Python中(大)1554 x 1554矩阵的行列式 [英] Determinant of a (large) 1554 x 1554 matrix in Python

问题描述

我需要在python中以单精度计算大的1554、1554值矩阵的行列式.这样做时，我会遇到运行时警告:

I need to calculate the determinant of a large 1554,1554 matrix of values with single precision in python. In doing so I encounter a runtime warning:

import numpy as np

from numpy import linalg as LA

a = np.random.random((1554, 1554))

b = np.random.random((1554, 1554))

c = np.dot(a,b)

det = LA.det(c)

RuntimeWarning:在det中遇到溢出 r = _umath_linalg.det(a，signature = signature)

RuntimeWarning: overflow encountered in det r = _umath_linalg.det(a, signature=signature)

关于如何解决此问题的任何想法?非常感谢！

Any ideas on how I can work around this problem? Many thanks!

这个问题的独特之处在于，它专门涉及以双精度计算大型矩阵的行列式，尽管此处包括一个可能的答案:

this question is unique in that it specifically refers to computing the determinant of large matrix in double precision, though a possible answer is included here: Can I get the matrix determinant using Numpy?

推荐答案

您可以使用以下关系: https://wikimedia.org/api/rest_v1/media/math/render/svg/f6404a766d86e9d78a5c4f82e05de37469a5f8e9

You can use this relation: https://wikimedia.org/api/rest_v1/media/math/render/svg/f6404a766d86e9d78a5c4f82e05de37469a5f8e9

来自 https://en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant

因此，将矩阵除以平均值，然后计算行列式以避免溢出.以后，您可以将均值乘以n(一轴的长度)的幂次

So divide your matrix by the mean and then compute the determinant to avoid overflow. Later you can multiply with the mean to the power of n (length of one axis)

我不确定平均值是否是理想选择.这更多是一个数学问题

edit: I'm not sure if the mean is the ideal choice though. This is more a math question

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