對比梯度下降和正規方程解性能

現在用導數的方式模擬線性迴歸中的梯度下降法

首先再來回顧一下梯度下降法的基礎

• 梯度下降法不是一個機器學習算法, 而是一個搜索算法
• 梯度下降法用在監督學習中
• 梯度下降法的過程: 對比模型輸出值和樣本值的差異不斷調整本省權重, 直到最後模型輸出值和樣本標籤差別達到理想誤差範圍內
• 梯度下降法的超參是梯度, 梯度過大會導致跳過最優或局部最優解, 梯度過小會過於消耗計算資源

現在寫一個sklearn中的正規方程的線性迴歸模型 沒有實現fit方法

import numpy as np
from .metrics import r2_score

class LinearRegression:

    def __init__(self):
        """初始化Linear Regression模型"""
        self.coef_ = None
        self.intercept_ = None
        self._theta = None

    def predict(self, X_predict):
        """給定待預測數據集X_predict，返回表示X_predict的結果向量"""
        assert self.intercept_ is not None and self.coef_ is not None, \
            "must fit before predict!"
        assert X_predict.shape[1] == len(self.coef_), \
            "the feature number of X_predict must be equal to X_train"

        X_b = np.hstack([np.ones((len(X_predict), 1)), X_predict])
        return X_b.dot(self._theta)

    def score(self, X_test, y_test):
        """根據測試數據集 X_test 和 y_test 確定當前模型的準確度"""

        y_predict = self.predict(X_test)
        return r2_score(y_test, y_predict)

    def __repr__(self):
        return "LinearRegression()"

因爲這裏是用線性迴歸做測試, 線性迴歸的正規方程解如下

此公式網上搜的, 推導過程可以看下這位老兄的博客 https://blog.csdn.net/chenlin41204050/article/details/78220280 (未經允許直接引用了哈)

實現正規方程解的fit如下:

    def fit_normal(self, X_train, y_train):
        """根據訓練數據集X_train, y_train訓練Linear Regression模型"""
        assert X_train.shape[0] == y_train.shape[0], \
            "the size of X_train must be equal to the size of y_train"
        X_b = np.hstack([np.ones((len(X_train), 1)), X_train])
        self._theta = np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y_train)
        self.intercept_ = self._theta[0]
        self.coef_ = self._theta[1:]
        return self

要用梯度下降法, 就得找到損失函數, 然後計算梯度
下面是多元線性迴歸的瞬時函數

線性迴歸的lose
接着實現上圖面試的梯度下降法的fit_gd如下

    def fit_gd(self, X_train, y_train, eta=0.01, n_iters=1e4):
        """根據訓練數據集X_train, y_train, 使用梯度下降法訓練Linear Regression模型"""
        assert X_train.shape[0] == y_train.shape[0], \
            "the size of X_train must be equal to the size of y_train"
        def J(theta, X_b, y):
            try:
                return np.sum((y - X_b.dot(theta)) ** 2) / len(y)
            except:
                return float('inf')
            
        def dJ(theta, X_b, y):
            return X_b.T.dot(X_b.dot(theta) - y) * 2. / len(y)
            
        def gradient_descent(X_b, y, initial_theta, eta, n_iters=1e4, epsilon=1e-8):
            theta = initial_theta
            cur_iter = 0
            while cur_iter < n_iters:
                gradient = dJ(theta, X_b, y)
                last_theta = theta
                theta = theta - eta * gradient
                if (abs(J(theta, X_b, y) - J(last_theta, X_b, y)) < epsilon):
                    break
                cur_iter += 1
            return theta
            
        X_b = np.hstack([np.ones((len(X_train), 1)), X_train])
        initial_theta = np.zeros(X_b.shape[1])
        self._theta = gradient_descent(X_b, y_train, initial_theta, eta, n_iters)
        self.intercept_ = self._theta[0]
        self.coef_ = self._theta[1:]
        return self

上面的方法都封裝在一個LinearRegression.py中

現在開始測試
準備一些隨機數據, 維度爲5000, 樣本數量1000

m = 1000
n = 5000

big_X = np.random.normal(size=(m,n))
true_theta = np.random.uniform(0.0, 100.0, size=n+1)
big_y = big_X.dot(true_theta[1:]) + true_theta[0] + np.random.normal(0, 10.0, size=m)
print(big_X.shape, big_y.shape)

%run LinearRegression.py

輸出:

(1000, 5000) (1000,)

使用正規方程解訓練線性迴歸

big_reg1 = LinearRegression()
%time big_reg1.fit_normal(X_train, y_train)
big_reg1.score(X_test, y_test)

輸出:

CPU times: user 26.8 s, sys: 921 ms, total: 27.7 s
Wall time: 11.1 s
LinearRegression()

使用梯度下降法訓練

big_reg2 = LinearRegression()
%time big_reg2.fit_gd(big_X, big_y)

輸出:

CPU times: user 7.44 s, sys: 97.2 ms, total: 7.54 s
Wall time: 4.15 s
LinearRegression()

結論

• 梯度下降法在無法求出直接最優解的情況下 , 可以搜索到相對較優的解
• 梯度下降法在數據量較大的情況下, 可以避免類似於正規方程解因爲數據規模變大帶來的長時間運算(也是相對性的)