Question

Matlab使用稀疏命令计算对角矩阵的倒数需要0.02秒。

P = diag(1:10000);
P = sparse(P);
tic;
A = inv(P);
toc

然而，对于Python代码，它需要永远 - 几分钟。

import numpy as np
import time

startTime = time.time()
P = np.diag(range(1,10000))
A = np.linalg.inv(P)
runningTime = (time.time()-startTime)/60
print "The script was running for %f minutes" % runningTime

我尝试使用Scipy.sparse模块，但它没有帮助。运行时间下降，但只有40秒。

import numpy as np
import time
import scipy.sparse as sps
import scipy.sparse.linalg as spsl

startTime = time.time()
P = np.diag(range(1,10000))
P_sps = sps.coo_matrix(P)
A = spsl.inv(P_sps)
runningTime = (time.time()-startTime)/60
print "The script was running for %f minutes" % runningTime

是否可以像在Matlab中一样快地运行代码？

Answer 1

这是答案。当您在matlab中为稀疏矩阵运行inv时，matlab会检查矩阵的不同属性以优化计算。对于稀疏对角矩阵，您可以运行以下代码来查看matlab正在做什么

n = 10000;
a = diag(1:n);
a = sparse(a);
I = speye(n,n);
spparms('spumoni',1);
ainv = inv(a);
spparms('spumoni',0);

Matlab将打印以下内容：

sp\: bandwidth = 0+1+0.
sp\: is A diagonal? yes.
sp\: do a diagonal solve.

所以matlab只反转对角线。

Scipy如何反转矩阵？我们在这里有code：

...
from scipy.sparse.linalg import spsolve
...

def inv(A):
    """
    Some comments...
    """
    I = speye(A.shape[0], A.shape[1], dtype=A.dtype, format=A.format)
    Ainv = spsolve(A, I)
    return Ainv

和spsolve

    # Cover the case where b is also a matrix
    Afactsolve = factorized(A)
    tempj = empty(M, dtype=int)
    x = A.__class__(b.shape)
    for j in range(b.shape[1]):
        xj = Afactsolve(squeeze(b[:, j].toarray()))
        w = where(xj != 0.0)[0]
        tempj.fill(j)
        x = x + A.__class__((xj[w], (w, tempj[:len(w)])),
                            shape=b.shape, dtype=A.dtype)

，即scipy factorize A然后求解一组线性系统，其中右侧是坐标向量（形成单位矩阵）。对矩阵中的所有解进行排序，我们得到初始矩阵的逆矩阵。

如果matlab被利用矩阵的对角线结构，但scipy不是（当然scipy也使用矩阵的结构，但效率较低，至少对于例子而言），matlab应该是更快。

修改可以肯定的是，正如@ P.Escondido提出的那样，我们将尝试对矩阵A进行微小的修改，以便在矩阵不是对角线时跟踪matlab过程：

n = 10000; a = diag(1:n); a = sparse(a); ainv = sparse(n,n); spparms('spumoni',1); a(100,10) = 500; a(10,1000) = 200; ainv = inv(a); spparms('spumoni',0);

打印出以下内容：

sp\: bandwidth = 90+1+990. sp\: is A diagonal? no. sp\: is band density (0.00) > bandden (0.50) to try banded solver? no. sp\: is A triangular? no. sp\: is A morally triangular? yes. sp\: permute and solve. sp\: sprealloc in sptsolve: 10000 10000 10000 15001

Answer 2

splu()如何，它更快但需要一个密集阵列并返回密集阵列：

创建一个随机矩阵：

import numpy as np
import time
import scipy.sparse as sps
import scipy.sparse.linalg as spsl
from numpy.random import randint
N = 1000
i = np.arange(N)
j = np.arange(N)
v = np.ones(N)

i2 = randint(0, N, N)
j2 = randint(0, N, N)
v2 = np.random.rand(N)

i = np.concatenate((i, i2))
j = np.concatenate((j, j2))
v = np.concatenate((v, v2))

A = sps.coo_matrix((v, (i, j)))
A = A.tocsc()

%time B = spsl.inv(A)

按splu()计算逆矩阵：

%%time
lu = spsl.splu(A)
eye = np.eye(N)
B2 = lu.solve(eye)

检查结果：

np.allclose(B.todense(), B2.T)

这是％时间输出：

inv: 2.39 s
splv: 193 ms

Answer 3

您正在使用软件中的关键信息：矩阵是对角线这一事实使得它非常容易反转：您只需反转其对角线的每个元素：

P = np.diag(range(1,10000))
A = np.diag(1.0/np.arange(1,10000))

当然，这仅适用于对角矩阵......

Answer 4

如果您尝试使用，结果会更好：

import numpy as np
import time
import scipy.sparse as sps
import scipy.sparse.linalg as spsl

P = np.diag(range(1,10000))
P_sps = sps.coo_matrix(P)
startTime = time.time()
A = spsl.inv(P_sps)
runningTime = (time.time()-startTime)/60
print "The script was running for %f minutes" % runningTime

现在您可以与您的matlab脚本进行比较。

是否有可能像在Matlab中一样快地计算Python中的稀疏矩阵的逆？

4 个答案: