Scikit-learn 和 Yellowbrick 给出不同的分数

Question

我正在使用 sklearn 计算分类器的平均精度和 roc_auc，并使用 yellowbrick 绘制 roc_auc 和精度召回曲线。问题是这些包在两个指标上给出了不同的分数，我不知道哪个是正确的。

使用的代码：

import numpy as np
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
from yellowbrick.classifier import ROCAUC
from yellowbrick.classifier import PrecisionRecallCurve
from sklearn.datasets import make_classification
from sklearn.metrics import roc_auc_score
from sklearn.metrics import average_precision_score

seed = 42

# provides de data
X, y = make_classification(n_samples=1000, n_features=2, n_redundant=0,
                           n_informative=2, random_state=seed)

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)

clf_lr = LogisticRegression(random_state=seed)
clf_lr.fit(X_train, y_train)

y_pred = clf_lr.predict(X_test)
roc_auc = roc_auc_score(y_test, y_pred)
avg_precision = average_precision_score(y_test, y_pred)
print(f"ROC_AUC: {roc_auc}")
print(f"Average_precision: {avg_precision}")
print('='*20)

# visualizations
viz3 = ROCAUC(LogisticRegression(random_state=seed))
viz3.fit(X_train, y_train) 
viz3.score(X_test, y_test)
viz3.show()
viz4 = PrecisionRecallCurve(LogisticRegression(random_state=seed))
viz4.fit(X_train, y_train)
viz4.score(X_test, y_test)
viz4.show()

代码产生以下输出：

如上所示，指标根据包给出不同的值。在 print 语句中是 scikit-learn 计算的值，而在图中出现注释的是 yellowbrick 计算的值。

Answer 1

由于您使用 scikit-learn 的 predict 方法，您的预测 y_pred 是硬 class 成员资格，而不是概率：

np.unique(y_pred)
# array([0, 1])

但是对于ROC和Precision-Recall计算，应该不是；您传递给这些方法的预测应该是概率，而不是硬 classes。来自 average_precision_score docs:

y_score: array, shape = [n_samples] or [n_samples, n_classes]

Target scores, can either be probability estimates of the positive class, confidence values, or non-thresholded measure of decisions (as returned by “decision_function” on some classifiers).

其中 non-thresholded 表示 不难 classes。 roc_auc_score (docs).

的情况类似

使用以下代码更正此问题，使 scikit-learn 结果与 Yellowbrick return 编辑的结果相同：

y_pred = clf_lr.predict_proba(X_test)     # get probabilities
y_prob = np.array([x[1] for x in y_pred]) # keep the prob for the positive class 1
roc_auc = roc_auc_score(y_test, y_prob)
avg_precision = average_precision_score(y_test, y_prob)
print(f"ROC_AUC: {roc_auc}")
print(f"Average_precision: {avg_precision}")

结果：

ROC_AUC: 0.9545954595459546
Average_precision: 0.9541994473779806

由于 Yellowbrick 在内部（并且透明地）处理所有这些计算细节，因此它不会遭受此处制作的手动 scikit-learn 过程中的错误。

请注意，在二进制情况下（如此处），您可以（并且应该）使用 binary=True 参数使您的绘图不那么混乱：

viz3 = ROCAUC(LogisticRegression(random_state=seed), binary=True) # similarly for the PrecisionRecall curve

而且，与人们的直觉预期相反，至少对于二进制情况，ROCAUC 的 score 方法不会 return 的 AUC，但准确度在 docs:

中指定

viz3.score(X_test, y_test)
# 0.88

# verify this is the accuracy:

from sklearn.metrics import accuracy_score
accuracy_score(y_test, clf_lr.predict(X_test))
# 0.88

Scikit-learn 和 Yellowbrick 给出不同的分数

Scikit-learn and Yellowbrick giving different scores

python

machine-learning

scikit-learn

yellowbrick