numpy vs mldivide,"\" Matlab运算符 [英] Numpy vs mldivide,"\" matlab operator
问题描述
A \ B
提供了一种特殊的解决方案,而numpy.linalg.lstsq却没有.
A \ B
in matlab gives a special solution while numpy.linalg.lstsq doesn't.
A = [1 2 0; 0 4 3];
b = [8; 18];
c_mldivide = A \ b
c_mldivide =
0
4
0.66666666666667
c_lstsq = np.linalg.lstsq([[1 ,2, 0],[0, 4, 3]],[[8],[18]])
print c_lstsq
c_lstsq = (array([[ 0.91803279],
[ 3.54098361],
[ 1.27868852]]), array([], dtype=float64), 2, array([ 5.27316304,1.48113184]))
- matlab中的mldivide
A \ B
如何提供特殊的解决方案? - 此解决方案对实现计算精度有用吗?
- 为什么这个解决方案很特别?您如何在numpy中实现呢?
- How does mldivide
A \ B
in matlab give a special solution? - Is this solution usefull in achieving computational accuracy?
- Why is this solution special and how might you implement it in numpy?
推荐答案
对于像您这样的不确定系统(等级小于变量数),mldivide
返回具有尽可能多的零值.哪个变量将被设置为零取决于其任意选择.
For under-determined systems such as yours (rank is less than the number of variables), mldivide
returns a solution with as many zero values as possible. Which of the variables will be set to zero is up to its arbitrary choice.
相反,在这种情况下,lstsq
方法返回最小范数的解:也就是说,在无穷大的精确解中,它将选择平方和最小的那个.的变量.
In contrast, the lstsq
method returns the solution of minimal norm in such cases: that is, among the infinite family of exact solutions it will pick the one that has the smallest sum of squares of the variables.
因此,Matlab的特殊"解决方案在某种程度上是任意的:在此问题中,可以将三个变量中的任何一个设置为零.实际上,NumPy提供的解决方案更为特殊:存在一个独特的最小范数解决方案
So, the "special" solution of Matlab is somewhat arbitrary: one can set any of the three variables to zero in this problem. The solution given by NumPy is in fact more special: there is a unique minimal-norm solution
哪种解决方案更适合您的目的取决于您的目的.解的非唯一性通常是重新考虑您的方程式方法的原因.但是,正如您所问的那样,这里是产生Matlab类型解决方案的NumPy代码.
Which solution is better for your purpose depends on what your purpose is. The non-uniqueness of solution is usually a reason to rethink your approach to the equations. But since you asked, here is NumPy code that produces Matlab-type solutions.
import numpy as np
from itertools import combinations
A = np.matrix([[1 ,2, 0],[0, 4, 3]])
b = np.matrix([[8],[18]])
num_vars = A.shape[1]
rank = np.linalg.matrix_rank(A)
if rank == num_vars:
sol = np.linalg.lstsq(A, b)[0] # not under-determined
else:
for nz in combinations(range(num_vars), rank): # the variables not set to zero
try:
sol = np.zeros((num_vars, 1))
sol[nz, :] = np.asarray(np.linalg.solve(A[:, nz], b))
print(sol)
except np.linalg.LinAlgError:
pass # picked bad variables, can't solve
对于您的示例,它输出三个特殊"解决方案,最后一个是Matlab选择的解决方案.
For your example it outputs three "special" solutions, the last of which is what Matlab chooses.
[[-1. ]
[ 4.5]
[ 0. ]]
[[ 8.]
[ 0.]
[ 6.]]
[[ 0. ]
[ 4. ]
[ 0.66666667]]
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