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bundles / scipy 1.17.1 / scipy / linalg / _basic / solve_circulant

function

scipy.linalg._basic:solve_circulant

source: /scipy/linalg/_basic.py :1148

Signature

def   solve_circulant ( c ,  b ,  singular = raise ,  tol = None ,  caxis = -1 ,  baxis = 0 ,  outaxis = 0 )

Summary

Solve the equation C @ x = b for x, where C is a circulant matrix defined by c.

Extended Summary

C is the circulant matrix associated with the vector c.

The system is solved by doing division in Fourier space. The calculation is 

x = ifft(fft(b) / fft(c))

where fft and ifft are the fast Fourier transform and its inverse, respectively. For a large vector c, this is much faster than solving the system with the full circulant matrix.

Parameters

c : array_like

The coefficients of the circulant matrix.

b : array_like

Right-hand side matrix in a x = b.

singular : str, optional

This argument controls how a near singular circulant matrix is handled. If singular is "raise" and the circulant matrix is near singular, a LinAlgError is raised. If singular is "lstsq", the least squares solution is returned. Default is "raise".

tol : float, optional

If any eigenvalue of the circulant matrix has an absolute value that is less than or equal to tol, the matrix is considered to be near singular. If not given, tol is set to 

tol = abs_eigs.max() * abs_eigs.size * np.finfo(np.float64).eps

where abs_eigs is the array of absolute values of the eigenvalues of the circulant matrix.

caxis : int

When c has dimension greater than 1, it is viewed as a collection of circulant vectors. In this case, caxis is the axis of c that holds the vectors of circulant coefficients.

baxis : int

When b has dimension greater than 1, it is viewed as a collection of vectors. In this case, baxis is the axis of b that holds the right-hand side vectors.

outaxis : int

When c or b are multidimensional, the value returned by solve_circulant is multidimensional. In this case, outaxis is the axis of the result that holds the solution vectors.

Returns

x : ndarray

Solution to the system C x = b.

Raises

: LinAlgError

If the circulant matrix associated with c is near singular.

Notes

For a 1-D vector c with length m, and an array b with shape (m, ...),

solve_circulant(c, b)

returns the same result as

solve(circulant(c), b)

where solve and circulant are from scipy.linalg.

Examples

import numpy as np
from scipy.linalg import solve_circulant, solve, circulant, lstsq
✓
c = np.array([2, 2, 4])
b = np.array([1, 2, 3])
solve_circulant(c, b)
✓
Compare that result to solving the system with `scipy.linalg.solve`: 
solve(circulant(c), b)
✓
A singular example: 
c = np.array([1, 1, 0, 0])
b = np.array([1, 2, 3, 4])
✓
Calling ``solve_circulant(c, b)`` will raise a `LinAlgError`. For the least square solution, use the option ``singular='lstsq'``: 
solve_circulant(c, b, singular='lstsq')
✗
Compare to `scipy.linalg.lstsq`: 
x, resid, rnk, s = lstsq(circulant(c), b)
✓
x
✗
A broadcasting example: Suppose we have the vectors of two circulant matrices stored in an array with shape (2, 5), and three `b` vectors stored in an array with shape (3, 5). For example, 
c = np.array([[1.5, 2, 3, 0, 0], [1, 1, 4, 3, 2]])
b = np.arange(15).reshape(-1, 5)
✓
We want to solve all combinations of circulant matrices and `b` vectors, with the result stored in an array with shape (2, 3, 5). When we disregard the axes of `c` and `b` that hold the vectors of coefficients, the shapes of the collections are (2,) and (3,), respectively, which are not compatible for broadcasting. To have a broadcast result with shape (2, 3), we add a trivial dimension to `c`: ``c[:, np.newaxis, :]`` has shape (2, 1, 5). The last dimension holds the coefficients of the circulant matrices, so when we call `solve_circulant`, we can use the default ``caxis=-1``. The coefficients of the `b` vectors are in the last dimension of the array `b`, so we use ``baxis=-1``. If we use the default `outaxis`, the result will have shape (5, 2, 3), so we'll use ``outaxis=-1`` to put the solution vectors in the last dimension. 
x = solve_circulant(c[:, np.newaxis, :], b, baxis=-1, outaxis=-1)
x.shape
np.set_printoptions(precision=3)  # For compact output of numbers.
✓
x
✗
Check by solving one pair of `c` and `b` vectors (cf. ``x[1, 1, :]``): 
solve_circulant(c[1], b[1, :])
✓

See also

circulant

circulant matrix

Aliases

  • scipy.linalg.solve_circulant

Referenced by

This package