how can I find the probabilities of the ecdf function of each duplicate values in y ?

I have a vector that has dublicate values (y). I need to extract each value probabitiy even if its duplicate. here is my code.

num_off_time=[2 2 3 3 1];
y=num_off_time;
[f,x]=ecdf(y);
cdfplot(y)

Answer 1

[the cyclist]

Do you mean you are trying to get these values?

histcounts(num_off_time,'Normalization','probability')

ans =

    0.2000    0.4000    0.4000

[talal alqahtani]

I already have those probabilities but it comes from a duplicate values in y like (2,3) .So, if I have Five channels and each channel have a value like this y=[2 2 3 3 1] So ch1=2 , ch2=2 , ch3=3 , ch4=3 , ch5=1 . if I apply the ecdf function I will only get one probability for both duplicated values. even though its the same probability but i need them separated and belong to its corresponding channel.

[the cyclist]

I'm not sure I understand what you want. For example,

num_off_time=[2 2 2 2 2 2 2 2 2 3 3 1];

will definitely give a different plot from

num_off_time=[2 2 3 3 1];

because it does take into account the number of 2's.

Do you just you mean you want to see the individual jumps in the plot?

Maybe you could give a couple more input examples, and the output you would expect.

[talal alqahtani]

I would expect if Y=[2 2 3 3 1], The output for example would be like this: [0.30 0.30 0.50 0.50 0.10] instead of [0.30 0.50 0.10]. I hope this is clear . Thank you .

[the cyclist]

It's not clear to me where you are getting your values of [0.3 0.5 0.1], because those don't seem to be probabilities related to your input vector.

So, here's what it seems like you want to do:

1. Find the total probability of each channel.
2. For each channel, output its probability, indexed according to the original input vector.

This code does that. I added an extra "3" to your input, to show that 2 and 3 give different output.

num_off_time=[2 2 3 3 3 1];

[unique_num_off_time,~,idxFromUniqueBackToAll] = unique(num_off_time);

probability_of_unique = histcounts(num_off_time,[unique_num_off_time Inf],'Normalization','probability');

probability_by_channel = probability_of_unique(idxFromUniqueBackToAll);

This gives

probability_by_channel =

    0.3333    0.3333    0.5000    0.5000    0.5000    0.1667