系数比之前多了一个分母m
批量梯度下降法，同上一篇方法，下面看随机梯度法，随机梯度通过一个样本更新所有w，类似笔记一
import pandas as pd
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
df = pd.read_csv('https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data', header=None)
y = df.iloc[0:100, 4].values
y = np.where(y=='Iris-setosa',1,-1)
x = df.iloc[0:100, [0,2]].values
x_std = np.copy(x)
x_std[:, 0] = (x[:,0]-x[:,0].mean())/x[:,0].std()
x_std[:, 1] = (x[:,1]-x[:,1].mean())/x[:,1].std()
class Perceptron():
def __init__(self, eta, X, Y, N):
self.eta = eta
self.X = X
self.Y = Y
self.N = N
self.w = [0]*len(X[0])
self.w0 = 0
self.m = len(X)
self.n = len(X[0])
def _shuffle(self, X, y):
"""Shuffle training data"""
r = np.random.permutation(len(y))
return X[r], y[r]
def output_y(self, x):
z = np.dot(x,self.w)+self.w0
return z
def training(self):
self.errors = []
for times in xrange(self.N):
error = 0
self.X, self.Y = self._shuffle(self.X, self.Y)
for i in xrange(self.m):
delta_y = self.Y[i]-self.output_y(self.X[i])
error += 0.5*delta_y**2
self.w0 += self.eta*delta_y
self.w += self.eta*delta_y*self.X[i]
self.errors.append(error/self.m)
per = Perceptron(0.01, x_std, y, 16)
per.training()
print per.w0,per.w
def f(x, y):
z = per.w0+np.dot(per.w,zip(x,y))
z = np.where(z>0,1,-1)
return z
n = 200
mx = np.linspace(-3, 3, n)
my = np.linspace(-2, 2, n)
# 生成网格数据
X, Y = np.meshgrid(mx, my)
fig, axes = plt.subplots(1,2)
axes0, axes1 = axes.flatten()
axes0.plot(per.errors, marker='o')
axes0.set_title('errors')
axes1.contourf(X, Y, f(X, Y), 2, alpha = 0.75, cmap = plt.cm.RdBu)
axes1.scatter(x_std[:,0][0:50], x_std[:, 1][0:50],s=80,edgecolors='r', marker='o')
axes1.scatter(x_std[:,0][50:100], x_std[:, 1][50:100], marker='x', color='g')
axes1.annotate(r'versicolor',xy=(5.5,4.5),xytext=(4.5,5.5),arrowprops=dict(arrowstyle='->', facecolor='blue'))
axes1.annotate(r'setosa',xy=(5.8,2),xytext=(6.5,3),arrowprops=dict(arrowstyle='->', facecolor='blue'))
fig.set_size_inches(12.5, 7.5)
plt.subplots_adjust(left=0.01, right= 0.9, bottom=0.1, top=0.5)
plt.show()
[5]:http://latex.codecogs.com/gif.latex?\mathbf{J(w)}=\frac{1}{2m}\sum{(y{(i)}-\phi(z{(i)}))}^2
