solve_circulant(c, b, singular='raise', tol=None, caxis=-1, baxis=0, outaxis=0)¶
Solve C x = b for x, where C is a circulant matrix.
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.
The coefficients of the circulant matrix.
Right-hand side matrix in
a x = b.
- singularstr, optional
This argument controls how a near singular circulant matrix is handled. If singular is “raise” and the circulant matrix is near singular, a
LinAlgErroris raised. If singular is “lstsq”, the least squares solution is returned. Default is “raise”.
- tolfloat, 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.
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.
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.
When c or b are multidimensional, the value returned by
solve_circulantis multidimensional. In this case, outaxis is the axis of the result that holds the solution vectors.
Solution to the system
C x = b.
If the circulant matrix associated with c is near singular.
For a one-dimensional vector c with length m, and an array b with shape
returns the same result as
New in version 0.16.0.
>>> 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) array([ 0.75, -0.25, 0.25])
Compare that result to solving the system with
>>> solve(circulant(c), b) array([ 0.75, -0.25, 0.25])
A singular example:
>>> c = np.array([1, 1, 0, 0]) >>> b = np.array([1, 2, 3, 4])
solve_circulant(c, b)will raise a
LinAlgError. For the least square solution, use the option
>>> solve_circulant(c, b, singular='lstsq') array([ 0.25, 1.25, 2.25, 1.25])
>>> x, resid, rnk, s = lstsq(circulant(c), b) >>> x array([ 0.25, 1.25, 2.25, 1.25])
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=-1to put the solution vectors in the last dimension.
>>> x = solve_circulant(c[:, np.newaxis, :], b, baxis=-1, outaxis=-1) >>> x.shape (2, 3, 5) >>> np.set_printoptions(precision=3) # For compact output of numbers. >>> x array([[[-0.118, 0.22 , 1.277, -0.142, 0.302], [ 0.651, 0.989, 2.046, 0.627, 1.072], [ 1.42 , 1.758, 2.816, 1.396, 1.841]], [[ 0.401, 0.304, 0.694, -0.867, 0.377], [ 0.856, 0.758, 1.149, -0.412, 0.831], [ 1.31 , 1.213, 1.603, 0.042, 1.286]]])
Check by solving one pair of c and b vectors (cf.
x[1, 1, :]):
>>> solve_circulant(c, b[1, :]) array([ 0.856, 0.758, 1.149, -0.412, 0.831])