Module « numpy.linalg »
Signature de la fonction lstsq
def lstsq(a, b, rcond='warn')
Description
lstsq.__doc__
Return the least-squares solution to a linear matrix equation.
Computes the vector x that approximatively 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 :math:`|| b - a x ||`.
Parameters
----------
a : (M, N) array_like
"Coefficient" matrix.
b : {(M,), (M, K)} array_like
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 : float, 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`.
.. versionchanged:: 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``.
Returns
-------
x : {(N,), (N, K)} ndarray
Least-squares solution. If `b` is two-dimensional,
the solutions are in the `K` columns of `x`.
residuals : {(1,), (K,), (0,)} 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),) ndarray
Singular values of `a`.
Raises
------
LinAlgError
If computation does not converge.
See Also
--------
scipy.linalg.lstsq : Similar function in SciPy.
Notes
-----
If `b` is a matrix, then all array results are returned as matrices.
Examples
--------
Fit a line, ``y = mx + c``, through some noisy data-points:
>>> 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
(1.0 -0.95) # may vary
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()
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