思维导图学习笔记
自己参考BoBo老师课程讲解实现：
# -*- coding: utf-8 -*-
import numpy as np
from metrics import r2_score
class LinearRegression(object):
def __int__(self):
self.coef_ = None # 表示系数
self.intercept_ = None # 表示截距
self._theta = None # 过程计算值，不需要暴露给外面
def fit(self, X_train, y_train):
"""根据训练数据集X_train, y_train训练Linear Regression模型"""
assert X_train is not None and y_train is not None, "训练集X和Y不能为空"
assert X_train.shape[0] == y_train.shape[0], "训练集X和Y的样本数要相等"
# np.linalg.inv(X) 表示求X的逆矩阵
# 不能忘了X要增加一列，第一列数据为0
ones = np.ones(shape=(len(X_train), 1))
X_train = np.hstack((ones, X_train))
self._theta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
self.intercept_ = self._theta[0]
self.coef_ = self._theta[1:]
def _predict(self, X):
return X.dot(self.coef_.T) + self.intercept_
def predict1(self, X_test):
"""给定待预测数据集X_test，返回表示X_test的结果向量"""
assert X_test.shape[1] == self.coef_.shape[0], '测试集X的特征值个数不对'
return np.array([self._predict(X) for X in X_test])
def predict(self, X_test):
"""给定待预测数据集X_test，返回表示X_test的结果向量"""
assert X_test.shape[1] == self.coef_.shape[0], '测试集X的特征值个数不对'
ones = np.ones(shape=(len(X_test), 1))
X_test = np.hstack((ones, X_test))
return X_test.dot(self._theta)
def scores(self, X_test, y_test):
"""根据测试数据集 X_test 和 y_test 确定当前模型的准确度"""
assert X_test.shape[0] == y_test.shape[0], '测试集X和Y的个数不相等'
return r2_score(y_test, self.predict1(X_test))
测试脚本：
# -*- encoding: utf-8 -*-
from sklearn.datasets import load_boston
from model_selection import train_test_split
from linearregression import LinearRegression
boston = load_boston()
X = boston.data
y = boston.target
X = X[y < 50]
y = y[y < 50]
X_train, X_test, y_train, y_test = train_test_split(X, y, seed=666)
lrg = LinearRegression()
lrg.fit(X_train, y_train)
# 为什么求出来的theta.shape== (13L,)
print lrg.coef_
print lrg.intercept_
print (lrg.scores(X_test, y_test))
运行结果：
[-1.18919477e-01 3.63991462e-02 -3.56494193e-02 5.66737830e-02
-1.16195486e+01 3.42022185e+00 -2.31470282e-02 -1.19509560e+00
2.59339091e-01 -1.40112724e-02 -8.36521175e-01 7.92283639e-03
-3.81966137e-01]
34.16143549624022
0.8129802602658537
实现总结：
1、上面的准确率是81.29%
2、在实现过程中，fit方法里面，忘了在X_train训练集中添加1的列向量，导致计算出来的系数矩阵参数不对
3、在pridect中，有两种实现方法，可以直接利用中间计算出来的theta值，也可以用X_test.dot(系数矩阵) + 截距举证
4、 np.hstack这个方法，参数值一个tuple，同时要注意相互之间的顺序，越是前面的矩阵则在合并后矩阵越靠前
5、在numpy中实现一个矩阵的逆矩阵用到的方法是np.linalg.inv方法
sklearn中的回归实现
# -*- encoding: utf-8 -*-
from sklearn.datasets import load_boston
from sklearn.neighbors import KNeighborsRegressor
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import train_test_split, GridSearchCV
def test_regression():
boston = load_boston()
X = boston.data
y = boston.target
X = X[y < 50]
y = y[y < 50]
X_train, X_test, y_train, y_test = train_test_split(X, y)
# sklearn中的利用LinearRegression
lrg = LinearRegression()
lrg.fit(X_train, y_train)
print ('LinearRegression:', lrg.score(X_test, y_test))
# 利用KNN回归
knnrg = KNeighborsRegressor()
# 利用网格搜索找到最好的KNN参数
param_grid = [
{
'weights': ['uniform'],
'n_neighbors': [i for i in range(1, 11)]
},
{
'weights': ['distance'],
'n_neighbors': [i for i in range(1, 11)],
'p': [i for i in range(1, 6)]
}
]
grid_search = GridSearchCV(knnrg, param_grid, n_jobs=-1, verbose=1)
grid_search.fit(X_train, y_train)
print 'GridSearchCV:', grid_search.best_estimator_.score(X_test, y_test)
print 'GridSearchCV Best Params:', grid_search.best_params_
if __name__ == '__main__':
test_regression()
运行结果：
LinearRegression: 0.7068690903842936
Fitting 3 folds for each of 60 candidates, totalling 180 fits
[Parallel(n_jobs=-1)]: Done 180 out of 180 | elapsed: 1.3s finished
GridSearchCV: 0.6994655719704079
GridSearchCV Best Params: {'n_neighbors': 10, 'weights': 'distance', 'p': 1}
sklearn中可以看到出来，线性回归还是要比KNN回归要好。此外也要注意，在使用网格搜索的时候，不能有grid_search.score（）方法，而是用grid_search.best_estimator_.score方法，因为两个实现不一样。
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