Average displacement from the diagonal method for delay time in traditional embedding

[Datseris]

In the traditional embedding scheme where one calculates independently delay time and embedding dimension, there is another method to compute optical delay time, besides automutual information minima.

This is the "average displacement of the time-delayed vectors from the phase space diagonal" from:

Paper: https://www.delucafoundation.org/download/bibliography/de-luca/061.pdf

Implementing this is easy, and shouldn't take more than 10 lines of code!

[hkraemer]

I'll go for it after my defense and vacation :) It is on my list anyway. Expect end of October.

[dynamic-queries]

Are the tags on this issue recent ? (i.e. Is this a issue that still needs contribution?)

[Datseris]

This open issue lists a feature request: something we'd like to have in the library. Someone needs to contribute this to be in the library, so I guess the answer to your question is "yes" :)

The recency of the tags does not matter. If this feature was in the library this issue would have been closed.

[denizhanpak]

Hey! I have a question and would appreciate some guidance.
In the paper the equation presupposes a given dimensionality (m) in equation 12. Although it seems that regardless of the dimension one should get a similar optimal lag, the authors note that different m will give slightly different results. Considering the point of this part of the library is to segment dimensionality and lag identification should I simply assume a specific dimension and allow dimension as an optional argument or should I call one of the dimensionality estimators in the script or both?

[Datseris]

The optimal dimension methods require already an optimal lag... So you'd go in circles. I would say, let this be a keyword argument to be put in by the user.

[denizhanpak]

Awesome! I think I got it. I will just add a test and send a pull request

[Datseris]

there was some work here: https://github.com/JuliaDynamics/DelayEmbeddings.jl/pull/120/files if someone can finish it.