十一周numpy

生成高斯分布矩阵和托普利兹矩阵可以使用python的numpy库和scipy库的函数实现。

首先是生成两个矩阵

import numpy as np
from scipy import linalg

# 生成高斯分布矩阵
A = np.random.normal(loc= 100, scale= 100, size = (200,500))

x = [i for i in range(1,501)]
x.reverse()
y = [i for i in range(500, 1000)]

# 生成托普利兹矩阵
B = linalg.toeplitz(x, y)

print("A = ")
print(A)
print("B = ")
print(B)

生成的结果如图

exercise 1

import numpy as np
from scipy import linalg

A = np.random.normal(loc= 100, scale= 100, size = (200,500))
x = [i for i in range(1,501)]
x.reverse()
y = [i for i in range(500, 1000)]
B = linalg.toeplitz(x, y)

A = np.mat(A)
B = np.mat(B)

print("A = ")
print(A)
print("B = ")
print(B)

print("A + A = ")
print(A + A)
print("A *  A's transpose = ")
print(A * np.transpose(A))
print("AB = ")
print(A * B)

x = input("Please input a constant x ")   
x = int(x)
I = np.mat(np.eye(500, 500), dtype = int)  #生成单位矩阵
print("A(B - xI) = ")
print(A * (B - x * I))

先生成A、B矩阵，然后使用numpy里面定义的矩阵运算方法计算出A+A，A乘A的转置，AB，A（B-xI）

运行结果：

Exercise 2

Bx = b， B^(-1) Bx= B^(-1) b ，x= B^(-1) b

只要求出B的逆和b相乘就可以算出x

import numpy as np
from scipy import linalg

A = np.random.normal(loc= 100, scale= 100, size = (200,500))
x = [i for i in range(1,501)]
y = [i for i in range(1,501)]  #之前的B矩阵没有逆，所以换了一个
B = linalg.toeplitz(x, y)

A = np.mat(A)
B = np.mat(B)

print("B = ")
print(B)

b = np.array(range(500))
print('b = ')
print(b)
b.shape = (500,1)  # 转换成列向量
print("x = ")
print(B.I * b)

结果如下

x 有500 行，这里就不上全图了

Exercise 3

import numpy as np
from scipy import linalg

A = np.random.normal(loc= 100, scale= 100, size = (200,500))
x = [i for i in range(1,501)]
y = [i for i in range(1,501)]
B = linalg.toeplitz(x, y)

A = np.mat(A)
B = np.mat(B)

A_F = np.linalg.norm(A, 'fro')  
print("the Frobenius norm:", A_F)  

B_F = np.linalg.norm(B, np.inf)  
print("the infinity norm:", B_F)  
          
#find the largest and smallest singular values of B  
large = np.linalg.norm(B, 2)  
small = np.linalg.norm(B, -2)  
print("the largest singular:", large)  
print("the smallest singular:", small)

运行结果：

the Frobenius norm: 44756.26340030189
the infinity norm: 125250.0
the largest singular: 87334.52045641867
the smallest singular: 0.5000049348345748

Exercise 4

import time
Z = np.random.normal(loc = 100, scale = 50,size = (200, 200))   
num = 0  
u_k = np.ones(200)  
v_k_norm = 0  
v_k = np.zeros(200)  
      
start = time.clock()  
while(True):  
    
    v_k = np.dot(Z, u_k)  
     
    v_k_norm_temp = v_k_norm  
    v_k_norm = np.linalg.norm(v_k)  
     
    u_k = v_k / v_k_norm  
    num += 1  
    if(abs(v_k_norm_temp - v_k_norm) < 0.0005):  
        break;  
end = time.clock()  
      
print("The largest eigenvalue:", v_k_norm)  
print("The corresponding eigenvector:", u_k)  
#How many iterations are needed till convergence  
print("iterations number : ", num)  
#Optional: use the time.clock() method to compare computation time when varying n.  
print("computation time :", end - start)

运行情况

Exercise 5

p = 0.6 
C = np.random. binomial(1, p, (200, 200))  
      
#Compute the singular values of C  
large = np.linalg.norm(C, 2)  
small = np.linalg.norm(C, -2)
  
print("The smallest singular:", small)  
print("The largest singular:", large)  
      
# the relationship between n, p and the largest singular value 
print("n * p:", 200*p)
if abs(200*p - large) < 1:
    print("the largest singular is nearly equal to n * p")

运行结果

Exercise 6

import numpy as np

def fun_closest(A, z):  
    
    B, C = A[A>z], A[A<=z]  
    top, bot = 0, 0  

    if(len(B)):  
        top = np.argmin(B)  
    else:  
        return C[np.argmax(C)]  
    if(len(C)):  
        bot = np.argmax(C)  
    else:  
        return B[top]  
          
    if(abs(B[top]-z) < abs(C[bot]-z)):  
        return B[top]  
    else:  
        return C[bot] 

A = np.random.normal(loc= 100, scale= 100, size = (200,500))

print("the closest value to 1000 is :", fun_closest(A, 1000))
print("the closest value to 100 is :", fun_closest(A, 100))
print("the closest value to 1 is :", fun_closest(A, 1))
print("the closest value to -1000 is :", fun_closest(A, -1000))

测试情况;

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