validator error

[ABarcaru]

Hi, I am trying to use the python application of this script.

nmi = NormalizedMI(normalize_method='joint', k=5 , verbose=False, n_dims = 1)
nmi.fit_transform(X)

and I am hitting the same error:

"BeartypeCallHintReturnViolation: Function normi._estimators.kraskov_estimator() return (np.float64(0.0), np.float64(2.084952429900486), np.float64(1.0187914941452727), np.float64...)) violates type hint tuple[typing.Annotated[typing.Union[float, numpy.floating], Is[lambda arr: bool(np.all(arr > 0))]], typing.Union[float, numpy.floating], typing.Union[float, numpy.floating], typing.Union[float, numpy.floating]], as tuple index 0 item <class "numpy.float64"> np.float64(0.0) violates validator Is[lambda arr: bool(np.all(arr > 0))]:
False == Is[lambda arr: bool(np.all(arr > 0))]."

My X is a numpy array 56x20, dtype float. It works in two other cases of the same size and same instantiation but fails in this one.
What could be the problem ?
test_array.csv

[braniii]

@ABarcaru thank you so much for taking the time to open this issue. This issue might be linked to #6, but looking at the provided test_array I am not sure if it is related. I will try to address #6 in the next weeks and hope that it will solve your problem as well. Sorry, for that.

[PerryL-y]

In the kraskov_estimator function, mutual information mi is finally rectified as np.max([digamma_N + digamma_k - digamma_nx - digamma_ny, 0]), which means it may return 0. So, maybe it should be type-annotated as non-negative float instead of positive float?

Indeed, when I test the normalizedMI class with independent random sequences, it correctly estimates 0 mutual information between random sequences, but throw error due to the beartype type checking

[jessLryan]

@braniii I have been getting a similar error when calling fit or fit_transform. I am not able to share the data I am working with, but have included the error details. There are 2 features, with the same dimensions, containing floats. Thank you!

BeartypeCallHintReturnViolation: Function normi._estimators.kraskov_estimator() return (0.0, -33.75037660839038, -21.184763217160228, -13.703348491372974) violates type hint tuple[typing.Annotated[typing.Union[float, numpy.floating], beartype.vale.Is[lambda arr: bool(np.all(arr > 0))]], typing.Union[float, numpy.floating], typing.Union[float, numpy.floating], typing.Union[float, numpy.floating]], as tuple index 0 item <class "numpy.float64"> 0.0 violates validator beartype.vale.Is[lambda arr: bool(np.all(arr > 0))]:
False == beartype.vale.Is[lambda arr: bool(np.all(arr > 0))].

[jessLryan]

@braniii from an initial look (and quite a basic understanding of what is happening here), it seems that the MI value returned by the kraskov_estimator function is required to be a PositiveFloat, which in turn is required to be of type IsStrictlyPositive, so an MI value of 0.0 results in an error - should this requirement be positive, and not strictly positive?

[braniii]

@ABarcaru @jessLryan sorry for the inconvenience. I guess that the limitation is the assumption of local constant density in the KSG estimator (which this method is based on). I will try to have a look into it.

[jessLryan]

presumably, if MI is zero, then NMI also ought to be zero?

[agheeraert]

@braniii and others, a quick and dirty fix that worked for me is to replace the return call from the kraskov_estimator function in src/normi/_estimators.py from

430     return (
431         np.max([digamma_N + digamma_k - digamma_nx - digamma_ny, 0),  # mi
432         digamma_N - digamma_k + (dx + dy) * mean_log_eps,  # hxy
433         digamma_N - digamma_nx + dx * mean_log_eps,  # hx
434         digamma_N - digamma_ny + dy * mean_log_eps,  # hy
435     )

to

430     return (
431         np.max([digamma_N + digamma_k - digamma_nx - digamma_ny, 1e-10),  # mi
432         digamma_N - digamma_k + (dx + dy) * mean_log_eps,  # hxy
433         digamma_N - digamma_nx + dx * mean_log_eps,  # hx
434         digamma_N - digamma_ny + dy * mean_log_eps,  # hy
435     )

I used to work a lot (not anymore) with the ksg estimator for MI estimation and it was often that we needed those dirty fixes on real-world data.