Probability density function for binary data

I have data for the index of each point: 0 for the point outside particle, 1 for the points inside particle.
How to calculate calculate the PDF for this case? If you know that my domaine points is in [200 67].
And I want to calculate the PDF in a fixed x-point for example 80.

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Image Analyst
Image Analyst el 2 de Oct. de 2021
Not sure what the PDF means in this context. You can compute the chance the particle will be a 0 or a 1. This would give a PDF with only two values on the x axis.
p(1) = sum(data == 0) / numel(data);
p(2) = sum(data == 1) / numel(data);
Is that what you want? Or do you have (x,y) locations for each point and want some sort of 2-D probability that the "1" will land at some location or radius?

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Tesla
Tesla el 4 de Oct. de 2021
Thank you for the answer.
I mean by PDF the Probability density function.
Attached an example of my data. The data for the whole domain (67X335).
And I want a 2-D PDF that the "1" will land at some location, for me I am interested in a fixed x, for example in X=50. And See how the PDF is changing along Y axis.
The attached data is for 1 iterations, and I have around 1000 iterations. You can see the picture. This data is already done by Matlab.
So the PDF has to be the average for the whole data iterations.
Image Analyst
Image Analyst el 4 de Oct. de 2021
The workbook contains just one of those 67x335 matrices. The values are either -1, 0, or 1. Explain what the rows and columns represent.
Tesla
Tesla el 5 de Oct. de 2021
The workbook is just for 1 iteration, and its 67x335 matrices which meand 67 point in Y-axis and 335 point in X-axis.
When the values ares 1 means we are in inside the cell, and 0 means outside the cell. for -1 its means outside the domain of my calculations.
Tesla
Tesla el 7 de Oct. de 2021
Any clue?
Image Analyst
Image Analyst el 7 de Oct. de 2021
Editada: Image Analyst el 7 de Oct. de 2021
So the row number is y and the column index is x? So I can get the one's (x,y) coordinates by doing
[y, x] = find(theMatrix);
Correct? If so, why don't you just add the matrix at each iteration to a master, growing, cumulative matrix and then divide that matrix by the number of iterations to get the pdf?
Tesla
Tesla el 7 de Oct. de 2021
Yes correct. How can I do that?
Tesla
Tesla el 8 de Oct. de 2021
Thank you,
But indeed my data are already stored on matlab variable, you can see the screenshot.
I am just worring how to use your suggestion to calculate the pdf.
If ca is your cell array:
numCells = length(ca);
sumArray = zeros(67, 335);
for k = 1 : numCells
thisCell = ca{k}; % A single 67x335 matrix.
sumArray = sumArray + thisCell;
end
meanArray = sumArray / numCells;
Does that do what you want?
Tesla
Tesla el 8 de Oct. de 2021
The code is giving the total pdf for the whole domain. I want the pdf in a specified x-point.
For example in x=30. You see?
Why isn't it just
% Get probability for various y values along the x=30 line.
pdf30 = meanArray(:, 30);
Tesla
Tesla el 8 de Oct. de 2021
Thank you very much, it works!

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