第十三周Scipy

import numpy as np
from scipy.optimize import leastsq

m = 20
n = 12

A = np.random.randint(1, 100, size = (m , n))
b = np.random.randint(1, 100, size = (m, 1))
x0 = np.random.randint(1, 100, size = (1, n))
A = np.mat(A)
b = np.mat(b)

def func(x):
    x = x.reshape(n , 1)
    x = np.mat(x)
    
    r = A * x - b
    r = np.array(r)
    
    return np.array(r.T).reshape(m)

l = leastsq(func, x0)

x = l[0].reshape(n, 1)
print('x = ', x)
print('residal = ', A * np.mat(x) - b)
print('norm = ', np.linalg.norm(A * np.mat(x) - b))

运行

x =  [[ 0.80420686]
 [ 0.62469073]
 [-0.19798721]
 [-0.24153484]
 [-0.09265393]
 [ 0.76398557]
 [ 0.22034202]
 [-0.17669824]
 [ 0.04719834]
 [-0.06067864]
 [-0.64575209]
 [ 0.19693909]]
residal =  [[ -3.33044561]
 [  9.29616319]
 [ 11.38514723]
 [-22.06835924]
 [  5.98869136]
 [ 11.43222131]
 [-20.964829  ]
 [  9.52452761]
 [ 10.66786616]
 [-15.4983632 ]
 [-13.98776675]
 [ 13.83945544]
 [-30.41646446]
 [ 15.19362674]
 [  6.14056642]
 [  3.24390932]
 [-13.7308669 ]
 [ 21.8667791 ]
 [-19.56388707]
 [  8.93166629]]
norm =  66.97451994397635


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(program exited with code: 0)

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求函数f(x)的最大值

import numpy as np
import math
import scipy.optimize as sc

f = lambda x : -math.pow( math.sin(x-2), 2) * math.pow(math.e , -x*x)

b = sc.minimize(f, 0)

print('在 x = ' , b.x[0], ' 处有最大值 ', -b.fun)

在函数前加上一个负号后求的最小值的相反数就是原函数的最大值。

运行结果

在 x =  0.21624131913960304  处有最大值  0.9116854118471545

import numpy as np
from scipy.spatial.distance import pdist

n = 5
m = 4
X = []
for i in range(1, m + 1):
    x =  [i for y in range(n)]
    X.append(x)
    
X = np.array(X)
print(X)
d = pdist(X, metric = 'euclidean')
count = 0
for i in range(1, m):
    for j in range(i + 1, m+1):
        print(X[i-1],' and ',X[j-1], ' distance = ', d[count])
        count += 1

运行结果

[[1 1 1 1 1]
 [2 2 2 2 2]
 [3 3 3 3 3]
 [4 4 4 4 4]]
[1 1 1 1 1]  and  [2 2 2 2 2]  distance =  2.23606797749979
[1 1 1 1 1]  and  [3 3 3 3 3]  distance =  4.47213595499958
[1 1 1 1 1]  and  [4 4 4 4 4]  distance =  6.708203932499369
[2 2 2 2 2]  and  [3 3 3 3 3]  distance =  2.23606797749979
[2 2 2 2 2]  and  [4 4 4 4 4]  distance =  4.47213595499958
[3 3 3 3 3]  and  [4 4 4 4 4]  distance =  2.23606797749979


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(program exited with code: 0)

请按任意键继续. . .

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