In [6]:

    

import numpy as np
from scipy import linalg
import matplotlib.pyplot as plt
%matplotlib inline
from time import process_time


    

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def example(m, n):
    A = np.vander(np.linspace(-1., 1., m), n)
    b = A.dot(np.ones(n))
    return A, b


    

In [8]:

    

def solve_lsq(A, b, method='qr'):
    
    if method == 'qr':
        Q, R = linalg.qr(A,mode='economic')
        k = R.shape[0]
        y = np.dot(Q.transpose(),b)
        x = linalg.solve_triangular(R,y[:k])
        return x
    
    elif method == 'normal':
        try:
            ATA = np.dot(A.transpose(),A)
            ATb = np.dot(A.transpose(),b)
            L = linalg.cholesky(ATA,lower=True)
            y = linalg.solve_triangular(L,ATb,lower=True)
            x = linalg.solve_triangular(L.transpose(),y)
        except LinAlgError:
            A = 10.0*A
            ATA = np.dot(A.tranpose(),A)
            ATb = np.dot(A,b)
            L = linalg.cholesky(ATA,lower=True)
            y = linalg.solve_triangular(L,ATb,lower=True)
            x = linalg.solve_triangular(L.transpose(),y)
        return x
    
    elif method == 'svd':
        U, S, VT = linalg.svd(A)
        S = 1/S
        S = np.diag(S)
        Sigma = np.zeros(A.transpose().shape)
        Sigma[:S.shape[0],:S.shape[1]] = S
        x = np.dot(U.transpose(),b)
        x = np.dot(Sigma,x)
        x = np.dot(VT.transpose(),x)
        return x
    else:
        print("doof")


    

In [13]:

    

m = 100
times = np.zeros(m)
for n in range(1, m):
    A, b = example(m, n)
    t_start = process_time()
    x = solve_lsq(A, b, method='svd')
    times[n] = process_time() - t_start


    

In [14]:

    

plt.plot(times)


    

    
Out[14]:





[<matplotlib.lines.Line2D at 0x7f7eed1961d0>]






    

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