numpy.linalg.slogdet(a)
[source]
Compute the sign and (natural) logarithm of the determinant of an array.
If an array has a very small or very large determinant, then a call to det
may overflow or underflow. This routine is more robust against such issues, because it computes the logarithm of the determinant rather than the determinant itself.
Parameters: |
a : (…, M, M) array_like Input array, has to be a square 2-D array. |
---|---|
Returns: |
sign : (…) array_like A number representing the sign of the determinant. For a real matrix, this is 1, 0, or -1. For a complex matrix, this is a complex number with absolute value 1 (i.e., it is on the unit circle), or else 0. logdet : (…) array_like The natural log of the absolute value of the determinant. If the determinant is zero, then -Inf. In all cases, the determinant is equal to |
See also
New in version 1.8.0.
Broadcasting rules apply, see the numpy.linalg
documentation for details.
New in version 1.6.0.
The determinant is computed via LU factorization using the LAPACK routine z/dgetrf.
The determinant of a 2-D array [[a, b], [c, d]]
is ad - bc
:
>>> a = np.array([[1, 2], [3, 4]]) >>> (sign, logdet) = np.linalg.slogdet(a) >>> (sign, logdet) (-1, 0.69314718055994529) >>> sign * np.exp(logdet) -2.0
Computing log-determinants for a stack of matrices:
>>> a = np.array([ [[1, 2], [3, 4]], [[1, 2], [2, 1]], [[1, 3], [3, 1]] ]) >>> a.shape (3, 2, 2) >>> sign, logdet = np.linalg.slogdet(a) >>> (sign, logdet) (array([-1., -1., -1.]), array([ 0.69314718, 1.09861229, 2.07944154])) >>> sign * np.exp(logdet) array([-2., -3., -8.])
This routine succeeds where ordinary det
does not:
>>> np.linalg.det(np.eye(500) * 0.1) 0.0 >>> np.linalg.slogdet(np.eye(500) * 0.1) (1, -1151.2925464970228)
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https://docs.scipy.org/doc/numpy-1.14.2/reference/generated/numpy.linalg.slogdet.html