Scikit-learn 逻辑回归的性能比 Python 中自己编写的逻辑回归差

Posted 2023-03-12

【中文标题】Scikit-learn 逻辑回归的性能比 Python 中自己编写的逻辑回归差

【英文标题】：Scikit-learn's logistic regression is performing poorer than self-written logistic regression in Python

【发布时间】：2018-11-18 02:05:48

【问题描述】：

我已经在 python 中编写了逻辑回归代码，并将其结果与 Scikit-learn 的逻辑回归进行了比较。后来在一个简单的一维样本数据上表现更差，如下所示：

我的后勤

import pandas as pd

import numpy as np

def findProb(xBias, beta):

    z = []
    for i in range(len(xBias)):
        z.append(xBias.iloc[i,0]*beta[0] + xBias.iloc[i,1]*beta[1])
    prob = [(1/(1+np.exp(-i))) for i in z]
    return prob

def calDerv(xBias, y, beta, prob):

    derv = []
    for i in range(len(beta)):
        helpVar1 = 0
        for j in range(len(xBias)):
            helpVar2 = prob[j]*xBias.iloc[j,i] - y[j]*xBias.iloc[j,i]
            helpVar1 = helpVar1 + helpVar2
        derv.append(helpVar1/len(xBias))
    return derv

def updateBeta(beta, alpha, derv):

    for i in range(len(beta)):
        beta[i] = beta[i] - derv[i]*alpha
    return beta

def calCost(y, prob):

    cost = 0
    for i in range(len(y)):
        if y[i] == 1: eachCost = -y[i]*np.log(prob[i])
        else: eachCost = -(1-y[i])*np.log(1-prob[i])
        cost = cost + eachCost
    return cost

def myLogistic(x, y, alpha, iters):

    beta = [0 for i in range(2)]
    bias = [1 for i in range(len(x))]
    xBias = pd.DataFrame('bias': bias, 'x': x)
    for i in range(iters):
        prob = findProb(xBias, beta)
        derv = calDerv(xBias, y, beta, prob)
        beta = updateBeta(beta, alpha, derv)
    return beta

比较小样本数据的结果

input = list(range(1, 11))

labels = [0,0,0,0,0,1,1,1,1,1]

print("\nmy logistic")

learningRate = 0.01

iterations = 10000

beta = myLogistic(input, labels, learningRate, iterations)

print("coefficients: ", beta)

print("decision boundary is at x = ", -beta[0]/beta[1])

decision = -beta[0]/beta[1]

predicted = [0 if i < decision else 1 for i in input]

print("predicted values: ", predicted)

输出：0、0、0、0、0、1、1、1、1、1

print("\npython logistic")

from sklearn.linear_model import LogisticRegression

lr = LogisticRegression()

input = np.reshape(input, (-1,1))

lr.fit(input, labels)

print("coefficient = ", lr.coef_)

print("intercept = ", lr.intercept_)

print("decision = ", -lr.intercept_/lr.coef_)

predicted = lr.predict(input)

print(predicted)

输出：0、0、0、1、1、1、1、1、1、1

【参考方案1】：

您的实现没有正则化项。 LinearRegression 估计器默认包括具有逆强度C = 1.0 的正则化。当您将C 设置为更高的值时，即削弱正则化，决策边界会更接近5.5：

for C in [1.0, 1000.0, 1e+8]:
    lr = LogisticRegression(C=C)
    lr.fit(inp, labels)
    print(f'C = C, decision boundary @ (-lr.intercept_/lr.coef_[0])[0]')

输出：

C = 1.0, decision boundary @ 3.6888430562595116
C = 1000.0, decision boundary @ 5.474229032805065
C = 100000000.0, decision boundary @ 5.499634348989383

【参考方案2】：

自定义函数或任何逻辑函数将取决于以下内容-

学习率 - alpha 迭代次数

因此，通过对学习率和迭代次数进行一些调整，可以找到大致相等的权重。

如需进一步分析，您可以参考此链接 - https://medium.com/@martinpella/logistic-regression-from-scratch-in-python-124c5636b8ac