python - 为 Python 的 Scipy 线性规划寻找严格大于零的解决方案的方法

Question

Scipy NNLS 执行此操作:

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

如果我寻求，有什么替代方法可以做到这一点严格非零解(即 x > 0)？

这是我使用 Scipy 的 NNLS 的 LP 代码:

import numpy as np
from numpy import array
from scipy.optimize import nnls

def by_nnls(A=None, B=None):
    """ Linear programming by NNLS """
    #print "NOF row = ", A.shape[0]
    A = np.nan_to_num(A)
    B = np.nan_to_num(B)

    x, rnorm = nnls(A,B)
    x = x / x.sum()
    # print repr(x)
    return x

B1 = array([  22.133,  197.087,   84.344,    1.466,    3.974,    0.435,
          8.291,   45.059,    5.755,    0.519,    0.   ,   30.272,
         24.92 ,   10.095])
A1 = array([[   46.35,    80.58,    48.8 ,    80.31,   489.01,    40.98,
           29.98,    44.3 ,  5882.96],
       [ 2540.73,    49.53,    26.78,    30.49,    48.51,    20.88,
           19.92,    21.05,    19.39],
       [ 2540.73,    49.53,    26.78,    30.49,    48.51,    20.88,
           19.92,    21.05,    19.39],
       [   30.95,  1482.24,   100.48,    35.98,    35.1 ,    38.65,
           31.57,    87.38,    33.39],
       [   30.95,  1482.24,   100.48,    35.98,    35.1 ,    38.65,
           31.57,    87.38,    33.39],
       [   30.95,  1482.24,   100.48,    35.98,    35.1 ,    38.65,
           31.57,    87.38,    33.39],
       [   15.99,   223.27,   655.79,  1978.2 ,    18.21,    20.51,
           19.  ,    16.19,    15.91],
       [   15.99,   223.27,   655.79,  1978.2 ,    18.21,    20.51,
           19.  ,    16.19,    15.91],
       [   16.49,    20.56,    19.08,    18.65,  4568.97,    20.7 ,
           17.4 ,    17.62,    25.51],
       [   33.84,    26.58,    18.69,    40.88,    19.17,  5247.84,
           29.39,    25.55,    18.9 ],
       [   42.66,    83.59,    99.58,    52.11,    46.84,    64.93,
           43.8 ,  7610.12,    47.13],
       [   42.66,    83.59,    99.58,    52.11,    46.84,    64.93,
           43.8 ,  7610.12,    47.13],
       [   41.63,   204.32,  4170.37,    86.95,    49.92,    87.15,
           51.88,    45.38,    42.89],
       [   81.34,    60.16,   357.92,    43.48,    36.92,    39.13,
         1772.07,    68.43,    38.07]])

用法:

In [9]: by_nnls(A=A1,B=B1)
Out[9]:
array([ 0.70089761,  0.        ,  0.06481495,  0.14325696,  0.01218972,
        0.        ,  0.02125942,  0.01906576,  0.03851557]

注意上面的零解。

Answer 1

如果你真的确定你想要严格正解，你可以使用lsq_linear在最新的 scipy 版本中可用。它比 nnls 允许更多地控制边界。

In [37]: from scipy.optimize import lsq_linear

In [38]: lsq_linear(A1, B1, bounds=(0.001, np.inf))
Out[38]: 
 active_mask: array([ 0, -1,  0,  0, -1, -1,  0,  0,  0])
        cost: 3784.3150152135881
         fun: array([ -0.06189388, -56.45892624,  56.28407376,   2.97647016,
         0.46847016,   4.00747016,  18.24947887, -18.51852113,
         0.19599207,   7.32663679,  15.0829264 , -15.1890736 ,
        -0.14570891,  -0.24341795])
     message: 'The first-order optimality measure is less than `tol`.'
         nit: 17
  optimality: 5.4491449547056092e-11
      status: 1
     success: True
           x: array([ 0.05506904,  0.001     ,  0.00501077,  0.01112669,  0.001     ,
        0.001     ,  0.00154812,  0.00147833,  0.00300156])