Counting the occurance of a unique number in an array - MATLAB - arrays

Question

I have an array that looks something like...
1 0 0 1 2 2 1 1 2 1 0
2 1 0 0 0 1 1 0 0 2 1
1 2 2 1 1 1 2 0 0 1 0
0 0 0 1 2 1 1 2 0 1 2
however my real array is (50x50).
I am relatively new to MATLAB and need to be able to count the amount of unique values in each row and column, for example there is four '1's in row-2 and three '0's in column-3. I need to be able to do this with my real array.
It would help even more if these quantities of unique values were in arrays of their own also.
PLEASE use simple language, or else i will get lost, for example if representing an array, don't call it x, but perhaps column_occurances_array... for me please :)

Answer 1

What I would do is iterate over each row of your matrix and calculate a histogram of occurrences for each row. Use histc to calculate the occurrences of each row. The thing that is nice about histc is that you are able to specify where the bins are to start accumulating. These correspond to the unique entries for each row of your matrix. As such, use unique to compute these unique entries.
Now, I would use arrayfun to iterate over all of your rows in your matrix, and this will produce a cell array. Each element in this cell array will give you the counts for each unique value for each row. Therefore, assuming your matrix of values is stored in A, you would simply do:
vals = arrayfun(#(x) [unique(A(x,:)); histc(A(x,:), unique(A(x,:)))], 1:size(A,1), 'uni', 0);
Now, if we want to display all of our counts, use celldisp. Using your example, and with the above code combined with celldisp, this is what I get:
vals{1} =
0 1 2
3 5 3
vals{2} =
0 1 2
5 4 2
vals{3} =
0 1 2
3 5 3
vals{4} =
0 1 2
4 4 3
What the above display is saying is that for the first row, you have 3 zeros, 5 ones and 3 twos. The second row has 5 zeros, 4 ones and 2 twos and so on. These are just for the rows. If you want to do these for columns, you have to modify your code slightly to operate along columns:
vals = arrayfun(#(x) [unique(A(:,x)) histc(A(:,x), unique(A(:,x)))].', 1:size(A,2), 'uni', 0);
By using celldisp, this is what we get:
vals{1} =
0 1 2
1 2 1
vals{2} =
0 1 2
2 1 1
vals{3} =
0 2
3 1
vals{4} =
0 1
1 3
vals{5} =
0 1 2
1 1 2
vals{6} =
1 2
3 1
vals{7} =
1 2
3 1
vals{8} =
0 1 2
2 1 1
vals{9} =
0 2
3 1
vals{10} =
1 2
3 1
vals{11} =
0 1 2
2 1 1
This means that in the first column, we see 1 zero, 2 ones and 1 two, etc. etc.

Answer 2

I absolutely agree with rayryeng! However, here is some code which might be easier to understand for you as a beginner. It is without cell arrays or arrayfuns and quite self-explanatory:
%% initialize your array randomly for demonstration:
numRows = 50;
numCols = 50;
yourArray = round(10*rand(numRows,numCols));
%% do some stuff of what you are asking for
% find all occuring numbers in yourArray
occVals = unique(yourArray(:));
% now you could sort them just for convinience
occVals = sort(occVals);
% now we could create a matrix occMat_row of dimension |occVals| x numRows
% where occMat_row(i,j) represents how often the ith value occurs in the
% jth row, analoguesly occMat_col:
occMat_row = zeros(length(occVals),numRows);
occMat_col = zeros(length(occVals),numCols);
for k = 1:length(occVals)
occMat_row(k,:) = sum(yourArray == occVals(k),2)';
occMat_col(k,:) = sum(yourArray == occVals(k),1);
end