如何使用numpy的np.linalg.solve求解超定最小二乘系统？

Question

我有矩阵a和矩阵b

A = [[536.08008658   1.        ]
     [580.40909091   1.        ]
     [624.73809524   1.        ]
     [671.83766234   1.        ]
     [718.93722944   1.        ]
     [760.495671     1.        ]
     [810.36580087   1.        ]
     [857.46536797   1.        ]]

b = [[210.82467532]
    [188.66017316]
    [172.03679654]
    [166.495671  ]
    [169.26623377]
    [177.57792208]
    [196.97186147]
    [224.67748918]]

如何找到矩阵x，它是一个2x1矩阵

我尝试了lsq = np.linalg.solve(A, b)，但出现了以下错误：

---------------------------------------------------------------------------
LinAlgError                               Traceback (most recent call last)
<ipython-input-18-e0f5641243e6> in <module>
     17 
     18 # a, b = np.linalg.lstsq(A, y, rcond=None)[0]
---> 19 lsq = np.linalg.solve(A, y)
     20 print(lsq)
     21 

<__array_function__ internals> in solve(*args, **kwargs)

~/opt/anaconda3/lib/python3.8/site-packages/numpy/linalg/linalg.py in solve(a, b)
    379     a, _ = _makearray(a)
    380     _assert_stacked_2d(a)
--> 381     _assert_stacked_square(a)
    382     b, wrap = _makearray(b)
    383     t, result_t = _commonType(a, b)

~/opt/anaconda3/lib/python3.8/site-packages/numpy/linalg/linalg.py in _assert_stacked_square(*arrays)
    202         m, n = a.shape[-2:]
    203         if m != n:
--> 204             raise LinAlgError('Last 2 dimensions of the array must be square')
    205 
    206 def _assert_finite(*arrays):

LinAlgError: Last 2 dimensions of the array must be square

任何关于如何用这种特定方法解决这一问题的想法都会有所帮助。谢谢

Answer 1

您为该作业调用了错误的函数。这就是求解一个正常的线性系统，它需要与变量数量相同的方程（即，它必须是平方的）。你要找的是^{}，它返回线性矩阵方程的最小二乘解

import numpy as np

A = [[536.08008658,  1.        ],
     [580.40909091,  1.        ],
     [624.73809524,  1.        ],
     [671.83766234,  1.        ],
     [718.93722944,  1.        ],
     [760.495671  ,  1.        ],
     [810.36580087,  1.        ],
     [857.46536797,  1.        ]]

b = [[210.82467532],
     [188.66017316],
     [172.03679654],
     [166.495671  ],
     [169.26623377],
     [177.57792208],
     [196.97186147],
     [224.67748918]]


lsq = np.linalg.lstsq(A, b, rcond=None)
print(lsq)

Answer 2

如果您必须使用np.linalg.solve，那么您可以按如下操作。但是，使用专门的最小二乘法程序会更好（更快、更准确）

（对不起，这不是python，但我不懂python） 这是“正常方程”方法

C = A'*A and z = A'*y (here A' is the transpose of A)
solve C*x = z for x