numpy.linalg.lstsq

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

Computes the vector x that approximately solves the equation a @ x = b. The equation may be under-, well-, or over-determined (i.e., the number of linearly independent rows of a can be less than, equal to, or greater than its number of linearly independent columns). If a is square and of full rank, then x (but for round-off error) is the “exact” solution of the equation. Else, x minimizes the Euclidean 2-norm \(||b - ax||\). If there are multiple minimizing solutions, the one with the smallest 2-norm \(||x||\) is returned.

“Coefficient” matrix.

Ordinate or “dependent variable” values. If b is two-dimensional, the least-squares solution is calculated for each of the K columns of b.

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.

Changed in version 1.14.0: If not set, a FutureWarning is given. The previous default of -1 will use the machine precision as rcond parameter, the new default will use the machine precision times max(M, N). To silence the warning and use the new default, use rcond=None, to keep using the old behavior, use rcond=-1.

Least-squares solution. If b is two-dimensional, the solutions are in the K columns of x.

Sums of squared residuals: Squared Euclidean 2-norm for each column in b - a @ x. If the rank of a is < N or M >> 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:

Plot the data along with the fitted line: