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scipy.optimize.

# nnls

scipy.optimize.nnls(A, b, maxiter=None, *, atol=None)

  Solve `argmin_x || Ax - b ||_2` for `x>=0`.

  This problem, often called as NonNegative Least Squares, is a convex optimization problem with convex constraints. It typically arises when the `x` models quantities for which only nonnegative values are attainable; weight of ingredients, component costs and so on.

  Parameters:
    A(m, n) ndarray

      Coefficient array

    b(m,) ndarray, float

      Right-hand side vector.

    maxiter: int, optional

      Maximum number of iterations, optional. Default value is `3 * n`.

    atol: float

      Tolerance value used in the algorithm to assess closeness to zero in the projected residual `(A.T @ (A x - b)` entries. Increasing this value relaxes the solution constraints. A typical relaxation value can be selected as `max(m, n) * np.linalg.norm(a, 1) * np.spacing(1.)`. This value is not set as default since the norm operation becomes expensive for large problems hence can be used only when necessary.

  Returns:
    x ndarray

      Solution vector.

    rnorm float

      The 2-norm of the residual, `|| Ax-b ||_2`.

  See also

  `lsq_linear`

    Linear least squares with bounds on the variables

  Notes

  The code is based on [2] which is an improved version of the classical algorithm of [1]. It utilizes an active set method and solves the KKT (Karush-Kuhn-Tucker) conditions for the non-negative least squares problem.

  References

  [1]

  : Lawson C., Hanson R.J., “Solving Least Squares Problems”, SIAM, 1995, DOI:10.1137/1.9781611971217

  [2]

  : Bro, Rasmus and de Jong, Sijmen, “A Fast Non-Negativity- Constrained Least Squares Algorithm”, Journal Of Chemometrics, 1997, DOI:10.1002/(SICI)1099-128X(199709/10)11:5<393::AID-CEM483>3.0.CO;2-L

  Examples

      >>> import numpy as np
      >>> from scipy.optimize import nnls
      ...
      >>> A = np.array([[1, 0], [1, 0], [0, 1]])
      >>> b = np.array([2, 1, 1])
      >>> nnls(A, b)
      (array([1.5, 1. ]), 0.7071067811865475)

      >>> b = np.array([-1, -1, -1])
      >>> nnls(A, b)
      (array([0., 0.]), 1.7320508075688772)