkeon Jeon on 7 Dec 2019
each cell can be one of these three : 'AA', 'AB', 'BB'
so possible rows are
'AA' 'AA'
'AA' 'AB'
'AA' 'BB'
'AB' 'AA'
'AB' 'AB'
'AB' 'BB'
'BB' 'AA'
'BB' 'AB'
'BB' 'BB'
i already used unique(cell2mat(total_people),'rows') to find all possible options but i don't know how to run a loop to count how many of each unique rows.
Kangkeon Jeon on 7 Dec 2019
if total_people{i,1} == 'AA' && total_people{i,2} == 'AA'
AAAA = AAAA + 1;
end
in a for-loop like that to count each option but it gives
"Operands to the || and && operators must be convertible to logical scalar values." error
per isakson on 7 Dec 2019
Doc says: The 'rows' option does not support cell arrays.

dpb on 7 Dec 2019
Edited: dpb on 8 Dec 2019
I agree w/ Per, use strings since TMW didn't see fit to extend unique to cell arrays and either is far better than char() arrays for almost every purpose...
% build a sample dataset...
s=string({'AA' 'AA'
'AA' 'AB'
'AA' 'BB'
'AB' 'AA'
'AB' 'AB'
'AB' 'BB'
'BB' 'AA'
'BB' 'AB'
'BB' 'BB'});
s=repmat(s,5,1); % duplicate to have more than one of each possible
s=s(randperm(45,30),:); % pick a random subset so not all the same...
% preliminaries done, now for the work...
[~,ia,ic]=unique(s,'rows'); % return locations of which set for all rows in s
n=histc(ic,unique(ic)); % count the number in each bin...
For the particular randomized sample I ran here at command line the above returns--
>> [s(ia,:) unique(ic) n]
ans =
9×4 string array
"AA" "AA" "1" "3"
"AA" "AB" "2" "4"
"AA" "BB" "3" "3"
"AB" "AA" "4" "4"
"AB" "AB" "5" "4"
"AB" "BB" "6" "3"
"BB" "AA" "7" "3"
"BB" "AB" "8" "4"
"BB" "BB" "9" "2"
>>
where used the penchant for string class to turn everything it touches into strings to be able to display disparate types together...the unique string pairs followed by the bin number for each and then the count for the respective bin.
To verify results are correct, here's the particular s array...
>> s
s =
30×2 string array
"BB" "AB"
"AA" "AA"
"BB" "AB"
"AB" "AB"
"AB" "AA"
"AB" "AB"
"AA" "AB"
"AB" "AA"
"AA" "AB"
"AB" "BB"
"BB" "BB"
"AA" "BB"
"AA" "AB"
"BB" "AA"
"AA" "AB"
"BB" "AB"
"AA" "BB"
"BB" "AA"
"BB" "BB"
"AB" "AA"
"AB" "AB"
"AB" "BB"
"BB" "AA"
"AB" "AA"
"AB" "AB"
"AA" "AA"
"AA" "BB"
"AA" "AA"
"AB" "BB"
"BB" "AB"
>>

Image Analyst on 7 Dec 2019
There are more compact (read cryptic) ways, but this simple and easy-to-understand (for a beginner) way using a for loop and switch statement works well:
% Define the cell array:
ca = {
'AA' 'AA'
'AA' 'AB'
'AA' 'BB'
'AB' 'AA'
'AB' 'AB'
'AB' 'BB'
'BB' 'AA'
'BB' 'AB'
'BB' 'BB'}
numRows = size(ca, 1);
counts = zeros(numRows, 1);
% Count how many of the unique strings there are
for row = 1 : size(ca, 1)
str = [ca{row, 1}, ca{row, 2}]
switch upper(str)
case 'AAAA'
counts(1) = counts(1) + 1; % Increment count for this patterm
case 'AAAB'
counts(2) = counts(2) + 1; % Increment count for this patterm
case 'AABB'
counts(3) = counts(3) + 1; % Increment count for this patterm
case 'ABAA'
counts(4) = counts(4) + 1; % Increment count for this patterm
case 'ABAB'
counts(5) = counts(5) + 1; % Increment count for this patterm
case 'ABBB'
counts(6) = counts(6) + 1; % Increment count for this patterm
case 'BBAA'
counts(7) = counts(7) + 1; % Increment count for this patterm
case 'BBAB'
counts(8) = counts(8) + 1; % Increment count for this patterm
case 'BBBB'
counts(9) = counts(9) + 1; % Increment count for this patterm
end
end
counts % Report to command window.