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I am wondering if any work has been done on machine learning with high dimensional spherical coordinates for tabular data.

Because with tabular data, upon encoding the data for machine learning, we will end up with a vector of the input dimension n. So then the machine learning model for classification or regression will be from $R^n$ to $R^{c}$ for classification, or R for regression. So, does that mean if we take the data points which are points in $R^{n}$, we could also re-write the same data point into a different coordinate system for $R^n$ to $R^c$ or R. But I have not been able to find any resources about this.

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You may want to look at Directional statistics. The problems that can arise with spherical coordinates are intrinsic as well as coordinate singularities. If your data lives on a sphere, you can try to work in space which contains the sphere. In this case your coordinates will be simple, but you will have problems with enforcing constraints of 'data on a sphere'. You can also try to work in a space of the same dimension as the sphere surface. Then the problems will be that your manifold will not be globally isomorphic to $\mathbb{R}^n$.

For 3d spheres, with 2d surfaces, because this problem occurs so often, Spherical Harmonics have been developed. These are polynomials that allow to represent any function on a spherical surface. It might be that you can use something like that.

A more practical solution, however, might be to embrace the fact that your manifold is globally not $\mathbb{R}^n$, even though locally it looks like it. You can then plan straight away that all your data will be broken into segments of local/global. You can use things like Covariant Drivative to differentiate, and things like Metric Tensor, to keep track of distance.

For the special case of 3d half-sphere, one can also have specialized coordinate-transform-based methods e.g. Stereographic coordinates for simpler far-field radiation analysis. These can be extended to higher dimensions, but it is not a free lunch - half-sphere can be deformed into $\mathbb{R}^2$, but a full sphere cannot.

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