numpy.linalg.solve的右侧超过三个维度 [英] numpy.linalg.solve with right-hand side of more than three dimensions
问题描述
我正在尝试求解一个3x3矩阵a和任意形状(3, ...)的右侧b的方程组.如果b具有一维或二维，则numpy.linalg.solve可以解决问题.不过，它可以分解为更多尺寸:
I'm trying to solve an equation system with a 3x3 matrix a and a right hand side b of arbitrary shape (3, ...). If b has one or two dimensions, numpy.linalg.solve does the trick. It breaks down for more dimensions though:
import numpy
a = numpy.random.rand(3, 3)
b = numpy.random.rand(3)
numpy.linalg.solve(a, b) # okay
b = numpy.random.rand(3, 4)
numpy.linalg.solve(a, b) # okay
b = numpy.random.rand(3, 4, 5)
numpy.linalg.solve(a, b) # ERR
ValueError: solve: Input operand 1 has a mismatch in its core
dimension 0, with gufunc signature (m,m),(m,n)->(m,n) (size 5 is
different from 3)
我希望形状为(3, 4, 5)的输出数组sol的解决方案对应于右侧b[:, i, j]为sol[:, i, j].
I would have expected an output array sol of shape (3, 4, 5) with the solution corresponding to the right-hand side b[:, i, j] is sol[:, i, j].
关于如何最好地解决此问题的任何提示?
Any hint on how to best work around this?
推荐答案
暂时将b整形为(3, 20)，求解线性系统，然后将结果数组整形为b的原始形状(3，4， 5):
Temporarily reshape b to (3, 20), solve the linear system, and then reshape the resultant array to original shape of b (3, 4, 5):
In [34]: a = numpy.random.rand(3, 3)
In [35]: b = numpy.random.rand(3, 4, 5)
In [36]: x = numpy.linalg.solve(a, b.reshape(b.shape[0], -1)).reshape(b.shape)
OR
使用b的第一个轴与第二个轴交换. "nofollow noreferrer"> np.swapaxes ，求解线性系统，然后还原轴:
Swap the first axis of b with the second using np.swapaxes, solve the linear system, and then restore the axes:
In [58]: x = np.swapaxes(np.linalg.solve(a, np.swapaxes(b, 0, 1)), 0, 1)
健全性检查:
Sanity Check:
In [38]: np.einsum('ij,jkl', a, x)
Out[38]:
array([[[ 0.44859955, 0.22967928, 0.74336067, 0.47440575, 0.53798895],
[ 0.80045696, 0.54138958, 0.89870834, 0.56862419, 0.28217437],
[ 0.02093982, 0.78534718, 0.77208236, 0.41568151, 0.95100661],
[ 0.03820421, 0.47067312, 0.71928294, 0.30852615, 0.64454321]],
[[ 0.31757072, 0.30527186, 0.36768759, 0.95869289, 0.86601996],
[ 0.60616508, 0.69927063, 0.53470332, 0.88906606, 0.76066344],
[ 0.95411847, 0.51116677, 0.29338398, 0.04418815, 0.96210206],
[ 0.23449429, 0.64159963, 0.7732404 , 0.4314741 , 0.81279619]],
[[ 0.6399571 , 0.57640652, 0.0186913 , 0.66304489, 0.83372239],
[ 0.28426522, 0.62367363, 0.37163699, 0.78217433, 0.90573787],
[ 0.91066088, 0.06699638, 0.43079394, 0.00263537, 0.399102 ],
[ 0.17711441, 0.48724858, 0.05526752, 0.34251648, 0.94059739]]])
In [39]: b
Out[39]:
array([[[ 0.44859955, 0.22967928, 0.74336067, 0.47440575, 0.53798895],
[ 0.80045696, 0.54138958, 0.89870834, 0.56862419, 0.28217437],
[ 0.02093982, 0.78534718, 0.77208236, 0.41568151, 0.95100661],
[ 0.03820421, 0.47067312, 0.71928294, 0.30852615, 0.64454321]],
[[ 0.31757072, 0.30527186, 0.36768759, 0.95869289, 0.86601996],
[ 0.60616508, 0.69927063, 0.53470332, 0.88906606, 0.76066344],
[ 0.95411847, 0.51116677, 0.29338398, 0.04418815, 0.96210206],
[ 0.23449429, 0.64159963, 0.7732404 , 0.4314741 , 0.81279619]],
[[ 0.6399571 , 0.57640652, 0.0186913 , 0.66304489, 0.83372239],
[ 0.28426522, 0.62367363, 0.37163699, 0.78217433, 0.90573787],
[ 0.91066088, 0.06699638, 0.43079394, 0.00263537, 0.399102 ],
[ 0.17711441, 0.48724858, 0.05526752, 0.34251648, 0.94059739]]])
使用 np.allclose() ，这样您就不必手动检查数字并检查，尤其是对于大型数组:
Use np.allclose() so that you don't have to manually going through the numbers and check, particularly for large arrays:
In [32]: b_ = np.einsum('ij,jkl', a, x)
In [33]: np.allclose(b, b_)
Out[33]: True
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