plot_precision_recall_curve函数绘制的平均精度是多少？

如何解决plot_precision_recall_curve函数绘制的平均精度是多少？

使用scikit中的plot_precision_recall_curve（）后，我想知道该函数使用的平均精度是多少。在文档中查找时，这是我找到的二进制目标：

# %%
# Compute the average precision score
# ...................................
from sklearn.metrics import average_precision_score
average_precision = average_precision_score(y_test,y_score)

print('Average precision-recall score: {0:0.2f}'.format(
      average_precision))

这是我的数据

clf_4 = svm.SVC()
clf_4.fit(X_train,y_train)
y_clf_4 = clf_4.predict(X_test)

y1_test = np.array([1,1,1]

y1_clf4 = np.array([0,1]

average_precision_5 = average_precision_score(y1_test,y1_clf4)
average_precision_5
Out: 0.5625

现在，我们使用plot_precision_recall_curve和X_test这样（与上面相同）：

X_test= np.array([[0.01167537,0.04676259,0.02145552,0.015625,0.,0.5,0.01020408,1.,0.        ],[0.00478415,0.01258993,0.06759886,0.09375,0.43421053,[0.01503446,0.04136691,0.02600806,0.13157895,0.02721088,[0.017396,0.04856115,0.07737383,0.046875,0.44736842,0.04421769,[0.0072882,0.01079137,0.07866155,0.078125,0.63157895,[0.00733909,0.0323741,0.0487578,0.02040816,[0.02579371,0.11151079,0.03639438,0.0625,0.53947368,0.02380952,[0.00203581,0.03417266,0.12611863,0.125,0.05263158,0.00680272,[0.00527275,0.03057554,0.0344563,0.03125,0.09210526,[0.00590385,0.02158273,0.05135926,0.00340136,[0.01910608,0.16366906,0.05917014,0.28947368,0.12244898,[0.12737045,0.13669065,0.07280827,0.46052632,0.07823129,[0.0537861,0.17446043,0.14109651,0.32894737,0.08843537,[0.01027066,0.05755396,0.06110172,0.30263158,0.01360544,[0.0085504,0.01978417,0.03185484,0.51315789,[0.02224122,0.05215827,0.06370968,0.47368421,0.04081633,[0.00896774,0.05035971,0.00974896,[0.03302084,0.07014388,0.00779787,0.25,0.03741497,[0.00630083,0.06115108,0.01495838,0.10526316,[0.00951741,0.03776978,0.13261576,0.140625,0.0170068,0.        ]])

现在，我们可以使用plot_precision_recall_curve函数并打印两个结果，它们是不同的：

disp = plot_precision_recall_curve(clf_4,X_test,y1_test)
disp.ax_.set_title(f'2-class Precision-Recall curve:{average_precision_5}')

那么差异从何而来？

解决方法

y_score的{​​{1}}参数需要是概率估计（或类似的连续得分），而不是硬分类结果。因此，您的average_precision_score不正确。