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scipy.linalg.

# lstsq

scipy.linalg.lstsq(a, b, cond=None, overwrite_a=False, overwrite_b=False, check_finite=True, lapack_driver=None)

  Compute least-squares solution to the equation `a @ x = b`.

  Compute a vector x such that the 2-norm `|b - A x|` is minimized.

  The documentation is written assuming array arguments are of specified “core” shapes. However, array argument(s) of this function may have additional “batch” dimensions prepended to the core shape. In this case, the array is treated as a batch of lower-dimensional slices; see Batched Linear Operations for details.

  Parameters:
    a(M, N) array_like

      Left-hand side array

    b(M,) or (M, K) array_like

      Right hand side array

    cond float, optional

      Cutoff for ‘small’ singular values; used to determine effective rank of a. Singular values smaller than `cond * largest_singular_value` are considered zero.

    overwrite_a bool, optional

      Discard data in a (may enhance performance). Default is False.

    overwrite_b bool, optional

      Discard data in b (may enhance performance). Default is False.

    check_finite bool, optional

      Whether to check that the input matrices contain only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.

    lapack_driver str, optional

      Which LAPACK driver is used to solve the least-squares problem. Options are `'gelsd'`, `'gelsy'`, `'gelss'`. Default (`'gelsd'`) is a good choice. However, `'gelsy'` can be slightly faster on many problems. `'gelss'` was used historically. It is generally slow but uses less memory.

      Added in version 0.17.0.

  Returns:
    x(N,) or (N, K) ndarray

      Least-squares solution.

    residues(K,) ndarray or float

      Square of the 2-norm for each column in `b - a x`, if `M > N` and `rank(A) == n` (returns a scalar if `b` is 1-D). Otherwise a (0,)-shaped array is returned.

    rank int

      Effective rank of a.

    s(min(M, N),) ndarray or None

      Singular values of a. The condition number of `a` is `s[0] / s[-1]`.

  Raises:
    LinAlgError

      If computation does not converge.

    ValueError

      When parameters are not compatible.

  See also

  `scipy.optimize.nnls`

    linear least squares with non-negativity constraint

  Notes

  When `'gelsy'` is used as a driver, residues is set to a (0,)-shaped array and s is always `None`.

  Examples

      >>> import numpy as np
      >>> from scipy.linalg import lstsq
      >>> import matplotlib.pyplot as plt

  Suppose we have the following data:

      >>> x = np.array([1, 2.5, 3.5, 4, 5, 7, 8.5])
      >>> y = np.array([0.3, 1.1, 1.5, 2.0, 3.2, 6.6, 8.6])

  We want to fit a quadratic polynomial of the form `y = a + b*x**2` to this data. We first form the “design matrix” M, with a constant column of 1s and a column containing `x**2`:

      >>> M = x[:, np.newaxis]**[0, 2]
      >>> M
      array([[  1.  ,   1.  ],
             [  1.  ,   6.25],
             [  1.  ,  12.25],
             [  1.  ,  16.  ],
             [  1.  ,  25.  ],
             [  1.  ,  49.  ],
             [  1.  ,  72.25]])

  We want to find the least-squares solution to `M.dot(p) = y`, where `p` is a vector with length 2 that holds the parameters `a` and `b`.

      >>> p, res, rnk, s = lstsq(M, y)
      >>> p
      array([ 0.20925829,  0.12013861])

  Plot the data and the fitted curve.

      >>> plt.plot(x, y, 'o', label='data')
      >>> xx = np.linspace(0, 9, 101)
      >>> yy = p[0] + p[1]*xx**2
      >>> plt.plot(xx, yy, label='least squares fit, $y = a + bx^2$')
      >>> plt.xlabel('x')
      >>> plt.ylabel('y')
      >>> plt.legend(framealpha=1, shadow=True)
      >>> plt.grid(alpha=0.25)
      >>> plt.show()