numpy.linalg.lstsq¶

numpy.linalg.
lstsq
(a, b, rcond='warn')[source]¶ Return the leastsquares solution to a linear matrix equation.
Solves the equation a x = b by computing a vector x that minimizes the Euclidean 2norm  b  a x ^2. 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 roundoff error) is the “exact” solution of the equation.
Parameters:  a : (M, N) array_like
“Coefficient” matrix.
 b : {(M,), (M, K)} array_like
Ordinate or “dependent variable” values. If b is twodimensional, the leastsquares solution is calculated for each of the K columns of b.
 rcond : float, optional
Cutoff 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, usercond=None
, to keep using the old behavior, usercond=1
.
Returns:  x : {(N,), (N, K)} ndarray
Leastsquares solution. If b is twodimensional, the solutions are in the K columns of x.
 residuals : {(1,), (K,), (0,)} ndarray
Sums of residuals; squared Euclidean 2norm 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 1dimensional, this is a (1,) shape array. Otherwise the shape is (K,). rank : int
Rank of matrix a.
 s : (min(M, N),) ndarray
Singular values of a.
Raises:  LinAlgError
If computation does not converge.
Notes
If b is a matrix, then all array results are returned as matrices.
Examples
Fit a line,
y = mx + c
, through some noisy datapoints:>>> 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 yaxis at, more or less, 1.
We can rewrite the line equation as
y = Ap
, whereA = [[x 1]]
andp = [[m], [c]]
. Now uselstsq
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] >>> print(m, c) 1.0 0.95
Plot the data along with the fitted line:
>>> import matplotlib.pyplot as plt >>> plt.plot(x, y, 'o', label='Original data', markersize=10) >>> plt.plot(x, m*x + c, 'r', label='Fitted line') >>> plt.legend() >>> plt.show()