(Week 11)Python-Numpy_exercises

import numpy as np
from scipy.linalg import toeplitz
A = np.random.normal(0,1,size=(200,500))
B = toeplitz(np.random.random((500)),np.random.random((500)))

[[-1.08139799  1.63626865 -1.19375553 ...  0.54016655  1.04225555
  -0.13933849]
 [-0.8328577   0.06145153 -0.41176367 ... -0.6699105   0.82843077
  -1.96984931]
 [ 0.84787184 -0.91498802  0.0050716  ... -1.22836145 -0.20244569
  -1.39372778]
 ...
 [-1.47544982 -0.63704547 -0.94399656 ...  0.9433685  -0.45693745
  -2.27505334]
 [-1.85360841  1.50529954 -0.09037166 ...  1.50275772  0.89649159
  -1.42276638]
 [-1.71975293  0.03681571  1.29264051 ... -0.45049806 -0.99695241
   0.30931893]]
[[0.77901348 0.53469863 0.58073141 ... 0.49282411 0.26355095 0.7060844 ]
 [0.49807365 0.77901348 0.53469863 ... 0.73799689 0.49282411 0.26355095]
 [0.58240378 0.49807365 0.77901348 ... 0.46331682 0.73799689 0.49282411]
 ...
 [0.37861052 0.91589719 0.19010318 ... 0.77901348 0.53469863 0.58073141]
 [0.78174467 0.37861052 0.91589719 ... 0.49807365 0.77901348 0.53469863]
 [0.30388698 0.78174467 0.37861052 ... 0.58240378 0.49807365 0.77901348]]

编写代码如下，分别打印出所求结果，定义函数compute计算值。

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

def compute(A, B, lamda):
    return A.dot(B - lamda * np.eye(5))

创建向量b并求解。

b = np.random.random(5)
x = np.linalg.solve(B,b)
#x = np.linalg.inv(B).dot(b)
print(x)

调用函数分别计算A的Frobenius范数（各元素平方和开方）和B的无穷范数（行绝对值和最大值），利用svd分解得到B的所有奇异值，分别打印最值。

normFro = np.linalg.norm(A, 'fro')
normInf = np.linalg.norm(B, np.inf)
u, s, vh = np.linalg.svd(B)
print(max(s), min(s))

import numpy as np

A = np.random.normal(size=(5,5))
v = np.ones(5)
epi = 1e-5
delta = 1
k = 0
while 1:
    u = v / max(v) #得到规范化向量
    p = v
    v = A.dot(u)
    delta = abs(max(v) - max(p))#比较是否达到精度
    k += 1
    if delta < epi: break
print(k)
print(max(v))       #最大分量为主特征值
print(v / max(v))   #规范化向量即为主特征值对应的特征向量

对于5*5矩阵，迭代次数大多为100以内，少数收敛速度慢或不收敛则需要100000次以上。

========================== RESTART: D:/python/c.py ==========================
20
2.4852607075818263
[ 1.          0.15200074  0.3943235   0.3377595  -0.07979286]
>>> 
========================== RESTART: D:/python/c.py ==========================
203360
1.3469029196104048
[ 0.99642544 -0.1414458   1.         -0.5099528  -0.78919265]
>>> 
========================== RESTART: D:/python/c.py ==========================
73
2.816940007693461
[ 1.          0.34605648 -0.04584567 -0.42430719  0.33601158]
>>> 
========================== RESTART: D:/python/c.py ==========================
47
1.9621261489192796
[ 0.15254321  0.99335278  1.         -3.51655808  0.12706556]
>>>

import numpy as np
from scipy.linalg import svd
import random
def get_c(n, p):
    C = np.zeros((n,n))
    for i in range(n):
        for j in range(n):
            C[i][j] = 1 if random.random() < p else 0
    return C
for i in range(11):
    C = get_c(10, 0.1 * i)
    u, s, vh = svd(C)
    print('p = ', 0.1 * i, 'max singular value is ', max(s))

由以下输出可看出，p越大，最大奇异值越大。

p =  0.0 max singular value is  0.0
p =  0.1 max singular value is  2.243169319983191
p =  0.2 max singular value is  2.668745545242216
p =  0.30000000000000004 max singular value is  3.31429371905275
p =  0.4 max singular value is  5.294757119546733
p =  0.5 max singular value is  6.653159594060823
p =  0.6000000000000001 max singular value is  7.115519446727156
p =  0.7000000000000001 max singular value is  7.426687969549355
p =  0.8 max singular value is  8.626913866418564
p =  0.9 max singular value is  9.180638085275236
p =  1.0 max singular value is  10.000000000000002
>>>

思路：先将A的每个元素减去z，取绝对值，利用argmin取得下标即可。

def get_closest(A, z):
    col = np.shape(A)[1]
    pos = np.argmin(abs(A - z))
    return round(A[pos // col][pos % col], 8)

A = np.random.randint(0, 10, size = (5,4))
print(A)
print(get_closest(A,5.6))

[[5 7 7 9]
 [6 5 1 5]
 [4 6 8 5]
 [6 3 1 5]
 [4 6 5 2]]
6
>>>

(Week 11)Python-Numpy_exercises

猜你喜欢