如何在numpy.linalg.solve中使用稀疏矩阵 [英] How to use a sparse matrix in numpy.linalg.solve

问题描述

我想为x

Ax = b

其中A是稀疏的，而b是正则列矩阵.但是，当我插入常规的np.linalg.solve(A,b)例程时，它给了我一个错误.但是，当我执行np.linalg.solve(A.todense(),b)时，效果很好.

Where A is sparse and b is just regular column matrix. However when I plug into the usual np.linalg.solve(A,b) routine it gives me an error. However when I do np.linalg.solve(A.todense(),b) it works fine.

我如何使用此线性求解仍保留A的sparseness?原因是A关于150 x 150很大，并且大约有50个这样的矩阵，因此我希望它尽可能长地保持稀疏.

How can I use this linear solve still preserving the sparseness of A?. The reason is A is quite large about 150 x 150 and there are about 50 such matrices and so keeping it sparse for as long as possible is the way I'd prefer it.

我希望我的问题有道理.我应该如何实现这一目标?

I hope my question makes sense. How should I go about achieving this?

推荐答案

np.linalg.solve 仅适用于类似数组的对象.例如，它可以在np.ndarray或np.matrix上工作(来自numpy文档的示例):

np.linalg.solve only works for array-like objects. For example it would work on a np.ndarray or np.matrix (Example from the numpy documentation):

import numpy as np

a = np.array([[3,1], [1,2]])
b = np.array([9,8])
x = np.linalg.solve(a, b)

或

import numpy as np

a = np.matrix([[3,1], [1,2]])
b = np.array([9,8])
x = np.linalg.solve(a, b)

或在A.todense()上，其中A=scipy.sparse.csr_matrix(np.matrix([[3,1], [1,2]]))会返回np.matrix对象.

or on A.todense() where A=scipy.sparse.csr_matrix(np.matrix([[3,1], [1,2]])) as this returns a np.matrix object.

要使用稀疏矩阵，您必须使用

To work with a sparse matrix, you have to use scipy.sparse.linalg.spsolve (as already pointed out by rakesh)

import numpy as np
import scipy.sparse
import scipy.sparse.linalg

a = scipy.sparse.csr_matrix(np.matrix([[3,1], [1,2]]))
b = np.array([9,8])
x = scipy.sparse.linalg.spsolve(a, b)

请注意，x仍然是np.ndarray，而不是稀疏矩阵.仅当您求解Ax = b时才返回稀疏矩阵，其中b是矩阵而不是向量.

Note that x is still a np.ndarray and not a sparse matrix. A sparse matrix will only be returned if you solve Ax=b, with b being a matrix and not a vector.

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