西瓜书习题3.4

问题描述

选择两个UCI数据集，比较10折交叉验证法和留一法所估计出的对率回归的错误率。

解答

选了一个wine的数据集，一共将近1600条数据，留一法快把电脑跑死机了。
Logistic 回归可以用 sklearn 库，这里用上一节写的函数，稍微修改一下学习率等。最终算的是正确率。
最终结果：

10折交叉验证准确率：0.7404
留一法准确率：0.7398

代码如下：

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.preprocessing import StandardScaler

def loss(X, y, beta): # 根据损失公式计算当前参数 beta 时打损失
    loss1 = np.sum(-y * (X.dot(beta)))
    loss2 = np.sum(np.log(1 + np.exp(X.dot(beta))))
    return loss1 + loss2

def grad(X, y, beta): # 根据梯度公式计算当前参数 beta 时的一阶梯度
    g = np.exp(X.dot(beta))
    grad1 = -y.dot(X) 
    grad2 = (g / (1.0 + g)).dot(X)
    return grad1 + grad2 

def score(X, y_true, beta): # 计算准确率
    y_pred = 1. / (1 + np.exp(-X.dot(beta)))
    y_pred = np.array(y_pred >= 0.5, dtype=np.float)
    acc = (y_pred == y_true).sum() / len(y_pred)
    return acc

def logistic_reg_fit(X, y): # 输入训练数据和标签，返回训练好的权重
    episilon = 1e-5  # 如果两次更新损失小于次，停止更新
    max_epoch = 500 # 最大迭代次数
    alpha = 0.0005     # 学习率

    beta = np.random.random(X.shape[1]) # 待学习参数

    for i in range(max_epoch):
        last_beta = beta.copy()
        grad_beta = grad(X, y, beta) # 计算梯度
        beta -= alpha * grad_beta # 梯度下降法

        if (abs(loss(X,y,beta) - loss(X,y,grad_beta))<episilon):
            break
    return beta

# 数据源： "http://archive.ics.uci.edu/ml/machine-learning-databases/wine-quality/winequality-red.csv"
wine = pd.read_csv("e:/data/uci/winequality-red.csv", sep=";")

X, y = wine[wine.columns[:-1]], wine[wine.columns[-1]]
# y:qulity 是酒的品质，为0-10之间的整数值，为了应用二分类，把0-5划分成类0，6-11为类1
y = y.apply(lambda label:0 if label<=5 else 1)

scaler = StandardScaler() # 对数据进行标准化
X = scaler.fit_transform(X) 
X = np.hstack([np.ones((len(X),1)), X]) # 最左面加1，截距项
print(X.shape, y.shape)

## 10 折交叉验证
from sklearn.model_selection import KFold

kf = KFold(n_splits=3)

accs = []
for train_idx, test_idx in kf.split(X):
    X_train, y_train = X[train_idx], y[train_idx]
    X_test, y_test = X[test_idx], y[test_idx]
    beta = logistic_reg_fit(X_train, y_train)
    acc = score(X_test, y_test, beta)
    accs.append(acc)
print(np.mean(accs))


## 留一法验证
from sklearn.model_selection import LeaveOneOut

onef = LeaveOneOut()

accs = []
for train_idx, test_idx in onef.split(X):
    X_train, y_train = X[train_idx], y[train_idx]
    X_test, y_test = X[test_idx], y[test_idx]
    beta = logistic_reg_fit(X_train, y_train)
    acc = score(X_test, y_test, beta)
    accs.append(acc)
print(np.mean(accs))