ddof parameter in FData.cov about scikit-fda

[vnmabus]

Note that this parameter should be also added to var.

[pcuestas]

I understand that ddof should also be 1 by default in var. This is going to change the default behavior of var, as it now normalizes by N.

[vnmabus]

That is right. I do not think that changing it will cause problems, as it is not a very used function.

[pcuestas]

Some tests are failing because of the modification in var.

================================== short test summary info ===================================
FAILED skfda/ml/clustering/_kmeans.py::skfda.ml.clustering._kmeans.FuzzyCMeans
FAILED skfda/tests/test_neighbors.py::TestNeighbors::test_score_functional_response - AssertionError: 
FAILED skfda/tests/test_scoring.py::TestScoreFunctionsGrid::test_all - AssertionError: 
FAILED skfda/tests/test_scoring.py::TestScoreFunctionGridBasis::test_all - AssertionError: 0.9859361757845293 != 0.992968013919044 within 2 places (0.00703183813451...
FAILED skfda/tests/test_scoring.py::TestScoreZeroDenominator::test_zero_r2 - AssertionError: 

I am going to write ddof=0 in the calls to var that appear in the code:

• scikit-fda/skfda/misc/scoring.py

  Line 77 in 415796b

  return var(x)

• scikit-fda/skfda/ml/clustering/_kmeans.py

  Line 150 in 415796b

  variance = fdata.var()

This solves the issues with the tests.

[vnmabus]

Do not "just" write ddof=0. It would be best to analyze the intended denominator in each place.

[pcuestas]

Sorry, I thought those functions were already intentionally using $1/N$ as the variance normalization.

First case

A function _var is defined:

scikit-fda/skfda/misc/scoring.py

Lines 70 to 82 in 415796b

	def _var(
	x: FDataGrid,
	weights: NDArrayFloat | None = None,
	) -> FDataGrid:
	from ..exploratory.stats import mean, var
	if weights is None:
	return var(x)
	return mean( # type: ignore[no-any-return]
	np.power(x - mean(x, weights=weights), 2),
	weights=weights,
	)

and it is only used in two places.

Here:

scikit-fda/skfda/misc/scoring.py

Lines 245 to 250 in 415796b

	num = _var(y_true - y_pred, weights=sample_weight)
	den = _var(y_true, weights=sample_weight)
	# Divisions by zero allowed
	with np.errstate(divide='ignore', invalid='ignore'):
	score = 1 - num / den

where the normalization coefficient is not relevant, due to the division in line 250.

And here:

scikit-fda/skfda/misc/scoring.py

Lines 1011 to 1020 in 415796b

	ss_res = mean(
	np.power(y_true - y_pred, 2),
	weights=sample_weight,
	)
	ss_tot = _var(y_true, weights=sample_weight)
	# Divisions by zero allowed
	with np.errstate(divide='ignore', invalid='ignore'):
	score: FDataGrid = 1 - ss_res / ss_tot

where the stats.var function is called iff sample_weight=None, in which case the variable ss_res is also normalized by N by mean, so I think ddof should be 0 in this first case.

Second case (

scikit-fda/skfda/ml/clustering/_kmeans.py

Line 150 in 415796b

variance = fdata.var()

)

Here the variance is used only to calculate the tolerance used by the K-Means algorithm to check for convergence:

scikit-fda/skfda/ml/clustering/_kmeans.py

Lines 149 to 153 in 415796b

	def _tolerance(self, fdata: Input) -> float:
	variance = fdata.var()
	mean_variance = np.mean(variance[0].data_matrix)
	return float(mean_variance * self.tol)

I believe that ddof=0 can be acceptable here but I do not have any arguments against ddof=1 other than the fact that the tests were built taking into account the previous definition of tolerance that used the variance with ddof=0.

[vnmabus]

I think that your analysis is correct.