Last 2 dimensions of the array must be square ошибка

#python #scipy #newtons-method

#python #scipy #метод ньютонов

Вопрос:

Вот мой код:

 def obj_function(x):
    val= 8*(x[0]**4)   3*(x[1]**2)- 6*x[0]*x[1]   2*x[1]
    return np.asmatrix(val)

def obj_function_prime(x):
    val=[32*(x[0]**3)-6*x[1], 6*x[1]-6*x[0] 2]
    return np.asmatrix(val)

def hess_obj_function(x):
    a= 96*(x[0]**2)
    b= -6
    c= -6
    d=  6
    val=[[a,b], [c,d]]
    return np.asmatrix(val)

tol= 10**(-6)
mu= 0.2
xk= np.asmatrix([1,1])
xold= np.asmatrix([0,0])

while np.linalg.norm(obj_function_prime(xk))> tol:
    xold=xk
    alpha=1.0
    while obj_function(xk- alpha*obj_function_prime(xk))> (obj_function(xk)- mu*alpha*np.dot(obj_function_prime(xk), obj_function(xk))):
        alpha=0.5* alpha
    xk= xk-alpha*obj_function_prime(xk)
    
print(" The minimum of the function occurs at", xk)

И это выдает мне ошибку:

Трассировка LinAlgError (последний последний вызов
)
в
20 xold= np.matrix([0,0])
21
—> 22 в то время как np.linalg.norm(obj_function_prime(xk))> tol:
23 xold= xk
24 альфа = 1.0

 <ipython-input-3-c0704e734bdd> in obj_function_prime(x)
      4 
      5 def obj_function_prime(x):
----> 6     val=[32*(x[0]**3)-6*x[1], 6*x[1]-6*x[0] 2]
      7     return np.matrix(val)
      8 

~anaconda3libsite-packagesnumpymatrixlibdefmatrix.py in __pow__(self, other)
    231 
    232     def __pow__(self, other):
--> 233         return matrix_power(self, other)
    234 
    235     def __ipow__(self, other):

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

~anaconda3libsite-packagesnumpylinalglinalg.py in matrix_power(a, n)
    620     a = asanyarray(a)
    621     _assert_stacked_2d(a)
--> 622     _assert_stacked_square(a)
    623 
    624     try:

~anaconda3libsite-packagesnumpylinalglinalg.py in _assert_stacked_square(*arrays)
    211         m, n = a.shape[-2:]
    212         if m != n:
--> 213             raise LinAlgError('Last 2 dimensions of the array must be square')
    214 
    215 def _assert_finite(*arrays):

LinAlgError: Last 2 dimensions of the array must be square

Что означает эта ошибка и как я могу ее устранить? Как мне преобразовать матрицу в квадратную, если она не определена таким образом?

1. Это означает, что последние два измерения должны быть одинаковыми. Полная трассировка может помочь точно определить, в чем проблема. Кроме того, я бы посоветовал избавиться от asmatrix вызовов. Попробуйте заставить это работать с обычными массивами numpy (они могут быть 2d, как np.matrix).

2. @hpaulj Я отредактировал вопрос и добавил трассировку стека. Пожалуйста, проверьте, спасибо!

3. Что x[0]**3 предполагается делать? Как вы думаете x[0] , что такое (форма, тип, dtype?). Соответствует ли это тому, что есть на самом деле?

Ответ №1:

Использование np.matrix вызывает проблемы.

Давайте создадим небольшую матрицу:

 In [273]: x = np.asmatrix([0,0])
In [274]: x
Out[274]: matrix([[0, 0]])

обратите внимание, что при индексации оно не меняется. Он выбирает первую «строку», но being np.matrix по-прежнему 2d.

 In [275]: x[0]
Out[275]: matrix([[0, 0]])

И ** np.matrix , по сути, матрица квадратная x@x .

 In [276]: x[0]**2
Traceback (most recent call last):
  File "<ipython-input-276-3908b7334115>", line 1, in <module>
    x[0]**2
  File "/usr/local/lib/python3.8/dist-packages/numpy/matrixlib/defmatrix.py", line 231, in __pow__
    return matrix_power(self, other)
  File "<__array_function__ internals>", line 5, in matrix_power
  File "/usr/local/lib/python3.8/dist-packages/numpy/linalg/linalg.py", line 621, in matrix_power
    _assert_stacked_square(a)
  File "/usr/local/lib/python3.8/dist-packages/numpy/linalg/linalg.py", line 204, in _assert_stacked_square
    raise LinAlgError('Last 2 dimensions of the array must be square')
LinAlgError: Last 2 dimensions of the array must be square

Я не собираюсь пытаться выяснить, что вы пытаетесь сделать, но очевидно, что использование asmatrix — это не путь — по крайней мере, не понимая, как ведет себя такой массив.

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Alternatively, the default kpss test is able to test stationarity on your training set without any problems:

In [99]: smodel = pm.auto_arima(Train['Count'], start_p=1, start_q=1,
    ...:                                test='kpss',
    ...:                                max_p=3, max_q=3, m=52,
    ...:                                start_P=0, seasonal=True,
    ...:                                d=None, D=1, trace=True,
    ...:                                error_action='ignore',
    ...:                                suppress_warnings=True,
    ...:                                stepwise=True)
Performing stepwise search to minimize aic
Fit ARIMA(1,0,1)x(0,1,1,52) [intercept=True]; AIC=nan, BIC=nan, Time=nan seconds
Fit ARIMA(0,0,0)x(0,1,0,52) [intercept=True]; AIC=52.914, BIC=51.687, Time=0.085 seconds
Fit ARIMA(1,0,0)x(1,1,0,52) [intercept=True]; AIC=56.852, BIC=54.398, Time=2.361 seconds
Near non-invertible roots for order (1, 0, 0)(1, 1, 0, 52); setting score to inf (at least one inverse root too close to the border of the unit circle: 0.999)
Fit ARIMA(0,0,1)x(0,1,1,52) [intercept=True]; AIC=nan, BIC=nan, Time=3.042 seconds
Fit ARIMA(0,0,0)x(0,1,0,52) [intercept=False]; AIC=51.935, BIC=51.321, Time=0.249 seconds
Fit ARIMA(0,0,0)x(1,1,0,52) [intercept=True]; AIC=54.913, BIC=53.072, Time=0.770 seconds
Fit ARIMA(0,0,0)x(0,1,1,52) [intercept=True]; AIC=54.885, BIC=53.044, Time=2.421 seconds
Near non-invertible roots for order (0, 0, 0)(0, 1, 1, 52); setting score to inf (at least one inverse root too close to the border of the unit circle: 0.997)
Fit ARIMA(0,0,0)x(1,1,1,52) [intercept=True]; AIC=56.884, BIC=54.429, Time=4.564 seconds
Near non-invertible roots for order (0, 0, 0)(1, 1, 1, 52); setting score to inf (at least one inverse root too close to the border of the unit circle: 0.999)
Fit ARIMA(1,0,0)x(0,1,0,52) [intercept=True]; AIC=54.905, BIC=53.064, Time=0.581 seconds
Fit ARIMA(0,0,1)x(0,1,0,52) [intercept=True]; AIC=nan, BIC=nan, Time=0.500 seconds
Fit ARIMA(1,0,1)x(0,1,0,52) [intercept=True]; AIC=nan, BIC=nan, Time=0.653 seconds
Total fit time: 24.058 seconds

Источник

Error in Python matrix calculation

In this case, if the matrix is not a square matrix, the following error will appear:

It is worth noting that when we use it in this way, the program will run normally

It seems that we have obtained the inverse of a non square matrix

Now let’s verify:

If you look at it carefully, it’s not a unit array. I think this method is not feasible, but with the idea of blog rigor, I decided to see the results of the square array

seeing here, I found that the inversion of square matrix also appeared abnormal. I can’t help but look carefully. The original term which should be 0 is very small. Friends who have studied numerical analysis should know that this is a non-zero problem caused by the rounding error of computer, so we can regard both as unit matrix

note: don’t convert the data type at will here, because converting the data type will result in the loss of data accuracy. For example, the unit matrix mentioned above will change when converting the data type, and a bit on the diagonal line may become 0. the unit matrix mentioned above will change when converting the data type

1 ответ

Лучший ответ

Предполагая, что A одинаково в обоих уравнениях, у вас есть 4 уравнения с 4 неизвестными:

A[0, 0] * u[0] + A[0][1] * u[1] = Au[0]
A[1, 0] * u[0] + A[1][1] * u[1] = Au[1]
A[0, 0] * v[0] + A[0][1] * v[1] = Av[0]
A[1, 0] * v[0] + A[1][1] * v[1] = Av[1]

Это можно перефразировать как матричное уравнение B a = Auv, где

a — значения A в векторе
Auv — конкатенированные векторы Au и Av,
B содержит значения u и v.

import numpy as np

u = [1, 1]
v = [1, -1]
Au = [3, 2]
Av = [-1, -2]
B = [
    [u[0], u[1], 0, 0],
    [0, 0, u[0], u[1]],
    [v[0], v[1], 0, 0],
    [0, 0, v[0], v[1]]]

Auv = np.concatenate([Au, Av])

A = np.linalg.solve(B, Auv).reshape(2, 2)

print("A:n", A)

Матрица A:

1 2
0 2

Igor
12 Сен 2020 в 09:06

Источник

Using np.matrix is causing problems.

Lets make a small matrix:

In [273]: x = np.asmatrix([0,0])
In [274]: x
Out[274]: matrix([[0, 0]])

note that when you index it, it doesn’t change. It selects the first ‘row’, but being np.matrix is still 2d.

In [275]: x[0]
Out[275]: matrix([[0, 0]])

And ** of np.matrix is matrix square, effectively [email protected].

In [276]: x[0]**2
Traceback (most recent call last):
  File "<ipython-input-276-3908b7334115>", line 1, in <module>
    x[0]**2
  File "/usr/local/lib/python3.8/dist-packages/numpy/matrixlib/defmatrix.py", line 231, in __pow__
    return matrix_power(self, other)
  File "<__array_function__ internals>", line 5, in matrix_power
  File "/usr/local/lib/python3.8/dist-packages/numpy/linalg/linalg.py", line 621, in matrix_power
    _assert_stacked_square(a)
  File "/usr/local/lib/python3.8/dist-packages/numpy/linalg/linalg.py", line 204, in _assert_stacked_square
    raise LinAlgError('Last 2 dimensions of the array must be square')
LinAlgError: Last 2 dimensions of the array must be square

I’m not going try to figure out what you are trying to do, but clearly using asmatrix is not the way to go — at least not without understanding how such an array behaves.

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python – LinAlgError: Last 2 dimensions of the array must be square

In case you still havent found an answer, or in case someone in the future has this question.

To solve Ax=b:

numpy.linalg.solve uses LAPACK gesv. As mentioned in the documentation of LAPACK, gesv requires A to be square:

LA_GESV computes the solution to a real or complex linear system of equations AX = B, where A is a square matrix and X and B are rectangular matrices or vectors. Gaussian elimination with row interchanges is used to factor A as A = PL*U , where P is a permutation matrix, L is unit lower triangular, and U is upper triangular. The factored form of A is then used to solve the above system.

If A matrix is not square, it means that you either have more variables than your equations or the other way around. In these situations, you can have the cases of no solution or infinite number of solutions. What determines the solution space is the rank of the matrix compared to the number of columns. Therefore, you first have to check the rank of the matrix.

That being said, you can use another method to solve your system of linear equations. I suggest having a look at factorization methods like LU or QR or even SVD. In LAPACK you can use getrs, in Python you can different things:

first do the factorization like QR and then feed the resulting matrices to a method like scipy.linalg.solve_triangular
solve the least-squares using numpy.linalg.lstsq

Also have a look here where a simple example is formulated and solved.

A square matrix is a matrix with the same number of rows and columns. The matrix you are doing is a 3 by 2. Add a column of zeroes to fix this problem.

python – LinAlgError: Last 2 dimensions of the array must be square

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$begingroup$

I’m trying to find the intersection of lines $y=a_1x+b_1$ and $y=a_2x+b_2$ using numpy.linalg.solve(). What I can’t get my head around is how to correctly make $A$ a square matrix for solve() to work. I’m familiar with solving linear equation systems, but there’s something here I don’t get.

What I’d like to do is:

def meeting_lines(a1, b1, a2, b2):
    a = np.array([[a1], [a2]])
    b = np.array([b1, b2])
    return np.linalg.solve(a, b)

def main():
    a1=1
    b1=4
    a2=3
    b2=2

    y, x = meeting_lines(a1, b1, a2, b2)

Where I expect $y=-3$ and $x=1$. However, this fails with numpy.linalg.LinAlgError: Last 2 dimensions of the array must be square.

Thank you very much for your help, trying to figure this out has messed up my day already!

asked Apr 1, 2019 at 12:47

$endgroup$

$begingroup$

You should formulate your lines as follows to have $(x, y)$ as unknowns:
$$begin{align}
left.begin{matrix}
a_1x-y=-b_1\
a_2x-y=-b_2
end{matrix}right}
rightarrow
overbrace{
begin{bmatrix}
a_1& -1\
a_2& -1
end{bmatrix}
}^{boldsymbol{a}}
overbrace{
begin{bmatrix}
x\
y
end{bmatrix}
}^{boldsymbol{x}}
=
overbrace{
begin{bmatrix}
-b_1\
-b_2
end{bmatrix}
}^{boldsymbol{b}}
end{align}$$
Therefore, the code should be:

import numpy as np

def meeting_lines(a1, b1, a2, b2):
    a = np.array([[a1, -1], [a2, -1]])
    b = np.array([-b1, -b2])
    return np.linalg.solve(a, b)

a1=1
b1=4
a2=3
b2=2
x, y = meeting_lines(a1, b1, a2, b2)
print(x, y)

which outputs:

1.0 5.0

answered Apr 1, 2019 at 13:52

EsmailianEsmailian

8,8672 gold badges27 silver badges45 bronze badges

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Last 2 dimensions of the array must be square ошибка

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Ответ №1:

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python – LinAlgError: Last 2 dimensions of the array must be square

python – LinAlgError: Last 2 dimensions of the array must be square

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