[Python知识库]2021-07-13

第一天打卡

1.画图

import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax3 = plt.axes(projection = '3d')
x1 = np.linspace(-100, 100, 256)
x2 = np.linspace(-100, 100, 256)
X1,X2 = np.meshgrid(x1,x2)
Z = (a - X1)**2 + b * (X2- X1*X1)**2
a = 0.1
b = 10
ax3.plot_surface(X1,X2,Z,cmap='rainbow')
plt.show()

a = 0.1 ,b = 10的图像

a=0.1,b=-10

a=0,b=0

a的正负号不会改变图像的大致形状，图像的开口方向主要受b的正负号影响

2.求全局最小值，牛顿法

import numpy as np
import matplotlib.pyplot as plt

def jacobian(x):
    return np.array([-400*x[0]*(x[1]-x[0]**2)-2*(1-x[0]),200*(x[1]-x[0]**2)])

def hessian(x):
    return np.array([[-400*(x[1]-3*x[0]**2)+2,-400*x[0]],[-400*x[0],200]])

X1=np.arange(-1.5,1.5+0.05,0.05)
X2=np.arange(-3.5,2+0.05,0.05)
[x1,x2]=np.meshgrid(X1,X2)
f=100*(x2-x1**2)**2+(1-x1)**2; # 给定的函数
plt.contour(x1,x2,f,20) # 画出函数的20条轮廓线


def newton(x0):

    print('初始点为:')
    print(x0,'\n')
    W=np.zeros((2,10**3))
    i = 1
    imax = 1000
    W[:,0] = x0 
    x = x0
    delta = 1
    alpha = 1

    while i<imax and delta>10**(-5):
        p = -np.dot(np.linalg.inv(hessian(x)),jacobian(x))
        x0 = x
        x = x + alpha*p
        W[:,i] = x
        delta = sum((x-x0)**2)
        print('第',i,'次迭代结果:')
        print(x,'\n')
        i=i+1
    W=W[:,0:i]  # 记录迭代点
    return W

x0 = np.array([-1.2,1])
W=newton(x0)

plt.plot(W[0,:],W[1,:],'g*',W[0,:],W[1,:]) # 画出迭代点收敛的轨迹
plt.show()