python - 梯度下降的多元线性回归

光环，

我是机器学习和Python的新手，我想通过梯度下降来预测Kaggle House Sales in King County dataset。

我将70％（15,000行）的训练和30％（6k行）的测试分开，我从19个中选择5个功能，但是存在性能问题，该算法花费了很多时间（超过11小时），内存为100％并且执行失败。

这是我的梯度下降课程：

class GradientDescent:

    X_train = []
    Y_train = []
    X_test  = []
    Y_test  = []
    lr = 0
    max_iter = 0
    theta = 0

    def __init__(self, X_train,Y_train,X_test,Y_test, lr=0.01, max_iter=100):
        self.X_train = X_train
        self.Y_train = Y_train
        self.X_test  = X_test
        self.Y_test  = Y_test
        self.lr = lr
        self.max_iter = max_iter
        self.theta = np.random.randn(X_train.shape[1], 1)
        print(self.theta)

    def costFunction(self,theta,X,y):
        "1/2m * E(h0-y)**2"
        m = len(y)
        y_pred = X.dot(theta)
        cost = (1/2*m) * np.sum(np.square(y_pred-y))

        return cost


    def estimate(self):
        m = len(self.Y_train)

        mse_hist = np.zeros(self.max_iter)

        #theta_hist = np.zeros(max_iter)
        i = 0
        while i < self.max_iter or mse_hist[i] > 0.01:
            y_pred = np.dot(self.X_train,self.theta)

            error = y_pred-self.Y_train
            self.theta = self.theta - (1/m)*self.lr*(self.X_train.T.dot((error)))
            mse_hist[i] = self.costFunction(self.theta,self.X_train, self.Y_train)

            #print(mse_hist[i])
            i+=1
        return (self.theta, mse_hist)


    def test(self):
        res = pd.DataFrame()
        for i,row in self.X_test.iterrows():
            price_pred = np.dot(row.values,self.theta)
            res = row
            res['price_actual'] = self.Y_test[i]
            res['price_predict'] = price_pred

        res['r2_score'] = r2_score(res['price_actual'].values, res['price_predict'])
        res.to_csv('output.csv')

有什么建议可以使它更好吗？

最佳答案

总的来说，该代码看起来不错，尽管我尚未对其进行测试。我能找到的唯一错误是您可能没有在while循环中增加i，因此循环永不退出。

关于python - 梯度下降的多元线性回归，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/56268136/