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I am actually not sure if I should post this here on Stackoverflow or it is better at CrossValidated but it could be moved if you think its not the right place :)

Briefly, I have X, Y coordinates of data for different IDs. These are cell track data and I wanted to try to calculate the fractal dimension of each individual track per ID. My data looks like this:

    structure(c("482.624", "483.577", "484.634", "486.883", "488.211", 
"493.759", "452.133", "450.953", "450.603", "450.424", "450.518", 
"445.979", "0-Si", "0-Si", "0-Si", "0-Si", "0-Si", "0-Si"), .Dim = c(6L, 
3L), .Dimnames = list(NULL, c("X", "Y", "ID")))

I found out that you can use the package fractaldim and I tried to use the function fd.estimate for 2d data but I cannot really figure out how to do it with individual IDs and not the whole data set.

Thanks for your help.

Kaye11
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2 Answers2

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I am not really sure if I understand your idea, but you need a matrix in order to use a method for 2d data. Maybe if you want an estimate for each id you should use other kind of method. In this case, each id it's only a point.

lesoleil14
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I think the problem here is that your data describes Cantor Dust, in fact (describes data points in the 2D space, so you are interested how far from 2D your fractal/Cantor Dust is). But the package fractaldim estimates how rough some 2D surface is. So since you are interested in estimating k for D = 2 - k, the fractaldim finds k for D = 2 + k, i.e., it measures how much is the 2D surface rugged and how much it approaches 3D. But you are interested how much dense/scarse your data projected to 2D are, so how much your data approaches 1D or even 0D form full 2D.

I am solving same problem and still haven't find the solution...

Fran
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  • This does not really answer the question. If you have a different question, you can ask it by clicking [Ask Question](https://stackoverflow.com/questions/ask). To get notified when this question gets new answers, you can [follow this question](https://meta.stackexchange.com/q/345661). Once you have enough [reputation](https://stackoverflow.com/help/whats-reputation), you can also [add a bounty](https://stackoverflow.com/help/privileges/set-bounties) to draw more attention to this question. - [From Review](/review/late-answers/30817730) – jared_mamrot Jan 20 '22 at 02:43