高階程式設計技術_課後作業（十四）
阿新 • • 發佈：2019-01-31
Numpy
下面為作業文件中Numpy這一章的內容
題目
解答
Exercise 9.1高斯隨機分佈矩陣的生成：
numpy.random.normal(loc=0.0, scale=1.0, size=None) ，其中loc為均值，scale為方差，size為輸出形狀乘法：numpy中*是矩陣元素逐個計算，np.dot(a,b)是按照矩陣乘法的運演算法則來運算toeplitz--生成託普利茲矩陣T=toeplitz(c,r)生成非對稱託普利茲矩陣，將c作為第一列，r作為第一行，若c(1)與r(1)不相等，則使用c(1)作為矩陣的第一個元素，同時列印一條警告資訊。
程式碼：
import numpy as np from scipy.linalg import toeplitz n = 200 m = 500 A = np.random.normal(size = (n,m)) print("A+A\n") print(A+A) print("AA'\n") print(np.dot(A, A.T)) print("A'A\n") print(np.dot(A.T, A)) c = np.random.normal(size = (1,500)) r = np.random.normal(size = (1,500)) B = toeplitz(c,r) print("AB\n") print(np.dot(A,B)) def A_B_I(r): A = np.random.normal(size = (n,m)) c = np.random.normal(size = (1,m)) r = np.random.normal(size = (1,m)) B = toeplitz(c,r) I = np.identity(m) return (np.dot(A, (B - r*I))) print("A(B-rI)\n") print(A_B_I(5))
結果截圖：
Exercise 9.2
程式碼：
import numpy as np
from scipy.linalg import toeplitz
m = 500
c = np.random.normal(size = (1,m))
r = np.random.normal(size = (1,m))
B = toeplitz(c,r)
b = np.random.permutation(m)
ans = np.linalg.solve(B, b)
print("Ans:\n")
print(ans)結果截圖：
Exercise 9.3
程式碼：
import numpy as np from scipy.linalg import toeplitz n = 200 m = 500 A = np.random.normal(size = (n,m)) c = np.random.normal(size = (1,500)) r = np.random.normal(size = (1,500)) B = toeplitz(c,r) frobenius = np.linalg.norm(A, "fro") infinity = np.linalg.norm(B, np.inf) e = np.linalg.eigvals(B) print("the Frobenius norm of A: ") print(frobenius) print("\nthe infinity norm of B") print(infinity) print("\nthe largest singular values of B: ") print(max(e)) print("\nthe smallest singular values of B: ") print(min(e))
結果截圖：
Exercise 9.4
程式碼：
import time import numpy as np from scipy.linalg import toeplitz for n in range(2, 6): start_CPU = time.clock() i = 1 eig = 0 Z = np.random.normal(1,1,size = (n,n)) u = np.random.randn(n) print("##Test of " + str(n)) while True: v = np.dot(Z, u) last = eig #eig = np.linalg.norm(v, np.inf) eig = v[np.argmax(np.abs(v))] u = v / eig if (i != 1) & (abs(last - eig) < (1/2)*(10**(-5))): break i = i + 1 print("迭代次數：") print(i) end_CPU = time.clock() print("\nThe time of " + str(n) + " is " + str(end_CPU - start_CPU)) print("\n") print("The lagest eigenvalue of Z") print(eig) print("\n") print("Its corresponding eigenvector") print(u) print("\n")
結果截圖：
Exercise 9.5
程式碼：
import time
import numpy as np
from scipy.linalg import svdvals
n = 4
p = .5
C = np.random.binomial(1, p, size = (n, n))
print(C)
lsv = svdvals(C)
print(lsv.max())結果截圖：
Maybe the largest singular value = n * p
Exercise 9.6
程式碼：
import numpy as np
def find_Closest_Element(A, z):
return A[np.argmin(np.abs(A-z))]
n = 5
A = np.random.normal(size = (n,1))
z = np.random.randint(1)
print(z)
ans = find_Closest_Element(A, z)
#ans = find_Closest_Element(ans, z)
print(A)
print(ans)
結果截圖：
