NumPy作业题

最新推荐文章于 2021-09-02 12:49:15 发布

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from scipy import linalg
import numpy as np

A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,500)))

print(A + A)
print(np.dot(A, A.T))
print(np.dot(A.T, A))
print(np.dot(A, B))

def calculate(x):
    return np.dot(A, B - x * np.eye(500))

from scipy import linalg
import numpy as np

A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,501)))

b = np.linspace(1, 500, 500)
print(np.linalg.solve(B, b))

from scipy import linalg
import numpy as np

A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,501)))

#求范数和奇异值
print(linalg.norm(A, 'fro'))
print(linalg.norm(B, np.inf))
x = linalg.svdvals(B).tolist()
print(max(x))
print(min(x))


from scipy import linalg
import numpy as np
import time 
A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,501)))

def Times(n):
    Z = np.random.normal(0, 1, n*n).reshape(n, n)#Z矩阵
    v = np.array(list(range(1, n+1))[::-1]) #初始特征向量
    a = linalg.norm(v, np.inf)#特征值
    temp = 0 
    times = 0 #次数
    t = time.clock() #时间
    while True:
        temp = a
        y = np.dot(Z, v)
        a = linalg.norm(y, np.inf)
        v = y / a
        times = times + 1
        if abs(temp - a) < 0.000001:
            break
    t = time.clock() - t
    return times, t, v, a

n = input()
result = Times(int(n))
print("Times:%d"%result[0])
print("Time:%f"%result[1])
print("Eigenvector:",end="")
print(result[2])
print("Eigenvalue:%f"%result[3])

from scipy import linalg
import numpy as np
import time 
A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,501)))

n = input()
p = input()
C = np.random.binomial(1, float(p), [int(n), int(n)])
x = linalg.svdvals(C).tolist()
print(max(x))

from scipy import linalg
import numpy as np
import time 
A = np.random.normal(0,1,100000).reshape(200,500)
B = linalg.toeplitz(list(range(1,501)))

def closest(A, z):
	a = min(A[A > z])#A矩阵中大于z的值的最小值
	b = max(A[A <= z])#A矩阵中小于等于z的值的最小值
	if abs(a - z) > abs(b - z):
		return b
	else:
		return a

z = input()
print(closest(A, float(z)))