Digits that repeat in an n by m matrix

If I have a multidimensional array of n by m and I want to count the number of digits that repeat in each row of the array.

Ex:

Columns 1 through 7

9 5 2 1 5 3 3

1 8 3 1 7 4 3

1 3 1 1 8 6 4

1 3 1 6 4 2 1

2 1 6 3 2 2 1

2 3 1 1 7 5 4

1 1 4 2 1 8 6

Columns 8 through 10

2 2 5

2 2 6

4 2 4

2 9 7

1 9 1

2 1 5

5 4 5

in the end i want:

first row i have : number 1 -> 1; 2-> 3; 3 -> 2,...

I try a few, but i don´t have any sucess..

someone can help me?

[Matt J]

in the end i want:

first row i have : number 1 -> 1; 2-> 3; 3 -> 2,...

That is not the number of digits that repeat in the first row (which would be 3). That is the number of repetitions of each digit.

[PEDRO ALEXANDRE Fernandes]

True. Sorry..

Answer 1

[Matt J]

A = [     9     5     2     1     5     3     3     2     2     5
     1     8     3     1     7     4     3     2     2     6
     1     3     1     1     8     6     4     4     2     4
     1     3     1     6     4     2     1     2     9     7
     2     1     6     3     2     2     1     1     9     1
     2     3     1     1     7     5     4     2     1     5
     1     1     4     2     1     8     6     5     4     5];


counts=histc(A',1:max(A(:))+1)'

counts = 7×10

1 3 2 0 3 0 0 0 1 0 2 2 2 1 0 1 1 1 0 0 3 1 1 3 0 1 0 1 0 0 3 2 1 1 0 1 1 0 1 0 4 3 1 0 0 1 0 0 1 0 3 2 1 1 2 0 1 0 0 0 3 1 0 2 2 1 0 1 0 0

[PEDRO ALEXANDRE Fernandes]

Fantastic. Works to perfection. I had already thought of this solution but I was missing something. Thank you very much

Answer 2

[KSSV]

A = [ 9     5     2     1     5     3     3];
x = unique(A);
N = numel(x);
count = zeros(N,1);
for k = 1:N
    count(k) = sum(A==x(k));
end
disp([ x(:) count ]);

     1     1
     2     1
     3     2
     5     2
     9     1

Answer 3

[Matt J]

A = [     9     5     2     1     5     3     3     2     2     5
     1     8     3     1     7     4     3     2     2     6
     1     3     1     1     8     6     4     4     2     4
     1     3     1     6     4     2     1     2     9     7
     2     1     6     3     2     2     1     1     9     1
     2     3     1     1     7     5     4     2     1     5
     1     1     4     2     1     8     6     5     4     5];


B=A==reshape(0:9,1,1,[]);

numRepetitions=sum(sum(B,2)>1,3)

numRepetitions = 7×1

3 3 2 2 2 3 3