【第十一周】numpy作业

Generate matrices A, with random Gaussian entries, B, a Toeplitz matrix, where A is n*m matrix and B is m*m matrix, for n=200, m=500.

Exercise 9.1:Matrix operation

Calculate A+A, AA',A'A and AB.Write a function that computes A(B-λI) for any λ.

from scipy.linalg import toeplitz
import numpy as np


###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))

def ex_one(A,B,c):
             print("Matrix A:\n",A)
             print("Starting to compute.......")
             print("Matrix A+A:\n",A+A)
             print("Matrix A.At:\n",[e-mail])
             print("Matrix At.A:\n",[e-mail])
             print("Matrix A.B:\n",[e-mail])
             
             print("Matrix A.(B-c*I):\n",[e-mail](B-c*np.eye(50)))
A = np.random.randn(20,50)
B = toeplitz(list(range(1,51)))
ex_one(A,B,2)

运行截图（部分）：

Exercise 9.2:Solving a linear system

Generate a vector b with m entries and solve Bx=b.

from scipy.linalg import toeplitz
import numpy as np

###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))


def ex_two(B,b):
    print("Bx=b solving for x....")
    print(np.linalg.solve(B,b))
    
A = np.random.randn(20,50)
B = toeplitz(list(range(1,51)))
b = np.array(list(range(1,51)))
ex_two(B,b)

运行截图：

Exercise 9.3:Norms 

Compute the Frobenius norm of A:||A||F and the infinity norm of B: ||B||∞.Also find the largest and smallest singular values of B.

from scipy.linalg import toeplitz
import numpy as np

###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))


def ex_three(A,B):
    print("The Frobenius norm of A:",np.linalg.norm(A))
    print("The infinity norm of B:",np.linalg.norm(B,np.inf))
    u,s,vh=np.linalg.svd(B)
    print("Max singular value:",max(s))
    print("Min singular value:",min(s))
    
A = np.random.randn(20,50)
B = toeplitz(list(range(1,51)))
b = np.array(list(range(1,51)))
ex_three(A,B)

运行截图：

Exercise 9.4: Power iteration

Generate a matrix Z, n × n, with Gaussian entries, and use the power iteration to find the largest eigenvalue and corresponding eigenvector of Z. How many iterations are needed till convergence? 

Optional: use the time.clock() method to compare computation time when varying n. 

from scipy.linalg import toeplitz
import numpy as np
import random
import time

###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))


def ex_four(A):
    begin = time.clock()
    esp = 0.000001
    count = 0
    
    b_k = np.random.rand(A.shape[1])
    a1=8
    a2=1
    
    while( abs(a2-a1)>esp ):
       b_k1 = np.dot(A,b_k)
       a1 = np.linalg.norm(b_k)
       
       b_k = b_k1/a1
       a2 = np.linalg.norm(b_k)
       
       count +=1

    end = time.clock()
    print(A)
    print("Iteration times:",count)
    print("time:",end-begin)
    
A = np.random.randn(20,20)
ex_four(A)

运行截图：

Exercise 9.5: Singular values
Generate an n × n matrix, denoted by C, where each entry is 1 with probability p and 0 otherwise. Use
the linear algebra library of Scipy to compute the singular values of  the linear algebra library of Scipy to compute the singular values of C. What can you say about the relationship between n, p and the largest singular value? 

from scipy.linalg import toeplitz
import numpy as np
import random

###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))


def ex_five(n,p):
    x = []
    for i in range(0,n):
        x.append([])
        for j in range(0,n):
           r=random.random()
           if(p<r):
             x[i].append(1)
           else:
             x[i].append(0)
    C = np.array(x)
    print("C is:\n",C)
    u,s,vh=np.linalg.svd(C)
    print("Max singular value:",max(s))


n=input("n=")
p=input("p=")
ex_five(int(n),float(p))

运行截图：

由不同的n、p输入，发现当n变大，p变小时max sigular value变大。

Exercise 9.6: Nearest neighbor
Write a function that takes a value z and an array A and finds the element in A that is closest to z. The function should return the closest value, not index.
Hint: Use the built-in functionality of Numpy rather than writing code to find this value manually. In particular, use brackets and argmin. 

from scipy.linalg import toeplitz
import numpy as np

###简化200*500 为20*50
###A = np.random.randn(200,500)
###B = toeplitz(list(range(1,501))


def ex_six(A,z):
    sub = [abs(A[i]-z) for i in range(0,A.size)]
    s=np.argmin(sub)
    return A[s]
    
A = np.random.randn(20,50)
B = toeplitz(list(range(1,51)))
b = np.array(list(range(1,51)))
C = np.array([1,2,3,4,7,8,9])
print("closest value:",ex_six(C,5))

运行截图：