Calculate probability of number appearing in a column

I have a 1000x296 matrix called Values. The rows represent trials of a simulation, and the rows represent the rank (1-296) of a particular element.

I want to create a matrix that returns the probability of a column having a particular value. I assume i will need a set of nested for loops to count the number of times each value appears in each column, and then divide by the number of rows (1000).

E.g.

The value in cell (1,1) will have the probability of a particular row of column 1 having value 1

The value in cell (1,15) will have the probability of a particular row of column 1 having value 15

The value in cell (200, 155) will have the probability of a particular row of column 200 having value 155.

Thanks!

[Walter Roberson]

accumarray() or https://www.mathworks.com/help/matlab/ref/histcounts2.html possibly.

[Emma Kuttler]

I'm not looking for the probability distribution, I'm looking for the simple probability of a value appearing in each column.

So more like counting (for example) the number of times the value 1 appears in column 1, dividing by the number of rows, and returning that value in position (1,1). Then counting the number of times 2 appears in column 1, dividing by the number of rows, and returning that in (2,1). Then repeating this for all the columns.

[Walter Roberson]

Neither accumarray nor histcounts2 calculate probability distributions. In its simplest form, accumarray just counts.

Answer 1

[Image Analyst]

Just loop over the array getting the histogram in each column.

Something like (untested)

numBins = 25; % Whatever you want.
edges = linspace(0, max(Values(:), numBins);
[rows, columns] = size(Values);
allCounts = zeros(numBins, columns);
for col = 1 : columns
    thisColumn = Values(:, col);
    counts = histcounts(thisColumns, edges);
    allCounts(:, col) = allCounts(:, col) + counts;
end
% Normalize
allCounts = allCounts / sum(allCounts(:));
bar3(allCounts); % Display

Answer 2

[Walter Roberson]

[rowidx, colidx] = ndgrid(1:size(Values,1), 1:size(Values,2));
probs = accumarray([colidx(:), Values(:)],1) ./ size(Values,1);;