项目文件夹

文件
ChG 5fbf6c0493 Add Data Valuation Issue Manager (#850)
Co-authored-by: Elías Snorrason <eliassno@gmail.com>
2023-11-06 17:04:57 +01:00

146 行
4.8 KiB
Python

# Copyright (C) 2017-2023 Cleanlab Inc.
# This file is part of cleanlab.
#
# cleanlab is free software: you can redistribute it and/or modify
# it under the terms of the GNU Affero General Public License as published
# by the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# cleanlab is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU Affero General Public License for more details.
#
# You should have received a copy of the GNU Affero General Public License
# along with cleanlab. If not, see <https://www.gnu.org/licenses/>.
from __future__ import annotations
from typing import (
TYPE_CHECKING,
Any,
ClassVar,
Dict,
List,
Optional,
Union,
)
import numpy as np
import pandas as pd
from scipy.sparse import csr_matrix
from cleanlab.datalab.internal.issue_manager import IssueManager
if TYPE_CHECKING: # pragma: no cover
import pandas as pd
from cleanlab.datalab.datalab import Datalab
class DataValuationIssueManager(IssueManager):
"""Manages data sample with low valuation."""
description: ClassVar[
str
] = """
Examples that contribute minimally to a model's training
receive lower valuation scores.
"""
issue_name: ClassVar[str] = "data_valuation"
issue_score_key: ClassVar[str]
verbosity_levels: ClassVar[Dict[int, List[str]]] = {
0: [],
1: [],
2: [],
3: ["average_data_valuation"],
}
DEFAULT_THRESHOLDS = 1e-6
def __init__(
self,
datalab: Datalab,
threshold: Optional[float] = None,
k: int = 10,
**kwargs,
):
super().__init__(datalab)
self.k = k
self.threshold = threshold if threshold is not None else self.DEFAULT_THRESHOLDS
def find_issues(
self,
**kwargs,
) -> None:
"""Calculate the data valuation score with a provided or existing knn graph.
Based on KNN-Shapley value described in https://arxiv.org/abs/1911.07128
The larger the score, the more valuable the data point is, the more contribution it will make to the model's training.
"""
knn_graph = self._process_knn_graph_from_inputs(kwargs)
labels = self.datalab.labels.reshape(-1, 1)
assert knn_graph is not None, "knn_graph must be already calculated by other issue managers"
assert labels is not None, "labels must be provided"
scores = _knn_shapley_score(knn_graph, labels)
self.issues = pd.DataFrame(
{
f"is_{self.issue_name}_issue": scores < self.threshold,
self.issue_score_key: scores,
},
)
self.summary = self.make_summary(score=scores.mean())
self.info = self.collect_info(self.issues)
def _process_knn_graph_from_inputs(self, kwargs: Dict[str, Any]) -> Union[csr_matrix, None]:
"""Determine if a knn_graph is provided in the kwargs or if one is already stored in the associated Datalab instance."""
knn_graph_kwargs: Optional[csr_matrix] = kwargs.get("knn_graph", None)
knn_graph_stats = self.datalab.get_info("statistics").get("weighted_knn_graph", None)
knn_graph: Optional[csr_matrix] = None
if knn_graph_kwargs is not None:
knn_graph = knn_graph_kwargs
elif knn_graph_stats is not None:
knn_graph = knn_graph_stats
if isinstance(knn_graph, csr_matrix) and kwargs.get("k", 0) > (
knn_graph.nnz // knn_graph.shape[0]
):
# If the provided knn graph is insufficient, then we need to recompute the knn graph
# with the provided features
knn_graph = None
return knn_graph
def collect_info(self, issues: pd.DataFrame) -> dict:
issues_info = {
"num_low_valuation_issues": sum(issues[f"is_{self.issue_name}_issue"]),
"average_data_valuation": issues[self.issue_score_key].mean(),
}
info_dict = {
**issues_info,
}
return info_dict
def _knn_shapley_score(knn_graph: csr_matrix, labels: np.ndarray) -> np.ndarray:
"""Compute the Shapley values of data points based on a knn graph."""
N = labels.shape[0]
scores = np.zeros((N, N))
dist = knn_graph.indices.reshape(N, -1)
k = dist.shape[1]
for i, y in enumerate(labels):
idx = dist[i][::-1]
ans = labels[idx]
scores[idx[k - 1]][i] = float(ans[k - 1] == y) / k
cur = k - 2
for j in range(k - 1):
scores[idx[cur]][i] = scores[idx[cur + 1]][i] + float(
int(ans[cur] == y) - int(ans[cur + 1] == y)
) / k * (min(cur, k - 1) + 1) / (cur + 1)
cur -= 1
return 0.5 * (np.mean(scores, axis=1) + 1)