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, 120], [30, 120, 90]]) b = np.array([67.5, 60]) x, rnorm = nnls(A,b) print x #[ 0. 0.17857143 0.42857143] # Now need to have this array sum to 1. What I want to do is to apply a constraint on x solution so that the it sums to 1. How can I do it?
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].