加速随机矩阵计算

Question

我正在创建随机Toeplitz矩阵来估计它们可逆的概率.我目前的代码是

import random
from scipy.linalg import toeplitz
import numpy as np
for n in xrange(1,25):
    rankzero = 0
    for repeats in xrange(50000):
        column = [random.choice([0,1]) for x in xrange(n)]
        row = [column[0]]+[random.choice([0,1]) for x in xrange(n-1)]
        matrix = toeplitz(column, row)
        if  (np.linalg.matrix_rank(matrix) < n):
            rankzero += 1
    print n, (rankzero*1.0)/50000

这可以加快吗？

我想增加值50000以获得更高的准确度,但目前这样做太慢了.

仅使用for n in xrange(10,14)节目进行性能分析

  400000    9.482    0.000    9.482    0.000 {numpy.linalg.lapack_lite.dgesdd}
  4400000    7.591    0.000   11.089    0.000 random.py:272(choice)
   200000    6.836    0.000   10.903    0.000 index_tricks.py:144(__getitem__)
        1    5.473    5.473   62.668   62.668 toeplitz.py:3(<module>)
   800065    4.333    0.000    4.333    0.000 {numpy.core.multiarray.array}
   200000    3.513    0.000   19.949    0.000 special_matrices.py:128(toeplitz)
   200000    3.484    0.000   20.250    0.000 linalg.py:1194(svd)
6401273/6401237    2.421    0.000    2.421    0.000 {len}
   200000    2.252    0.000   26.047    0.000 linalg.py:1417(matrix_rank)
  4400000    1.863    0.000    1.863    0.000 {method 'random' of '_random.Random' objects}
  2201015    1.240    0.000    1.240    0.000 {isinstance}
[...]

Answer 1

一种方法是通过缓存放置值的索引来节省重复调用 toeplitz() 函数的一些工作。以下代码比原始代码快约 30%。其余的性能是在排名计算中......而且我不知道对于具有0和1的托普利茨矩阵是否存在更快的排名计算。

（更新）如果用 scipy.linalg.det() == 0 替换matrix_rank，代码实际上会快4倍（行列式比小矩阵的排名计算更快）

import random
from scipy.linalg import toeplitz, det
import numpy as np,numpy.random

class si:
    #cache of info for toeplitz matrix construction
    indx = None
    l = None

def xtoeplitz(c,r):
    vals = np.concatenate((r[-1:0:-1], c))
    if si.indx is None or si.l != len(c):
        a, b = np.ogrid[0:len(c), len(r) - 1:-1:-1]
        si.indx = a + b
        si.l = len(c)
    # `indx` is a 2D array of indices into the 1D array `vals`, arranged so
    # that `vals[indx]` is the Toeplitz matrix.
    return vals[si.indx]

def doit():
    for n in xrange(1,25):
        rankzero = 0
        si.indx=None

        for repeats in xrange(5000):

            column = np.random.randint(0,2,n)
            #column=[random.choice([0,1]) for x in xrange(n)] # original code

            row = np.r_[column[0], np.random.randint(0,2,n-1)]
            #row=[column[0]]+[random.choice([0,1]) for x in xrange(n-1)] #origi

            matrix = xtoeplitz(column, row)
            #matrix=toeplitz(column,row) # original code

            #if  (np.linalg.matrix_rank(matrix) < n): # original code
            if  np.abs(det(matrix))<1e-4: # should be faster for small matrices
                rankzero += 1
        print n, (rankzero*1.0)/50000