高级编程技术第九次作业

exercise 9 Numpy

事先声明，这里用到函数前提是安装好numpy和scipy包并在开头引用如下：

from numpy import *
from scipy.linalg import toeplitz
import time

Generate matrices A, with random Gaussian entries, B, a Toeplitz matrix, where A ∈R^n × m and B ∈R^m × m, for n = 200, m = 500.

用随机数构造矩阵可以用numpy下的random.normal(size)函数构造符合标准正态分布的随机数的矩阵，size是要构造的随机列表的大小，如下几个例子

A=mat(random.normal(size=(200,500)))
b0=mat(random.normal(size=500))

上面一个是构造一个2维的列表，行数是200，列数是500，下面是构造一个一维的列表，长度为500.

而构造toeplitz矩阵可以用scipy.linalg下的toeplitz(c,r)函数，这个函数接收两个一维的列表，以前者c为第一行，后者r为第二行构造tooeplitz矩阵，如果两个列表c,r的第一个数不同，默认使用前者c的第一个数。

构造A,B的代码如下：

A=mat(random.normal(size=(200,500)))
b0=mat(random.normal(size=500))
b1=mat(random.normal(size=500))
B=toeplitz(b0,b1)

练习9-1

Matrix operations

Calculate A + A, AA^T,A^TA and AB. Write a function that computes A(B − λI) for any λ.

这里加减法用numpy下的矩阵加减法就可以了，而乘法就需要用numpy的dot（A,B）函数完成，A，B分别代表要相乘的两个矩阵。转置使用A.T即可完成。需要注意做乘法时要满足矩阵乘法的规则。而I单位矩阵可以用eye(size)函数构造，size代表的是单位矩阵的是size*size大的。

代码如下

def fun_e1(A,B,lamda):
    return dot(A,B-lamda*eye(500))
def e1(A,B):
    print("A+A=",A+A)
    print("A*At=",dot(A,A.T))
    print("At*A=",dot(A.T,A))
    lamda=5
    ans = fun_e1(A,B,lamda)
    print("A(B-lamda*I)=",ans)
    return ans

练习9-2

Solving a linear system

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

这题直接用numpy里的linalg的solve(B,b)函数就能解决,代码如下

def e2(B):
    b=mat(random.normal(size=(500,1)))
    print("b:",b)
    x=linalg.solve(B,b)
    print("Bx=b,the x is:",x)

练习9-3

Norms
Compute the Frobenius norm of A and the infinity norm of B. Also find the largest and smallest singular values of B.

这题用numpy里的linalg的norm(A，choose)函数就可以了,A代表要求范数的矩阵，无choose或choose为'fro'时，是计算frobenius范数，choose为numpy.inf时求无穷范数，为2求最大奇异值，-2求最小奇异值。代码如下：

def e3(A,B):
    print("Frobenius norm of A",linalg.norm(A,'fro'))
    print("Infiinity norm of B",linalg.norm(B,inf))
    print("Largest singular of B",linalg.norm(B,2))
    print("Smallest singular of B",linalg.norm(B,-2))

练习9-4

Exercise 9.4: Power iteration
Generate a matrix Z, n × n, with Gaussian entries, and usethe power iteration to find the largest
eigenvalue and corresponding eigenvector of Z. How many iterations are neededtill convergence?
Optional: use the time.clock() method to compare computation time when varying n.

生成矩阵可以用前面用过的random.normal(200,200)的函数，算largest eigenvalue和对应的eigenvector和迭代可以用幂迭代法来计算，所以需要生成全零和全一矩阵，可以分别用numpy里的zeros()和ones()函数来做用相邻两次迭代的向量的范数的差的绝对值作为判停条件，这里认定当它小于0.00005时为收敛，同时这里还要算迭代的次数，用time.clock()算经历了多少时间。代码如下：

def e4():
    times=0
    Z=random.standard_normal((200,200))
    Z0=ones(200)
    Z1=zeros(200)
    Znorm=0
    sta=time.clock()
    while(1):
        Z1=dot(Z,Z0)
        tem=Znorm
        Znorm=linalg.norm(Z1)
        Z0=Z1/Znorm
        times+=1
        if(abs(Znorm-tem)<0.00005):
            break
    sto=time.clock()
    print("Largest eigenvalue",Znorm)
    print("Corresponding eigenvector",Z0)
    print("Iterations",times)
    print("Time needed",sto-sta)

练习9-5

Exercise 9.5: Singular values
Generate an n × n matrix, denoted by C, where each entry is 1 withprobability p and 0 otherwise. Use

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?

C矩阵的生成可以用numpy里的binomial(1,p,(200,200))函数生成最大最小奇异值上面已经说了，这里直接给出代码。

def e5():
    p=0.5
    C=random.binomial(1,p,(200,200))
    print("Largest singular of C",linalg.norm(C,2))
    print("Smallest singular of C",linalg.norm(C,-2))
    print("The relationship between largest singular and n,p,n*p=",200*p,"≈largestsingular=",linalg.norm(C,2))

练习9-6

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 findthis value manually. In particular, use brackets and argmin.

首先生成数组A,也是用到random.normal的函数。然后可以用A[A>z]和A[A<=z]以z为界限分开A为两部分，然后可以用argmax()和argmin()找数组里最大数最小数对应的索引，通过一系列条件判断就可以找出来与z相差最小的部分了。代码：

def e6(z):
    A = random.normal(size=100000)
    AD=A[A>z]
    AX=A[A<=z]
    print(AD,AX)
    lea,lar=0,0
    if(len(AD)>0):
        lar=argmin(AD)
    else:
        aaa= AX[argmax(AX)]
        print("Closest value:", aaa)
        return 
    if(len(AX)>0):
        lea=argmax(AX)
    else:
        aaa= AD[lar]
        print("Closest value:", aaa)
        return 
    if(abs(AD[lar]-z)>abs(AX[lea]-z)):
        aaa= AX[lea]
        print("Closest value:", aaa)
        return 
    else:
        aaa=AD[lar]
        print("Closest value:",aaa)
        return

全部的代码：

from numpy import *
from scipy.linalg import toeplitz
import time





def fun_e1(A,B,lamda):
    return dot(A,B-lamda*eye(500))
def e1(A,B):
    print("A+A=",A+A)
    print("A*At=",dot(A,A.T))
    print("At*A=",dot(A.T,A))
    lamda=5
    ans = fun_e1(A,B,lamda)
    print("A(B-lamda*I)=",ans)
    return ans
def e2(B):
    b=mat(random.normal(size=(500,1)))
    print("b:",b)
    x=linalg.solve(B,b)
    print("Bx=b,the x is:",x)
def e3(A,B):
    print("Frobenius norm of A",linalg.norm(A,'fro'))
    print("Infiinity norm of B",linalg.norm(B,inf))
    print("Largest singular of B",linalg.norm(B,2))
    print("Smallest singular of B",linalg.norm(B,-2))
def e4():
    times=0
    Z=random.standard_normal((200,200))
    Z0=ones(200)
    Z1=zeros(200)
    Znorm=0
    sta=time.clock()
    while(1):
        Z1=dot(Z,Z0)
        tem=Znorm
        Znorm=linalg.norm(Z1)
        Z0=Z1/Znorm
        times+=1
        if(abs(Znorm-tem)<0.00005):
            break
    sto=time.clock()
    print("Largest eigenvalue",Znorm)
    print("Corresponding eigenvector",Z0)
    print("Iterations",times)
    print("Time needed",sto-sta)
def e5():
    p=0.5
    C=random.binomial(1,p,(200,200))
    print("Largest singular of C",linalg.norm(C,2))
    print("Smallest singular of C",linalg.norm(C,-2))
    print("The relationship between largest singular and n,p,n*p=",200*p,"≈largestsingular=",linalg.norm(C,2))
def e6(z):
    A = random.normal(size=100000)
    AD=A[A>z]
    AX=A[A<=z]
    lea,lar=0,0
    if(len(AD)>0):
        lar=argmin(AD)
    else:
        aaa= AX[argmax(AX)]
        print("Closest value:", aaa)
        return
    if(len(AX)>0):
        lea=argmax(AX)
    else:
        aaa= AD[lar]
        print("Closest value:", aaa)
        return
    if(abs(AD[lar]-z)>abs(AX[lea]-z)):
        aaa= AX[lea]
        print("Closest value:", aaa)
        return
    else:
        aaa=AD[lar]
        print("Closest value:",aaa)
        return
A=mat(random.normal(size=(200,500)))
b0=mat(random.normal(size=500))
b1=mat(random.normal(size=500))
B=toeplitz(b0,b1)
e1(A,B)
e2(B)
e3(A,B)
e4()
e5()
z=2
e6(z)

截图：

高级编程技术第九次作业

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