dpnp.linalg.lstsq

dpnp.linalg.lstsq(a, b, rcond=None)[source]

Return the least-squares solution to a linear matrix equation.

For full documentation refer to numpy.linalg.lstsq.

Parameters:

a ((M, N) {dpnp.ndarray, usm_ndarray}) -- "Coefficient" matrix.
b ({(M,), (M, K)} {dpnp.ndarray, usm_ndarray}) -- Ordinate or "dependent variable" values. If b is two-dimensional, the least-squares solution is calculated for each of the K columns of b.
rcond ({int, float, None}, optional) -- Cut-off ratio for small singular values of a. For the purposes of rank determination, singular values are treated as zero if they are smaller than rcond times the largest singular value of a. The default uses the machine precision times max(M, N). Passing -1 will use machine precision.

Returns:

x ({(N,), (N, K)} dpnp.ndarray) -- Least-squares solution. If b is two-dimensional, the solutions are in the K columns of x.
residuals ({(1,), (K,), (0,)} dpnp.ndarray) -- Sums of squared residuals: Squared Euclidean 2-norm for each column in b - a @ x. If the rank of a is < N or M <= N, this is an empty array. If b is 1-dimensional, this is a (1,) shape array. Otherwise the shape is (K,).
rank (int) -- Rank of matrix a.
s ((min(M, N),) dpnp.ndarray) -- Singular values of a.

Examples

Fit a line, y = mx + c, through some noisy data-points:

>>> import dpnp as np
>>> x = np.array([0, 1, 2, 3])
>>> y = np.array([-1, 0.2, 0.9, 2.1])

By examining the coefficients, we see that the line should have a gradient of roughly 1 and cut the y-axis at, more or less, -1.

We can rewrite the line equation as y = Ap, where A = [[x 1]] and p = [[m], [c]]. Now use lstsq to solve for p:

>>> A = np.vstack([x, np.ones(len(x))]).T
>>> A
array([[0., 1.],
       [1., 1.],
       [2., 1.],
       [3., 1.]])

>>> m, c = np.linalg.lstsq(A, y, rcond=None)[0]
>>> m, c
(array(1.), array(-0.95)) # may vary