【udacity】机器学习-回归

【udacity】机器学习-回归
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1.什么是回归？

regression
在监督学习中，包括了输入和输出的样本，在此基础上，我们能够通过新的输入来表示结果，映射到输出
输出包含了离散输出和连续输出

2.回归与函数逼近

回归并不是指向平均值回落，而是使用函数形式来逼近一堆数据点

3.线性回归

什么是线性方程？

线性方程就是直线方程，可以理解为

Y=mx+b

这里的m是斜率，b是截距，这是一个线性方程而不是平面方程

什么是回归分析？

回归分析是统计的概念。这里的想法是观察数据和构建一个方程，使我们可以为丢失的数据或未来数据的预测。

什么是线性回归？

线性回归是模型之间的线性关系因变量(Y)和自变量(X1、X2、X3等的关系)

Y=θ0+θ1X1+θ2X2+...+θnXn

序号统计学分数编程学分数数据科学分数

A 50 80 65

B 80 65 83

C 60 60 69

D 95 80 92

E 95 50 84

F 40 90 55

这里的y是一个输出变量，X1,X2...Xn是输入变量和θ0,θ1...θn被称为参数或者权重
所以在上面的得分数据集中，y是数据科学的评分。X1代表统计得分，X2是在编程得分
y=θ0+θ1x1+θ2x2

为什么这些θ就是所谓的权重
每个θ告诉我们如何重视想用的X在预测的输出。这表示，如果一个特定的θ相比其他值小，相应的X起着预测输出多大的作用。

为什么会出现错误
尽管能够根据线性方程预测，但是在现实世界中，情况是多元性的，不可能使用简单的线性方程就会实现预测，这会导致模型的错误。
所以我们要考虑误差
绝对误差综合：∑i=1m∣yi^−yi∣
误差平方的总和：21∑i=1m(y^−yi)2
我们会在梯度下降算法以后使用这个表达式来考虑优化模型
为什么梯度下降与误差平方是有用的？
梯度下降算法使用被最小化的函数的导数。

4.找到最佳拟合

f(x)=c
E(c)=∑i=1n(yi−c)2

5.多项式的阶

Xw≈y
XTXw≈XTy
(XTX)−1XTXw≈(XTX)−1XTy
w=(XTX)−1XTy

6.误差

出现误差的原因是多方面的

7.交叉验证

交叉验证的目的是为了泛化我们的机器学习模型，所以不能直接在测试集中直接建模

8.小结

1.回归的历史
2.模型的选择-过拟合、欠拟合，一般拟合、交叉验证
3.线性回归和多项式回归
4.平方误差下的最佳常数和过程中用的微积分
5.一般处理方法
6.表示法以及回归的应用

线性回归算法

解决回归问题

思想简答，实现容易

许多强大的非线性模型的基础

结果具有很好的可解释性

蕴含机器学习中的很多重要思想

一类机器学习算法的基本思路

目标：找到a和b，使得∑i=1m(y(i)−ax(i)−b)2即 损失函数(loss function) 尽可能小，当在算法中用拟合的程度来测量，即 效用函数(utility function)
典型的最小二乘法问题：最小化误差的平方

a=∑i=1m(x(i)−x)2∑i=1m(x(i)−x)(y(i)−y)b=y−ax

通过分析问题，确定问题的损失函数或者效用函数

通过最优化损失函数或者效用函数，获得机器学习模型

所有参数学习的算法都是为了最终求解某一个值的极值，学习模型最终的参数，找到相应的参数来最优化损失函数或者效用函数

线性回归

多项式回归

SVM

神经网络

逻辑回归

.............
最优化原理：在经典的传统算法里使用的也是最优化思路
凸优化：解决的是特殊的优化思路

最小二乘法

简单的线性回归算法

class SimpleLinearRegression1: def __init__(self): self.a_ = None self.b_ = None def fit(self,x_train,y_train): ''' 根据训练数据集x_train,y_train训练simple linear regression模型 :param x_train: :param y_train: :return: ''' assert x_train.ndim ==1 ,"simple linear regressor can only solve single feature training data." assert len(x_train) == len(y_train),'the size of x_train must be equal to the size of y_train' x_mean = np.mean(x_train) y_mean = np.mean(y_train) num = 0.0 d = 0.0 for x_i,y_i in zip(x_train,y_train): num += (x_i-x_mean)*(y_i-y_mean) d += (x_i - x_mean) **2 self.a_ = num /d self.b_ = y_mean-a*x_mean def predict(self, x_predict): ''' 给定待预测数据集x_predict,返回x_predict的结果向量 :param x_predict: :return: ''' assert x_predict.ndim ==1 ,"simple linear regressor can only solve single feature training data." assert self.a_ is not None and self.b_ is not None,'must fit before predict!' return np.array([self._predict(x) for x in x_predict]) def _predict(self,x_single): ''' 给定单个待预测数据x_single，返回x_single的预测结果值 :param x_single: :return: ''' return self.a_ * x_single + self.b_ def __repr__(self): return 'SimpleLinearRegression1()' reg1 = SimpleLinearRegression1() reg1.fit(x,y) y_hat1 = reg1.predict(x) plt.scatter(x,y) plt.plot(x,y_hat1,color='red') plt.show()

向量化运算

使用向量化运算比简单的线性计算运算速度上要快的多

class SimpleLinearRegression2: def __init__(self): self.a_ = None self.b_ = None def fit(self,x_train,y_train): ''' 根据训练数据集x_train,y_train训练simple linear regression模型 :param x_train: :param y_train: :return: ''' assert x_train.ndim ==1 ,"simple linear regressor can only solve single feature training data." assert len(x_train) == len(y_train),'the size of x_train must be equal to the size of y_train' x_mean = np.mean(x_train) y_mean = np.mean(y_train) num = (x_train-x_mean).dot(y_train-y_mean) d = (x_train-x_mean).dot(x_train-x_mean) self.a_ = num /d self.b_ = y_mean-a*x_mean def predict(self, x_predict): ''' 给定待预测数据集x_predict,返回x_predict的结果向量 :param x_predict: :return: ''' assert x_predict.ndim ==1 ,"simple linear regressor can only solve single feature training data." assert self.a_ is not None and self.b_ is not None,'must fit before predict!' return np.array([self._predict(x) for x in x_predict]) def _predict(self,x_single): ''' 给定单个待预测数据x_single，返回x_single的预测结果值 :param x_single: :return: ''' return self.a_ * x_single + self.b_ def __repr__(self): return 'SimpleLinearRegression2()' reg2 = SimpleLinearRegression2() reg2.fit(x,y) print(reg2.a_, reg2.b_) y_hat2 = reg2.predict(x) plt.scatter(x,y) plt.plot(x,y_hat2,color='red') plt.show()

性能测试，线性回归向量化运算比线性运算快约50倍

m = 1000000 big_x = np.random.random(size=m) big_y = big_x * 2.0 + 3.0 + np.random.normal(size=m) %timeit reg1.fit(big_x, big_y) %timeit reg2.fit(big_x, big_y) >>> 962 ms ± 60.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 25.9 ms ± 286 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) >>>

回归算法的评价

线性回归算法的评测

m1i=1∑m(ytest(i)−y^test(i))2

均方误差MSE(Mean Squared Error)

m1i=1∑m(ytest(i)−y^test(i))2=MEStest

均方根误差RMSE(Root Mean Squared Error)

m1i=1∑m∣ytest(i)−y^test(i)∣

平均绝对误差MAE(Mean Absolute Error)

R2=1−SStotalSSresidual

R2=1−∑(y(i)−y(i))2∑(y^(i)−y(i))2

衡量了自己的模型没有产生错误的指标

R2<=1

R2 越大越好。当我们的预测模型不犯任何错误时，R^2得到最大值1

当我们的模型等于基准模型时，R2 为0

如果R2<0,说明我们学习到的模型还不如基准模型。此时，很有可能我们的数据不存在任何线性关系。

sklearn中的线性回归算法中的score就是使用的R2评估标准

多元线性回归

数据有多少特征有多少维度，就会有多少个θ表示，θ0就是截距
目标找到θ0 , θ1,θ2 , ...,θn 使得∑i=1m(y(i)−y^(i))2尽可能小
即(y−Xb⋅θ)T(y−Xb⋅θ)尽可能小
多元线性回归的正规方程解(Normal Equation)
θ=(XbTXb)−1XbTy
问题：时间复杂度高 O(n3)优化O(n2.4)
有点：不需要对数据进行归一化处理
解决方案：
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Win a contest, win a challenge
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 SE 2014年4月3日
 SE 2014年4月2日
原文地址：https://www.cnblogs.com/pandaboy1123/p/10245549.html

序号	统计学分数	编程学分数	数据科学分数
A	50	80	65
B	80	65	83
C	60	60	69
D	95	80	92
E	95	50	84
F	40	90	55

【udacity】机器学习-回归

1.什么是回归？

2.回归与函数逼近

3.线性回归

4.找到最佳拟合

5.多项式的阶

6.误差

7.交叉验证

8.小结

线性回归算法

一类机器学习算法的基本思路

最小二乘法

简单的线性回归算法

向量化运算

回归算法的评价

多元线性回归