CODEWITHSUNDEEP
pythonnumpylinear-algebraleast-squares
Johnson Durant
Johnson Durant

Reputation: 59

How to use numpy's np.linalg.solve to solve over deteremined least square system?

I have a matrix A and matrix 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]]

How do i find matrix x which is a 2 x 1 matrix?

I tried lsq = np.linalg.solve(A, b) but I had this error:

---------------------------------------------------------------------------
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

Any thoughts as to how to solve this with that specific method would be helpful. Thanks.

Upvotes: 0

Views: 2132

Answers (2)

dmuir
dmuir

Reputation: 4431

If you absolutely have to use np.linalg.solve then you can do so as below. However it would be better (faster, less inaccurate to use a specialised least squares routine

(Sorry its not python, but I don't know python) This is the 'normal equations' approach

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

Upvotes: 0

penguin359
penguin359

Reputation: 1489

You are calling the wrong function for that job. That is solving a normal Linear System which requires the same number of equations as variables (i.e. it must be square). What you are looking for is numpy.linalg.lstsq which returns the least-squares solution to a linear matrix equation.

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)

Upvotes: 2

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