How to include constraint to Scipy NNLS function solution so that it sums to 1

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难免孤独
难免孤独 2020-12-15 12:19

I have the following code to solve non-negative least square. Using scipy.nnls.

import numpy as np
from scipy.optimize import nnls 

A = np.array([[60, 90,          


        
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  • I don't think you can use nnls directly as the Fortran code it calls doesn't allow extra constraints. However, the constraint that the equation sums to one can be introduced as a third equation, so your example system is of the form,

    60 x1 + 90  x2 + 120 x3 = 67.5
    30 x1 + 120 x2 +  90 x3 = 60
       x1 +     x2 +     x3 = 1
    

    As this is now a set of linear equations, the exact solution can be obtained from x=np.dot(np.linalg.inv(A),b) so that x=[0.6875, 0.3750, -0.0625]. This requires x3 to be negative. Therefore, there is no exact solution when x is positive to this problem.

    For an approximate solution where x is constrained to be positive, this can be obtained using,

    import numpy as np
    from scipy.optimize import nnls 
    
    #Define minimisation function
    def fn(x, A, b):
        return np.sum(A*x,1) - b
    
    #Define problem
    A = np.array([[60., 90., 120.], 
                  [30., 120., 90.],
                  [1.,  1.,   1. ]])
    
    b = np.array([67.5, 60., 1.])
    
    x, rnorm = nnls(A,b)
    
    print(x,x.sum(),fn(x,A,b))
    

    which gives, x=[0.60003332, 0.34998889, 0.] with a x.sum()=0.95.

    I think if you wanted a more general solution including sum constraints, you'd need to use minimise with explicit constraints/bounds in the following form,

    import numpy as np
    from scipy.optimize import minimize 
    from scipy.optimize import nnls 
    
    #Define problem
    A = np.array([[60, 90, 120], 
                  [30, 120, 90]])
    
    b = np.array([67.5, 60])
    
    #Use nnls to get initial guess
    x0, rnorm = nnls(A,b)
    
    #Define minimisation function
    def fn(x, A, b):
        return np.linalg.norm(A.dot(x) - b)
    
    #Define constraints and bounds
    cons = {'type': 'eq', 'fun': lambda x:  np.sum(x)-1}
    bounds = [[0., None],[0., None],[0., None]]
    
    #Call minimisation subject to these values
    minout = minimize(fn, x0, args=(A, b), method='SLSQP',bounds=bounds,constraints=cons)
    x = minout.x
    
    print(x,x.sum(),fn(x,A,b))
    

    which gives x=[0.674999366, 0.325000634, 0.] and x.sum()=1. From minimise, the sum is correct but the value of x is not quite right with np.dot(A,x)=[ 69.75001902, 59.25005706].

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