Posted in Python onJanuary 18, 2019
本文实例为大家分享了Python感知器算法实现的具体代码,供大家参考,具体内容如下
先创建感知器类:用于二分类
# -*- coding: utf-8 -*- import numpy as np class Perceptron(object): """ 感知器:用于二分类 参照改写 https://blog.csdn.net/simple_the_best/article/details/54619495 属性: w0:偏差 w:权向量 learning_rate:学习率 threshold:准则阈值 """ def __init__(self,learning_rate=0.01,threshold=0.001): self.learning_rate=learning_rate self.threshold=threshold def train(self,x,y): """训练 参数: x:样本,维度为n*m(样本有m个特征,x输入就是m维),样本数量为n y:类标,维度为n*1,取值1和-1(正样本和负样本) 返回: self:object """ self.w0=0.0 self.w=np.full(x.shape[1],0.0) k=0 while(True): k+=1 dJw0=0.0 dJw=np.zeros(x.shape[1]) err=0.0 for i in range(0,x.shape[0]): if not (y[i]==1 or y[i]==-1): print("类标只能为1或-1!请核对!") break update=self.learning_rate*0.5*(y[i]-self.predict(x[i])) dJw0+=update dJw+=update*x[i] err+=np.abs(0.5*(y[i]-self.predict(x[i]))) self.w0 += dJw0 self.w += dJw if np.abs(np.sum(self.learning_rate*dJw))<self.threshold or k>500: print("迭代次数:",k," 错分样本数:",err) break return self def predict(self,x): """预测类别 参数: x:样本,1*m维,1个样本,m维特征 返回: yhat:预测的类标号,1或者-1,1代表正样本,-1代表负样本 """ if np.matmul(self.w,x.T)+self.w0>0: yhat=1 else: yhat=-1 return yhat def predict_value(self,x): """预测值 参数: x:样本,1*m维,1个样本,m维特征 返回: y:预测值 """ y=np.matmul(self.w,x.T)+self.w0 return y
然后为Iris数据集创建一个Iris类,用于产生5折验证所需要的数据,并且能产生不同样本数量的数据集。
# -*- coding: utf-8 -*- """ Author:CommissarMa 2018年5月23日 16点52分 """ import numpy as np import scipy.io as sio class Iris(object): """Iris数据集 参数: data:根据size裁剪出来的iris数据集 size:每种类型的样本数量 way:one against the rest || one against one 注意: 此处规定5折交叉验证(5-cv),所以每种类型样本的数量要是5的倍数 多分类方式:one against the rest """ def __init__(self,size=50,way="one against the rest"): """ size:每种类型的样本数量 """ data=sio.loadmat("C:\\Users\\CommissarMa\\Desktop\\模式识别\\课件ppt\\PR实验内容\\iris_data.mat") iris_data=data['iris_data']#iris_data:原数据集,shape:150*4,1-50个样本为第一类,51-100个样本为第二类,101-150个样本为第三类 self.size=size self.way=way self.data=np.zeros((size*3,4)) for r in range(0,size*3): self.data[r]=iris_data[int(r/size)*50+r%size] def generate_train_data(self,index_fold,index_class,neg_class=None): """ index_fold:5折验证的第几折,范围:0,1,2,3,4 index_class:第几类作为正类,类别号:负类样本为-1,正类样本为1 """ if self.way=="one against the rest": fold_size=int(self.size/5)#将每类样本分成5份 train_data=np.zeros((fold_size*4*3,4)) label_data=np.full((fold_size*4*3),-1) for r in range(0,fold_size*4*3): n_class=int(r/(fold_size*4))#第几类 n_fold=int((r%(fold_size*4))/fold_size)#第几折 n=(r%(fold_size*4))%fold_size#第几个 if n_fold<index_fold: train_data[r]=self.data[n_class*self.size+n_fold*fold_size+n] else: train_data[r]=self.data[n_class*self.size+(n_fold+1)*fold_size+n] label_data[fold_size*4*index_class:fold_size*4*(index_class+1)]=1 elif self.way=="one against one": if neg_class==None: print("one against one模式下需要提供负类的序号!") return else: fold_size=int(self.size/5)#将每类样本分成5份 train_data=np.zeros((fold_size*4*2,4)) label_data=np.full((fold_size*4*2),-1) for r in range(0,fold_size*4*2): n_class=int(r/(fold_size*4))#第几类 n_fold=int((r%(fold_size*4))/fold_size)#第几折 n=(r%(fold_size*4))%fold_size#第几个 if n_class==0:#放正类样本 if n_fold<index_fold: train_data[r]=self.data[index_class*self.size+n_fold*fold_size+n] else: train_data[r]=self.data[index_class*self.size+(n_fold+1)*fold_size+n] if n_class==1:#放负类样本 if n_fold<index_fold: train_data[r]=self.data[neg_class*self.size+n_fold*fold_size+n] else: train_data[r]=self.data[neg_class*self.size+(n_fold+1)*fold_size+n] label_data[0:fold_size*4]=1 else: print("多分类方式错误!只能为one against one 或 one against the rest!") return return train_data,label_data def generate_test_data(self,index_fold): """生成测试数据 index_fold:5折验证的第几折,范围:0,1,2,3,4 返回值: test_data:对应于第index_fold折的测试数据 label_data:类别号为0,1,2 """ fold_size=int(self.size/5)#将每类样本分成5份 test_data=np.zeros((fold_size*3,4)) label_data=np.zeros(fold_size*3) for r in range(0,fold_size*3): test_data[r]=self.data[int(int(r/fold_size)*self.size)+int(index_fold*fold_size)+r%fold_size] label_data[0:fold_size]=0 label_data[fold_size:fold_size*2]=1 label_data[fold_size*2:fold_size*3]=2 return test_data,label_data
然后我们进行训练测试,先使用one against the rest策略:
# -*- coding: utf-8 -*- from perceptron import Perceptron from iris_data import Iris import numpy as np if __name__=="__main__": iris=Iris(size=50,way="one against the rest") correct_all=0 for n_fold in range(0,5): p=[Perceptron(),Perceptron(),Perceptron()] for c in range(0,3): x,y=iris.generate_train_data(index_fold=n_fold,index_class=c) p[c].train(x,y) #训练完毕,开始测试 correct=0 x_test,y_test=iris.generate_test_data(index_fold=n_fold) num=len(x_test) for i in range(0,num): maxvalue=max(p[0].predict_value(x_test[i]),p[1].predict_value(x_test[i]), p[2].predict_value(x_test[i])) if maxvalue==p[int(y_test[i])].predict_value(x_test[i]): correct+=1 print("错分数量:",num-correct,"错误率:",(num-correct)/num) correct_all+=correct print("平均错误率:",(num*5-correct_all)/(num*5))
然后使用one against one 策略去训练测试:
# -*- coding: utf-8 -*- from perceptron import Perceptron from iris_data import Iris import numpy as np if __name__=="__main__": iris=Iris(size=10,way="one against one") correct_all=0 for n_fold in range(0,5): #训练 p01=Perceptron()#0类和1类比较的判别器 p02=Perceptron() p12=Perceptron() x,y=iris.generate_train_data(index_fold=n_fold,index_class=0,neg_class=1) p01.train(x,y) x,y=iris.generate_train_data(index_fold=n_fold,index_class=0,neg_class=2) p02.train(x,y) x,y=iris.generate_train_data(index_fold=n_fold,index_class=1,neg_class=2) p12.train(x,y) #测试 correct=0 x_test,y_test=iris.generate_test_data(index_fold=n_fold) num=len(x_test) for i in range(0,num): vote0=0 vote1=0 vote2=0 if p01.predict_value(x_test[i])>0: vote0+=1 else: vote1+=1 if p02.predict_value(x_test[i])>0: vote0+=1 else: vote2+=1 if p12.predict_value(x_test[i])>0: vote1+=1 else: vote2+=1 if vote0==max(vote0,vote1,vote2) and int(vote0)==int(y_test[i]): correct+=1 elif vote1==max(vote0,vote1,vote2) and int(vote1)==int(y_test[i]): correct+=1 elif vote2==max(vote0,vote1,vote2) and int(vote2)==int(y_test[i]): correct+=1 print("错分数量:",num-correct,"错误率:",(num-correct)/num) correct_all+=correct print("平均错误率:",(num*5-correct_all)/(num*5))
实验结果如图所示:
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持三水点靠木。
python实现感知器算法(批处理)
- Author -
CommissarMa声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@