wehub-resource-sync
113 lines
4.2 KiB
Python
113 lines
4.2 KiB
Python
"""
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Helper functions used internally for outlier detection tasks.
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"""
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from typing import Optional
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import numpy as np
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from cleanlab.internal.constants import EPSILON
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def transform_distances_to_scores(
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avg_distances: np.ndarray, t: int, scaling_factor: float
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) -> np.ndarray:
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"""Returns an outlier score for each example based on its average distance to its k nearest neighbors.
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The transformation of a distance, :math:`d` , to a score, :math:`o` , is based on the following formula:
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.. math::
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o = \\exp\\left(-dt\\right)
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where :math:`t` scales the distance to a score in the range [0,1].
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Parameters
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----------
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avg_distances : np.ndarray
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An array of distances of shape ``(N)``, where N is the number of examples.
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Each entry represents an example's average distance to its k nearest neighbors.
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t : int
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A sensitivity parameter that modulates the strength of the transformation from distances to scores.
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Higher values of `t` result in more pronounced differentiation between the scores of examples
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lying in the range [0,1].
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scaling_factor : float
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A scaling factor used to normalize the distances before they are converted into scores. A valid
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scaling factor is any positive number. The choice of scaling factor should be based on the
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distribution of distances between neighboring examples. A good rule of thumb is to set the
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scaling factor to the median distance between neighboring examples. A lower scaling factor
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results in more pronounced differentiation between the scores of examples lying in the range [0,1].
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Returns
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-------
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ood_features_scores : np.ndarray
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An array of outlier scores of shape ``(N,)`` for N examples.
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Examples
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--------
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>>> import numpy as np
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>>> from cleanlab.outlier import transform_distances_to_scores
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>>> distances = np.array([[0.0, 0.1, 0.25],
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... [0.15, 0.2, 0.3]])
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>>> avg_distances = np.mean(distances, axis=1)
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>>> transform_distances_to_scores(avg_distances, t=1, scaling_factor=1)
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array([0.88988177, 0.80519832])
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"""
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# Map ood_features_scores to range 0-1 with 0 = most concerning
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return np.exp(-t * avg_distances / max(scaling_factor, EPSILON))
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def correct_precision_errors(
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scores: np.ndarray,
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avg_distances: np.ndarray,
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metric: str,
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C: int = 100,
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p: Optional[int] = None,
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):
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"""
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Ensure that scores where avg_distances are below the tolerance threshold get a score of one.
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Parameters
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----------
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scores :
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An array of scores of shape ``(N)``, where N is the number of examples.
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Each entry represents a score between 0 and 1.
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avg_distances :
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An array of distances of shape ``(N)``, where N is the number of examples.
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Each entry represents an example's average distance to its k nearest neighbors.
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metric :
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The metric used by the knn algorithm to calculate the distances.
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It must be 'cosine', 'euclidean' or 'minkowski', otherwise this function does nothing.
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C :
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Multiplier used to increase the tolerance of the acceptable precision differences.
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It is a multiplicative factor of the machine epsilon that is used to calculate the tolerance.
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For the type of values that are used in the distances, a value of 100 should be a sensible
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default value for small values of the distances, below the order of 1.
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p :
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This value is only used when metric is 'minkowski'.
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A ValueError will be raised if metric is 'minkowski' and 'p' was not provided.
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Returns
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-------
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fixed_scores :
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An array of scores of shape ``(N,)`` for N examples with scores between 0 and 1.
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"""
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if metric == "cosine":
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tolerance = C * np.finfo(np.float64).epsneg
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elif metric == "euclidean":
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tolerance = np.sqrt(C * np.finfo(np.float64).eps)
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elif metric == "minkowski":
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if p is None:
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raise ValueError("When metric is 'minkowski' you must specify the 'p' parameter")
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tolerance = (C * np.finfo(np.float64).eps) ** (1 / p)
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else:
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return scores
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candidates_mask = avg_distances < tolerance
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scores[candidates_mask] = 1
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return scores
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