寻找最高Precision对应的数值阈值：模型选择与结果异常问题

数据集 #

(26.5625,0)
(29.5625,0)
(30.390625,0)
(18.640625,0)
(27.984375,0)
(26.984375,0)
(25.703125,0)
(25.78125,0)
(32.09375,0)
(25.59375,0)
(27.703125,0)
(30.828125,0)
(23.578125,0)
(21.890625,0)
(25.734375,0)
(24.65625,0)
(27.46875,0)
(31.640625,0)
(26.53125,0)
(25.078125,0)
(30.65625,0)
(24.515625,0)
(25.21875,0)
(21.78125,0)
(28.984375,0)
(29.765625,0)
(27.171875,1)
(30.46875,1)
(35.3125,1)
(27.90625,1)
(34.9375,1)
(33.4375,1)
(30.90625,1)
(31.671875,1)
(32.40625,1)
(26.078125,1)
(31.171875,1)
(36.21875,1)
(35.0625,1)
(35.65625,1)
(36.65625,1)
(37.96875,1)
(31.953125,1)
(33.15625,1)
(37.34375,1)

有序标签与Precision #

ordered_labels: [1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

average precision: 0.7338

需求与问题 #

需要找到类似27.0的原始特征数值阈值，让尽可能多的1类样本排在阈值右侧，0类排在左侧，以此最大化Precision（比如完美排序[1,1,1,0,0,0]的Precision为1.0）。但当前用Logistic回归得到的是0.7这类概率阈值，而非目标数值阈值，询问是否应该改用Linear Regression或SVM。

当前输出结果 #

Precision: [0.33333333 0.         0.         1.        ]
Recall: [1. 0. 0. 0.]
Threshold: [0.13154558 0.7006058  0.72969373]

当前代码 #

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn.datasets import make_classification
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
from sklearn.metrics import precision_recall_curve
from ast import literal_eval

# Create a simple dataset
scores_labels_path = 'data.txt'
X, y = [], []
with open(scores_labels_path) as file:
    for line in file:
        line = literal_eval(line.rstrip())
        X.append(line[0])
        y.append(line[1])

X = np.array(X).reshape(-1, 1)
y = np.array(y)
# X1, y1 = make_classification(n_samples=1000, random_state=42)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.1, random_state=7)
lr = LogisticRegression(random_state=42)
lr.fit(X_train, y_train)
y_scores = lr.predict_proba(X_test)
precision, recall, threshold = precision_recall_curve(y_test, y_scores[:, 1])

print("Precision: {}".format(precision))
print("Recall: {}".format(recall))
print("Threshold: {}".format(threshold))

解决方案 #

1. 无需更换模型：从Logistic回归转换得到数值阈值 #

Logistic回归的预测概率是sigmoid(z)，其中z = w*x + b。当概率等于设定的阈值（比如默认0.5）时，z=0，此时对应的原始特征x值就是你要的数值阈值，计算公式为：

x_threshold = -截距b / 权重w

修改现有代码，加入以下部分即可得到数值阈值：

# 从训练好的Logistic模型中提取数值阈值
w = lr.coef_[0][0]
b = lr.intercept_[0]
x_threshold = -b / w
print(f"原始特征数值阈值: {x_threshold:.4f}")

2. 单特征场景更优方案：直接遍历所有可能阈值 #

因为你的数据是单特征分类任务，直接遍历所有样本的特征值作为阈值，计算每个阈值对应的Precision，能找到全局最优解，无需依赖任何模型：

# 全量数据寻找最优数值阈值
data = sorted(zip(X.flatten(), y), key=lambda item: item[0])
xs = [x for x, y_label in data]
ys = [y_label for x, y_label in data]

max_precision = 0.0
best_threshold = None

# 遍历所有可能的划分点
for idx in range(len(xs)):
    current_threshold = xs[idx]
    # 取所有x >= 当前阈值的样本作为预测为1的集合
    predicted_positives = ys[idx:]
    if not predicted_positives:
        continue
    current_precision = sum(predicted_positives) / len(predicted_positives)
    # 更新最优阈值
    if current_precision > max_precision:
        max_precision = current_precision
        best_threshold = current_threshold

print(f"最优数值阈值: {best_threshold:.4f}")
print(f"对应最高Precision: {max_precision:.4f}")

3. 为什么不选Linear Regression或SVM？ #

• Linear Regression：它是回归模型，输出连续值，不适合0-1分类任务，直接用其输出阈值划分会忽略分类任务的概率分布特性，效果远不如分类模型。
• SVM：虽然可以输出决策函数值并转换为数值阈值，但操作逻辑不如Logistic回归直观，且对于单特征场景，直接遍历阈值的方法效率更高、结果更明确。

内容的提问来源于stack exchange，提问作者AnaRhisT