高级编程技术（Python）作业13

Exercises for numpy


Solution：

from scipy.linalg import toeplitz
import numpy as np


def compute(A, B, lamda):
    temp = B - lamda * np.eye(500)
    return A @ temp


A = np.random.standard_normal((200, 500))
B = toeplitz(list(range(0, 500)))
ApA = A + A
AmAt = A @ A.T
AtmA = A.T @ A
AmB = A @ B
print("A =")
print(A)
print("\nB =")
print(B)
print("\nA + A =")
print(ApA)
print("\nAAt =")
print(AmAt)
print("\nAtA =")
print(AtmA)
print("\nAB =")
print(AmB)
print("\nA(B-Iλ) = ")
lamda = np.random.randint(1, 254)
print(compute(A, B, lamda))

Output：

A =
[[ 0.63527689 -0.46965411 -0.08144386 ...  0.99068757  0.24196114
   2.46171393]
 [ 0.16796722 -0.60395727  1.35046286 ...  1.17290856 -0.40837032
   0.89156049]
 [ 2.26451677  1.39258995  1.21977885 ...  0.08485108 -1.01412036
  -0.34401827]
 ...
 [-1.44848543 -0.53995438  1.19615986 ... -0.12222356  1.70176326
  -0.0234794 ]
 [-0.05456266  1.17431131  0.5153051  ...  0.15164733  0.5956557
  -2.02930834]
 [-0.53588379 -0.30198579 -0.25434934 ... -1.04304831  0.5700688
  -0.20671051]]

B =
[[  0   1   2 ... 497 498 499]
 [  1   0   1 ... 496 497 498]
 [  2   1   0 ... 495 496 497]
 ...
 [497 496 495 ...   0   1   2]
 [498 497 496 ...   1   0   1]
 [499 498 497 ...   2   1   0]]

A + A =
[[ 1.27055378 -0.93930821 -0.16288772 ...  1.98137514  0.48392227
   4.92342786]
 [ 0.33593445 -1.20791454  2.70092572 ...  2.34581713 -0.81674065
   1.78312097]
 [ 4.52903355  2.78517989  2.43955771 ...  0.16970215 -2.02824072
  -0.68803653]
 ...
 [-2.89697086 -1.07990876  2.39231972 ... -0.24444712  3.40352652
  -0.04695881]
 [-0.10912533  2.34862261  1.03061021 ...  0.30329466  1.1913114
  -4.05861668]
 [-1.07176758 -0.60397159 -0.50869867 ... -2.08609662  1.1401376
  -0.41342101]]

AAt =
[[519.54293441  14.3904712   -2.9505389  ... -21.90000768   2.02306948
   -3.41645362]
 [ 14.3904712  440.50130118 -33.85362696 ...  22.19478433 -23.30267606
  -11.96390157]
 [ -2.9505389  -33.85362696 493.58048886 ... -13.15129537   2.14798488
   54.67429124]
 ...
 [-21.90000768  22.19478433 -13.15129537 ... 516.97485101 -21.01454409
   -4.02739196]
 [  2.02306948 -23.30267606   2.14798488 ... -21.01454409 454.00573306
   16.10363589]
 [ -3.41645362 -11.96390157  54.67429124 ...  -4.02739196  16.10363589
  542.97716628]]

AtA =
[[184.71382733   0.6795867   16.36261    ... -26.47664379 -25.59999917
   -1.96203064]
 [  0.6795867  203.74206579   4.55367036 ... -11.39796967 -23.80483802
  -18.47184855]
 [ 16.36261      4.55367036 203.0411682  ...  20.43336526  11.52491627
   -3.95755536]
 ...
 [-26.47664379 -11.39796967  20.43336526 ... 176.76151718  12.5619039
  -13.31237111]
 [-25.59999917 -23.80483802  11.52491627 ...  12.5619039  231.9201409
   11.53468903]
 [ -1.96203064 -18.47184855  -3.95755536 ... -13.31237111  11.53468903
  262.08339928]]

AB =
[[ -7184.3574239   -7162.43392549  -7141.44973529 ...  -3069.82528224
   -3095.88557701  -3121.46194951]
 [ -6291.33132701  -6284.02075838  -6277.91810429 ...   2827.68763872
    2819.74662421   2810.98886905]
 [ -3179.20304615  -3142.99021659  -3103.99220715 ... -12571.04788375
  -12600.01540251 -12631.01116198]
 ...
 [ -5702.54085177  -5676.44978707  -5651.43863112 ...  -8701.20321744
   -8733.54782072  -8762.48889748]
 [-10470.5670167  -10440.26969865 -10407.62375799 ...  -4648.36126303
   -4675.90040113  -4702.24822783]
 [ -3061.39377938  -3074.01560292  -3087.24139805 ...   8802.08488601
    8812.90822537   8824.87170234]]

A(B-Iλ) = 
[[ -7279.64895714  -7091.98580961  -7129.23315648 ...  -3218.42841751
   -3132.17974739  -3490.7190393 ]
 [ -6316.52641052  -6193.42716775  -6480.4875335  ...   2651.751354
    2881.00217284   2677.25479597]
 [ -3518.88056222  -3351.87870853  -3286.95903519 ... -12583.77554526
  -12447.89734833 -12579.40842218]
 ...
 [ -5485.26803707  -5595.4566301   -5830.86260986 ...  -8682.86968343
   -8988.81230938  -8758.96698679]
 [-10462.38261708 -10616.41639449 -10484.91952362 ...  -4671.1083626
   -4765.24875629  -4397.85197701]
 [ -2981.0112107   -3028.71773383  -3049.08899754 ...   8958.54213287
    8727.39790535   8855.87827824]]

Solution：

b = np.array(list(range(0, 500)))
x = np.linalg.solve(B, b)

print("\n Solve Bx = b, and get x = \n", x)

Output：

 Solve Bx = b, and get x = 
 [ 1.00000000e+00 -1.20281567e-27  5.55111512e-17 -1.11022302e-16
 -2.22044605e-16  6.66133815e-16 -4.44089210e-16 -4.92308369e-27
  4.44089210e-16 -1.33226763e-15  1.77635684e-15 -8.88178420e-16
 ...
 -2.37289360e-27  2.97873878e-27 -2.32240650e-27 -1.08547261e-27
  2.27191941e-27 -1.91850972e-27  1.79229198e-27 -5.42736303e-28
 -1.51461294e-28 -1.88064440e-27 -2.84217094e-14  2.84217094e-14]

Solution：

Fro_norm_for_A = np.linalg.norm(A, 'fro')  
Inf_norm_for_B = np.linalg.norm(B, np.inf)  

u, s, vh = np.linalg.svd(B)  
largest_value = max(s)  
smallest_value = min(s)

print("\nlargest singular value is ", largest_value)
print("smallest singular value is ", smallest_value)

Output：

largest singular value is  86851.66898467168
smallest singular value is  0.50000493483471

Solution：

import time
import numpy as np


def power_iteration(Z, n):  
    start = time.clock()
    b_k = np.random.standard_normal(n)
    old_norm = 1000
    new_norm = 1000
    cnt = 0
    while(True):
        cnt += 1
        old_norm = new_norm
        b_k1 = b_k @ Z
        b_k1_norm = np.linalg.norm(b_k1, np.inf)
        b_k = b_k1 / b_k1_norm
        new_norm = b_k1_norm
        if(abs(old_norm-new_norm) < 0.0001):
            break
    end = time.clock()
    T = end - start
    print("the largest eigenvalue:", b_k1_norm)  
    print("the corresponding eigenvector:", b_k)
    print("the number of iterations:", cnt)
    print("computing time for n = %d is %.3fs" % (n, T))


n = 100
Z = np.random.standard_normal((n, n))
power_iteration(Z, n)

Outout：
n = 100:

the largest eigenvalue: 11.068088233359687
the corresponding eigenvector: [-0.2944245  -0.22318278  0.32930188  0.16652893 -0.18671432  0.1033079
 -0.43079565 -0.06757289 -0.23260499 -0.26774119 -0.02095388  0.01306529
  0.07476483  0.72441983 -0.051148   -0.05055061 -0.10833943 -0.13528726
 -0.12779153  0.24640859 -0.174878   -0.63527586  0.58981573  0.64681457
 -0.29897181 -0.48018419  0.05681275  1.         -0.00627435 -0.36424601
  0.0143298   0.10627736 -0.45038955 -0.19567988  0.21500549 -0.20055526
  0.10586666 -0.19353277  0.46092431  0.06506723  0.5968188   0.15199873
  0.7420006   0.46032591 -0.02483576 -0.2949899  -0.18934441 -0.80883885
 -0.24025505 -0.23677737 -0.70489411 -0.12813257 -0.33248621 -0.39345534
  0.40717082 -0.77745637 -0.55848406  0.16120735 -0.13881272  0.0977156
  0.0308597   0.53181834 -0.02230041  0.76887701 -0.20394658  0.11921737
 -0.10243752 -0.33448665  0.0029017  -0.10277001  0.01786354  0.20901145
 -0.5368839  -0.72974602 -0.20956522  0.32944598 -0.08409827 -0.19542284
  0.86377975 -0.34299062 -0.52208403 -0.14712398 -0.17957801  0.19866612
  0.03280761 -0.30087161 -0.12790014  0.03411339  0.54760192  0.55866826
  0.27828528  0.00197266 -0.11427112  0.84978421 -0.28409862 -0.3365838
  0.41414519 -0.22929057  0.22212446 -0.56226912]
the number of iterations: 1338
computing time for n = 100 is 0.095s

n = 1000

the largest eigenvalue: 31.451726620620093
the corresponding eigenvector: [-3.67723062e-01  2.29867064e-01  2.51742840e-01 -1.79455868e-02
  5.04008912e-01  4.12574268e-01  1.95029166e-01  1.15594965e-01
  3.74950571e-01 -7.71544262e-03  2.28021228e-01 -8.53313186e-02
  2.22506787e-01 -7.48276906e-01  5.83083702e-02 -1.81471910e-01
 ...
 -1.65448767e-01 -7.66347760e-01  3.25175204e-02 -3.44718183e-01
 -4.24661983e-01  3.13930755e-01  2.49665989e-01 -3.24210108e-01
  1.01066408e-01 -3.38146873e-01  4.17035416e-01 -2.08158123e-01]
the number of iterations: 14054
computing time for n = 1000 is 7.798s

注释：事实上，当n变大的时候也时间和迭代次数不一定会变大，这个迭代次数取决于计算的矩阵特性，和一开始b_k矩阵的特性，迭代次数较大可能意味着这个矩阵不收敛，因为并不是所有的正态分布的矩阵都是收敛的。

Solution：

import numpy as np


for n in [10, 20, 30]:  
    for p in [0.25, 0.50, 0.75]:  
        print("n = %d, p = %.2f" % (n,p))  
        C = np.random.binomial(1, p, (n, n))  
        u, s, vh = np.linalg.svd(C)
        print("the singular values of C are: ")
        for value in s:
            print("%.2f" % value, end=" ")
        maxvalue = max(s)  
        print("\nthe larset singular value is: ", maxvalue)
        print("n * p: ", n*p, end="\n\n")

Output：

n = 10, p = 0.25
the singular values of C are: 
3.34 2.08 1.94 1.34 1.32 0.75 0.67 0.37 0.22 0.00 
the larset singular value is:  3.3381035284614593
n * p:  2.5

n = 10, p = 0.50
the singular values of C are: 
5.46 2.73 2.55 1.67 1.35 1.21 0.90 0.52 0.30 0.00 
the larset singular value is:  5.45755368360496
n * p:  5.0

n = 10, p = 0.75
the singular values of C are: 
7.60 2.43 2.01 1.54 1.09 1.00 0.72 0.42 0.15 0.00 
the larset singular value is:  7.598226322714773
n * p:  7.5

n = 20, p = 0.25
the singular values of C are: 
6.25 3.66 3.14 2.79 2.56 2.50 2.31 2.13 1.87 1.74 1.59 1.39 1.26 1.09 0.89 0.61 0.39 0.26 0.19 0.03 
the larset singular value is:  6.2486372435890285
n * p:  5.0

n = 20, p = 0.50
the singular values of C are: 
10.26 4.06 3.61 3.35 3.17 2.93 2.56 2.30 2.24 2.07 1.69 1.36 1.26 1.01 0.93 0.74 0.61 0.26 0.23 0.06 
the larset singular value is:  10.259488286339908
n * p:  10.0

n = 20, p = 0.75
the singular values of C are: 
15.46 3.21 3.09 3.00 2.75 2.38 2.16 1.97 1.88 1.75 1.60 1.53 1.24 1.04 0.90 0.54 0.33 0.29 0.22 0.10 
the larset singular value is:  15.457854613602697
n * p:  15.0

n = 30, p = 0.25
the singular values of C are: 
8.17 4.55 4.23 3.97 3.64 3.35 3.24 2.88 2.79 2.60 2.55 2.46 2.26 2.05 1.96 1.80 1.54 1.53 1.23 1.17 1.10 0.97 0.90 0.73 0.66 0.43 0.36 0.34 0.15 0.10 
the larset singular value is:  8.174664215684738
n * p:  7.5

n = 30, p = 0.50
the singular values of C are: 
15.33 5.17 4.77 4.62 4.12 3.91 3.89 3.54 3.21 3.13 2.91 2.88 2.63 2.48 2.33 2.25 2.11 1.81 1.69 1.56 1.37 1.26 1.02 0.93 0.78 0.69 0.56 0.32 0.22 0.11 
the larset singular value is:  15.32571688028713
n * p:  15.0

n = 30, p = 0.75
the singular values of C are: 
22.14 4.47 4.33 3.95 3.91 3.52 3.35 3.31 3.04 2.85 2.73 2.57 2.50 2.01 1.93 1.82 1.69 1.52 1.40 1.36 1.25 1.05 1.00 0.91 0.85 0.61 0.45 0.29 0.20 0.01 
the larset singular value is:  22.13899094271687
n * p:  22.5

注释：当n 和 p逐渐增大的时候，最大的特征值也开始逐渐接近n*p

Solution：

import numpy as np


A = np.random.standard_normal((200, 500))
print("A =")
print(A)
z = input("\nPlease give a value to z: ")
z = float(z)
print("z = ", z)
print("the nearest neighbor in A is ", A.flatten()[np.argmin(np.abs(A-z))])

Output：

A =
[[ 0.16978732  0.90922799  1.57803628 ... -0.94090241 -0.90116503
   1.18149994]
 [-0.94495112  0.49088629  0.39733231 ... -0.39165062  0.67323871
  -2.73955901]
 [-0.68621472  0.67000602 -2.29200515 ...  0.00739803 -0.09910325
  -1.6808361 ]
 ...
 [ 0.63333291  1.00799629  0.23079636 ... -0.24659729 -0.35720556
   0.04880485]
 [-0.56683618  0.25756957 -1.29729843 ...  1.14639432 -1.60384547
   1.36067139]
 [-1.39205996 -0.4861746  -0.21603671 ...  1.95942107 -0.98680696
  -0.11168355]]

Please give a value to z: 5
z =  5.0
the nearest neighbor in A is  4.299794939351856

注释：我对”use brackets and argmin”的理解是使用一行带很多括号以及使用了argmin的代码解决这个问题。A.flatten()是将这个矩阵变为一维矩阵，得到index之后方便返回所对应的值。