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I might be misunderstanding your question, but this should do the trick:

I had this issue and solved it by re-installing the scipy package with pip install -U scipy in order to upgrade to version 1.7

In scipy's PyPI page, it looks like scipy doesn't support 3.10 as the meta says

This may be a rounding/precision issue. In the R implementation if you add a rounding option to the loss function:

Check again what the input arguments of solve_ivp are. The integration interval is given by the first two numbers in the t_span argument, so in your application most values in sol.sol(t) are obtained via wild extrapolation.

Correct that by giving the interval as [min(t),max(t)].

To get more compatible computations, explicitly set the error tolerances, as the default values need not be equal. For instance atol=1e-22, rtol=1e-9 so that only the relative tolerance has an effect.

It is curious how you use the args mechanism. It was only recently introduced to solve_ivp to be more compatible with odeint. I would not use it in either case here, as the definition of the parameter and its use is contained in a 3-line block. It has its uses where proper encapsulation and isolation from other code is a real concern.

I suspect the problem is that you are running a different version of SciPy than the one the documents are describing.

Turning a comment into an answer, BSpline.design_matrix is constructing what you are after, in the csr sparse format. It'll be available from scipy 1.8 when it is released. Until then, you can either grab the master branch of scipy, or use a workaround suggested by the docs (https://scipy.github.io/devdocs/reference/generated/scipy.interpolate.BSpline.design_matrix.html#scipy.interpolate.BSpline.design_matrix) :

You can use typing.Protocol to assert that the type implements __matmul__.

In 2D, regions are separated by a single line segment, thus always 2 points per ridge. In 3D and up, regions separation "plane segments" are typically triangular, but they can have 4+ edges, too.

For sceletonization purposes, one approach would be to show the outline of the separation region, skipping virtual (-1) points. So, [3, 0, -1] would translate to one line between points 3 and 0. [1, 0, 3, 2] will generate segments 1-0, 0-3, 3-2, 2-1. As an additional improvement, ridges with 4+ points can be further split into trianges, so in case of [1, 0, 3, 2] another segment would be 0-2 or 1-3.

I am still not sure if I got the question right, let me know if I didn't