Counting proportion of all rows taken by each unique row

Hi Everyone,

Suppose I have a matrix A, and I wish to determine the number of unique rows this matrix has. Then I could form B using:

B = unique(A,'rows')

So far, so good. This yields, say, 5 unique rows out of 100.

My problem is the following. Suppose, having found B, I wish to count for each of the 5 unique rows, how many rows are identical ... and then divided by the total number of rows in A, and sort. This will in this case give a 5 x 1 matrix summing to 100%.

How might this be done?

Regards,

Ulrik.

Answer 1

[Jan]

X = ceil(rand(10, 10) * 5);  % Test data
[B, I, J] = unique(X, 'rows');
n = size(X, 1);
H = histc(J, 1:n) / n

[Ulrik Nash]

I have applied this last line instead:

H = sort(histc(J, 1:n) / n, 'descend')

[Andrei Bobrov]

H = histc(J, 1:length(I)) / size(X,1)

[Sean de Wolski]

Andrei, why the double computation 'n'?

[Andrei Bobrov]

Hi Sean! I think so

X = ceil (rand (10, 10) * 5);% Test date

[B, I, J] = unique (X, 'rows');

H = histc (J, 1: length (I)) / size (X, 1)

Answer 2

[Andrei Bobrov]

more variant

[B, I, J] = unique(X, 'rows');
out = diff(find([1; diff(sort(J)); 1]))/length(J)