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Coursera机器学习基石作业二python实现

##机器学习基石作业二
Coursera机器学习基石作业二python实现

Coursera机器学习基石作业二python实现

下面的代码是17、18题的结合:

import numpy as np
import random
class decisonStump(object):
    def __init__(self,dimension,data_count,noise):
        self.dimension=dimension
        self.data_count=data_count
        self.noise=noise
    def generate_dataset(self):
        dataset=np.zeros((self.data_count,self.dimension+1))
        for i in range(self.data_count):
            x=random.uniform(-1,1)
            line=[]
            line.append(x)
            y=np.sign(x)*np.sign(random.uniform(0,1)-self.noise)
            line.append(y)
            dataset[i:]=line
        return dataset
    def get_theta(self,dataset):
        l=np.sort(dataset[:,0])
        theta=np.zeros((self.data_count,1))
        for i in range(self.data_count-1):
            theta[i]=(l[i]+l[i+1])/2
        theta[-1]=1
        return theta
    def question1718(self):
        sum_e_in = 0
        sum_e_out=0
        for i in range(5000):
            dataset = self.generate_dataset()
            theta=self.get_theta(dataset)
            e_in = np.zeros((2, self.data_count))
            for j in range(self.data_count):
                a=dataset[:,1]*np.sign(dataset[:,0]-theta[j])
                e_in[0][j] = (self.data_count - np.sum(a)) / (2 * self.data_count)  # 数组只有-1和+1,可直接计算出-1所占比例
                e_in[1][j] = (self.data_count - np.sum(-a)) / (2 * self.data_count)
            min0, min1 = np.min(e_in[0]), np.min(e_in[1])
            s=0
            theta_best=0
            if min0 < min1:
                s = 1
                theta_best = theta[np.argmin(e_in[0]),0]
                sum_e_in+=min0
            else:
                s = -1
                theta_best = theta[np.argmin(e_in[1]),0]
                sum_e_in+=min1
            e_out=0.5+0.3*s*(np.abs(theta_best)-1)
            sum_e_out+=e_out
        print(sum_e_in/5000,sum_e_out/5000)


if __name__=='__main__':
    decision=decisonStump(1,20,0.2)
    decision.question1718()

Coursera机器学习基石作业二python实现

Coursera机器学习基石作业二python实现

下面的代码是19、20题的结合:

import numpy as np

class decisonStump(object):
    def get_train_dataset(self,path):
        with open(path,'r') as f:
            rawData=f.readlines()
        dimension=len(rawData[0].strip().split(' '))-1
        data_count=len(rawData)
        data_set=np.zeros((data_count,dimension+1))
        for i in range(data_count):
            data_set[i:]=rawData[i].strip().split(' ')
        return data_set,dimension,data_count
    def get_theta(self,dataset):
        data_count=len(dataset)
        l=np.sort(dataset)
        theta=np.zeros((data_count,1))
        for i in range(data_count-1):
            theta[i]=(l[i]+l[i+1])/2
        theta[-1]=1
        return theta
    def question19(self):
        dataset,dimension,data_count=self.get_train_dataset('hw2_train.dat.txt')
        s1=[]
        theta_best1=[]
        E_in=[]
        for i in range(dimension):
            theta=self.get_theta(dataset[:,i])
            e_in = np.zeros((2, data_count))
            for j in range(data_count):
                a=dataset[:,-1]*np.sign(dataset[:,i]-theta[j])
                e_in[0][j] = (data_count - np.sum(a)) / (2 * data_count)  # 数组只有-1和+1,可直接计算出-1所占比例
                e_in[1][j] = (data_count - np.sum(-a)) / (2 * data_count)
            min0,min1=np.min(e_in[0,:]),np.min(e_in[1,:])
            if min0>=min1:
                s1.append(-1)
                theta_best1.append(theta[np.argmin(e_in[1])])
            else:
                s1.append(1)
                theta_best1.append(theta[np.argmin(e_in[0])])
            E_in.append(np.min(np.min(e_in)))
        minS=s1[np.argmin(E_in)]
        minTheta=theta_best1[np.argmin(E_in)]
        print(np.min(E_in))
        return minS,minTheta
    def question20(self):
        s,theta=self.question19()
        dataset, dimension, data_count = self.get_train_dataset('hw2_test.dat.txt')
        E_out=[]
        for i in range(dimension):
            a=dataset[:,-1]*np.sign(dataset[:,i]-theta)*s
            e_out=(data_count-np.sum(a))/(2*data_count)
            E_out.append(e_out)
        print(np.min(E_out))



if __name__=='__main__':
    decision=decisonStump()
    decision.question20()


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