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问题描述
scipy/numpy/...
的宇宙中是否存在用于矩阵高斯消除的标准方法?
有人通过Google找到了许多代码片段,但我希望尽可能使用受信任的"模块.
我终于发现,可以使用 LU分解完成此操作.这里的 U 矩阵表示线性系统的简化形式.from numpy import array
from scipy.linalg import lu
a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])
pl, u = lu(a, permute_l=True)
然后u
读取
array([[ 2., 4., 4., 4.],
[ 0., 2., 1., 2.],
[ 0., 0., 1., 1.],
[ 0., 0., 0., 0.]])
取决于系统的可溶性,该基质具有上部三角形或梯形结构.在上述情况下,由于矩阵仅具有等级3
,所以会出现零线.
Is there somewhere in the cosmos of scipy/numpy/...
a standard method for Gauss-elimination of a matrix?
One finds many snippets via google, but I would prefer to use "trusted" modules if possible.
解决方案
I finally found, that it can be done using LU decomposition. Here the U matrix represents the reduced form of the linear system.
from numpy import array
from scipy.linalg import lu
a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])
pl, u = lu(a, permute_l=True)
Then u
reads
array([[ 2., 4., 4., 4.],
[ 0., 2., 1., 2.],
[ 0., 0., 1., 1.],
[ 0., 0., 0., 0.]])
Depending on the solvability of the system this matrix has an upper triangular or trapezoidal structure. In the above case a line of zeros arises, as the matrix has only rank 3
.
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