CODEWITHSUNDEEP
matlab
Antonio Craveiro
Antonio Craveiro

Reputation: 121

How to calculate the "rest value" of a plot?

Didn't know how to paraphrase the question well.

Function for example:

Data:https://www.dropbox.com/s/wr61qyhhf6ujvny/data.mat?dl=0

In this case how do I calculate that the rest point of this function is ~1? I have access to the vector that makes the plot. I guess the mean is an approximation but in some cases it can be pretty bad.

Upvotes: 0

Views: 88

Answers (1)

rayryeng
rayryeng

Reputation: 104525

Under the assumption that the "rest" point is the steady-state value in your data and the fact that the steady-state value happens the majority of the times in your data, you can simply bin all of the points and use each unique value as a separate bin. The bin with the highest count should correspond to the steady-state value.

You can do this by a combination of histc and unique. Assuming your data is stored in y, do this:

%// Find all unique values in your data
bins = unique(y);

%// Find the total number of occurrences per unique value
counts = histc(y, bins);

%// Figure out which bin has the largest count
[~,max_bin] = max(counts);

%// Figure out the corresponding y value
ss_value = bins(max_bin);

ss_value contains the steady-state value of your data, corresponding to the most occurring output point with the assumptions I laid out above.

A minor caveat with the above approach is that this is not friendly to floating point data whose unique values are generated by floating point values whose decimal values beyond the first few significant digits are different.

Here's an example of your data from point 2300 to 2320:

>> format long g;
>> y(2300:2320)

ans =

          0.99995724232555
         0.999957488454868
         0.999957733165346
         0.999957976465197
         0.999958218362579
         0.999958458865564
         0.999958697982251
         0.999958935720613
         0.999959172088623
         0.999959407094224
         0.999959640745246
         0.999959873049548
         0.999960104014889
         0.999960333649014
         0.999960561959611
         0.999960788954326
          0.99996101464076
         0.999961239026462
         0.999961462118947
         0.999961683925704
         0.999961904454139

Therefore, what I'd recommend is to perhaps round so that the first 5 or so significant digits are maintained.

You can do this to your dataset before you continue:

num_digits = 5;
y_round = round(y*(10^num_digits))/(10^num_digits);

This will first multiply by 10^n where n is the number of digits you desire so that the decimal point is shifted over by n positions. We round this result, then divide by 10^n to bring it back to the scale that it was before. If you do this, for those points that were 0.9999... where there are n decimal places, these will get rounded to 1, and it may help in the above calculations.

However, more recent versions of MATLAB have this functionality already built-in to round, and you can just do this:

num_digits = 5;
y_round = round(y,num_digits);

Minor Note

More recent versions of MATLAB discourage the use of histc and recommend you use histcounts instead. Same function definition and expected inputs and outputs... so just replace histc with histcounts if your MATLAB version can handle it.

Using the above logic, you could also use the median too. If the majority of data is fluctuating around 1, then the median would have a high probability that the steady-state value is chosen... so try this too:

ss_value = median(y_round);

Upvotes: 2

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