import numpy as np import scipy.linalg import matplotlib.pyplot as plt #initialize n, m = 200, 500 A = np.mat(np.random.normal(size = (200, 500))) B = np.mat(scipy.linalg.toeplitz([np.random.normal(0, 1) for i in range(m)])) def f1(Lambda): return A * (B - Lambda * np.eye(m)) #Excercise 9.1 AaddA = A + A AAT = A * A.T ATA = A.T * A AB = A * B f1(1) #Exercise 9.2 b = [1 for i in range(m)] scipy.linalg.solve(B, b) #Exercise 9.3 FrobeniusNormOfA = scipy.linalg.norm(A, 'fro') infinityNormOfB = scipy.linalg.norm(B, np.inf) SmallestingularValueOfB = min(scipy.linalg.svdvals(B)) LargestingularValueOfB = max(scipy.linalg.svdvals(B)) #Exercise 9.4 Z = np.random.normal(size = (500, 500)) bk = np.random.normal(size = (500, )) i = 0 while 1: bk1 = np.dot(Z, bk) bk1 /= scipy.linalg.norm(bk1) i += 1 if abs(scipy.linalg.norm(bk, np.inf) - scipy.linalg.norm(bk1, np.inf)) < 10e-6: break bk = bk1 print(i) #Exercise 9.5 #Analyse the relationship between the N and the largest singular value p = 0.5 nValues, LargestingularValues = [], [] for i in range(10): N = (i + 1) * 50 nValues.append(N) C = [[1 if np.random.random() > p else 0 for k in range(N)] for j in range(N)] U, sigma, VT=scipy.linalg.svd(C) LargestingularValues.append(max(sigma)) plt.scatter(nValues, LargestingularValues, s = 100) plt.xlabel("N", fontsize=14) plt.ylabel("largest singular value", fontsize=14) plt.show() #Analyse the relationship between the p and the largest singular value N = 100 pValues, LargestingularValues = [], [] for i in range(10): p = (i + 1) * 0.1 pValues.append(p) C = [[1 if np.random.random() > p else 0 for k in range(N)] for j in range(N)] U, sigma, VT=scipy.linalg.svd(C) LargestingularValues.append(max(sigma)) plt.scatter(pValues, LargestingularValues, s = 100) plt.xlabel("p", fontsize=14) plt.ylabel("largest singular value", fontsize=14) plt.show() #Exercise 9.6 def f2(z, A): return A[np.argmin([abs(A[i] - z) for i in range(len(A))])] f2(0.5, [np.random.normal(0, 1) for i in range(500)])
numpy,scipy学习笔记
猜你喜欢
转载自blog.csdn.net/weixin_38196217/article/details/80386218
今日推荐
周排行