python实现感知器算法(批处理)

天池像两个鸡蛋融合在一起,上面大雾迷蒙,更添加了几分神秘的色彩。而且天池在长白山的最高点,所以海拔很高,就像耸入云端似的,都是大雾,所以被称为天池。旁边还有火山,火山的顶上灰蒙蒙的,全都被火山灰覆盖了。那里还有很多火山石,有的像狮子,有的像大象,有的像绵羊……数也数不清。

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。

标签: 感知器 python