replace zeros according to distribution of integers in each column

if I have some matrix x of randomly distributed values (value 0 to n = 5 or 10 in this case)

x = [binornd( 5, 0.05, 10000, 1 ), binornd( 10, 0.1, 10000, 1 )];

I want to create a second matrix where any zeros in x are replaced at random by a value from 1 to n according to the distribution when zero is excluded in that particular column. I want to do this by random sampling from the values in each column excluding zero.

Any ideas how to do this quickly (in reality the matrices are considerably bigger than this example...)

Answer 1

[Guillaume]

If I understand correctly, you want to replace all zeros in a column by random non-zero values from the same column? In which case:

replacementcount = sum(x == 0);
for column = 1:size(x, 2)
   validvalues = nonzeros(x(:, column));
   x(x(:, column) == 0, column) = validvalues(randperm(numel(validvalues), replacementcount(column)));
end

[J T]

Thanks for the quick reply.

Unfortunately I get an error:

Error using randperm
K must be less than or equal to N.

but I changed it to

replacementcount = sum(x == 0);
for column = 1:size(x, 2)
   validvalues = nonzeros(x(:, column));
   x(x(:, column) == 0, column) = datasample(validvalues, replacementcount(column), 'Replace', true);
end

and it seemed to work. So thanks!

I wonder if there's an easier way without the need for loops. If I can avoid then I really would like to speed up this code

[Guillaume]

The error with randperm is because you have more 0s in the column than non-zeros, something I had not considered.

Because the number of 0s can vary per column you cannot avoid the loop. Note that it's only looping over the columns whose number I understood is negligible compared to the number of rows.

[J T]

Thanks for the reply and help!

Answer 2

[Andriy Kavetsky]

You can try this n=10; or n=5; and x(x==0)=randi(n)