The problem of constrained linear least squares is usually stated in following way:
$$\min_x \; \frac12\|Cx-d\|_2^2 + \lambda \|x\|_2^2, $$ $$s.t. \;\; Ax \leq a, \;\;\; Bx = b. $$
using the fact that $\|x\|_2^2 = x^\top x$: $$\frac12\|Cx-d\|_2^2 + \lambda \|x\|_2^2 = \frac12 (Cx - d)^\top(Cx-d) + \lambda x^\top x =$$ $$=\frac12(x^\top C^\top - d^\top)(Cx-d) + \lambda x^\top I x = \frac12 x^\top (C^\top Cx + \lambda I)x - d^\top Cx + \frac12 d^\top d.$$ While minimizing over $x$ we don't care about $d^\top d$, taking $Q = C^\top C + \lambda I$, and $r=d^\top C$ we can rewrite our initial optimization problem as follows:$$\min_x \; \frac12 x^\top Qx + rx, $$ $$s.t. \;\; Ax \leq a, \;\;\; Bx = b. $$
Now it is obvious, that we have formulated a quadratic program, that can be solved by cvxopt.solvers.qp. Stacking box inequality constraints(Matlab notation lb and ub such that lb<=x<=ub) to matrix $A$ and vector $a$, we can fully emulate lsqlin and lsqnonneg with sparse and dense matrices. Testing lsqnonneg:Here is Matlab reference code:
C = [0.0372, 0.2869; 0.6861, 0.7071; ... 0.6233, 0.6245; 0.6344, 0.6170]; d = [0.8587, 0.1781, 0.0747, 0.8405]'; x = lsqnonneg(C, d) % output: % x = [0, 0.6929]And here is Python code:
import lsqlin import numpy as np C = np.array([[0.0372, 0.2869], [0.6861, 0.7071], \ [0.6233, 0.6245], [0.6344, 0.6170]]); d = np.array([0.8587, 0.1781, 0.0747, 0.8405]); ret = lsqlin.lsqnonneg(C, d, {'show_progress': False}) print ret['x'].T #output: #[ 2.50e-07 6.93e-01]So it works perfectly! Disregard tiny value of x[0]=2.5e-07 this can be fixed, by imposing small regularization. Testing lsqlin:
Matlab reference code:
C = [0.9501 0.7620 0.6153 0.4057 0.2311 0.4564 0.7919 0.9354 0.6068 0.0185 0.9218 0.9169 0.4859 0.8214 0.7382 0.4102 0.8912 0.4447 0.1762 0.8936]; d = [0.0578, 0.3528, 0.8131, 0.0098, 0.1388]'; A =[0.2027 0.2721 0.7467 0.4659 0.1987 0.1988 0.4450 0.4186 0.6037 0.0152 0.9318 0.8462]; b =[0.5251, 0.2026, 0.6721]'; lb = -0.1*ones(4,1); ub = 2*ones(4,1); x = lsqlin(C, d, A, b, [], [], lb, ub) % output: % x = [-0.1000, -0.1000, 0.2152, 0.3502]Python code:
import lsqlin import numpy as np C = np.array(np.mat('''0.9501,0.7620,0.6153,0.4057; 0.2311,0.4564,0.7919,0.9354; 0.6068,0.0185,0.9218,0.9169; 0.4859,0.8214,0.7382,0.4102; 0.8912,0.4447,0.1762,0.8936''')) A = np.array(np.mat('''0.2027,0.2721,0.7467,0.4659; 0.1987,0.1988,0.4450,0.4186; 0.6037,0.0152,0.9318,0.8462''')) d = np.array([0.0578, 0.3528, 0.8131, 0.0098, 0.1388]) b = np.array([0.5251, 0.2026, 0.6721]) lb = np.array([-0.1] * 4) ub = np.array([2] * 4) ret = lsqlin.lsqlin(C, d, 0, A, b, None, None, \ lb, ub, None, {'show_progress': False}) print ret['x'].T # output: # [-1.00e-01 -1.00e-01 2.15e-01 3.50e-01]Seems to work fine as well. For sparse matrices in my case(about hundreds columns, tens thousands rows, 3 non-zero elements in each row) both Matlab and Python provided same solutions, and Python implementation seemed to work a bit faster. However I can not provide any timings or memory consumption test. Just hope it can help somebody someday. lsqlin.py can be downloaded from here, or just copypasted from below. You can use numpy dense matrices, scipy sparse matrices or cvxopt matrices as inputs. Module is documented:
#!/usr/bin/python # See http://maggotroot.blogspot.ch/2013/11/constrained-linear-least-squares-in.html for more info ''' A simple library to solve constrained linear least squares problems with sparse and dense matrices. Uses cvxopt library for optimization ''' __author__ = 'Valeriy Vishnevskiy' __email__ = 'valera.vishnevskiy@yandex.ru' __version__ = '1.0' __date__ = '22.11.2013' __license__ = 'WTFPL' import numpy as np from cvxopt import solvers, matrix, spmatrix, mul import itertools from scipy import sparse def scipy_sparse_to_spmatrix(A): coo = A.tocoo() SP = spmatrix(coo.data, coo.row.tolist(), coo.col.tolist()) return SP def spmatrix_sparse_to_scipy(A): data = np.array(A.V).squeeze() rows = np.array(A.I).squeeze() cols = np.array(A.J).squeeze() return sparse.coo_matrix( (data, (rows, cols)) ) def sparse_None_vstack(A1, A2): if A1 is None: return A2 else: return sparse.vstack([A1, A2]) def numpy_None_vstack(A1, A2): if A1 is None: return A2 else: return np.vstack([A1, A2]) def numpy_None_concatenate(A1, A2): if A1 is None: return A2 else: return np.concatenate([A1, A2]) def get_shape(A): if isinstance(C, spmatrix): return C.size else: return C.shape def numpy_to_cvxopt_matrix(A): if A is None: return A if sparse.issparse(A): if isinstance(A, sparse.spmatrix): return scipy_sparse_to_spmatrix(A) else: return A else: if isinstance(A, np.ndarray): if A.ndim == 1: return matrix(A, (A.shape[0], 1), 'd') else: return matrix(A, A.shape, 'd') else: return A def cvxopt_to_numpy_matrix(A): if A is None: return A if isinstance(A, spmatrix): return spmatrix_sparse_to_scipy(A) elif isinstance(A, matrix): return np.array(A).squeeze() else: return np.array(A).squeeze() def lsqlin(C, d, reg=0, A=None, b=None, Aeq=None, beq=None, \ lb=None, ub=None, x0=None, opts=None): ''' Solve linear constrained l2-regularized least squares. Can handle both dense and sparse matrices. Matlab's lsqlin equivalent. It is actually wrapper around CVXOPT QP solver. min_x ||C*x - d||^2_2 + reg * ||x||^2_2 s.t. A * x <= b Aeq * x = beq lb <= x <= ub Input arguments: C is m x n dense or sparse matrix d is n x 1 dense matrix reg is regularization parameter A is p x n dense or sparse matrix b is p x 1 dense matrix Aeq is q x n dense or sparse matrix beq is q x 1 dense matrix lb is n x 1 matrix or scalar ub is n x 1 matrix or scalar Output arguments: Return dictionary, the output of CVXOPT QP. Dont pass matlab-like empty lists to avoid setting parameters, just use None: lsqlin(C, d, 0.05, None, None, Aeq, beq) #Correct lsqlin(C, d, 0.05, [], [], Aeq, beq) #Wrong! ''' sparse_case = False if sparse.issparse(A): #detects both np and cxopt sparse sparse_case = True #We need A to be scipy sparse, as I couldn't find how #CVXOPT spmatrix can be vstacked if isinstance(A, spmatrix): A = spmatrix_sparse_to_scipy(A) C = numpy_to_cvxopt_matrix(C) d = numpy_to_cvxopt_matrix(d) Q = C.T * C q = - d.T * C nvars = C.size[1] if reg > 0: if sparse_case: I = scipy_sparse_to_spmatrix(sparse.eye(nvars, nvars,\ format='coo')) else: I = matrix(np.eye(nvars), (nvars, nvars), 'd') Q = Q + reg * I lb = cvxopt_to_numpy_matrix(lb) ub = cvxopt_to_numpy_matrix(ub) b = cvxopt_to_numpy_matrix(b) if lb is not None: #Modify 'A' and 'b' to add lb inequalities if lb.size == 1: lb = np.repeat(lb, nvars) if sparse_case: lb_A = -sparse.eye(nvars, nvars, format='coo') A = sparse_None_vstack(A, lb_A) else: lb_A = -np.eye(nvars) A = numpy_None_vstack(A, lb_A) b = numpy_None_concatenate(b, -lb) if ub is not None: #Modify 'A' and 'b' to add ub inequalities if ub.size == 1: ub = np.repeat(ub, nvars) if sparse_case: ub_A = sparse.eye(nvars, nvars, format='coo') A = sparse_None_vstack(A, ub_A) else: ub_A = np.eye(nvars) A = numpy_None_vstack(A, ub_A) b = numpy_None_concatenate(b, ub) #Convert data to CVXOPT format A = numpy_to_cvxopt_matrix(A) Aeq = numpy_to_cvxopt_matrix(Aeq) b = numpy_to_cvxopt_matrix(b) beq = numpy_to_cvxopt_matrix(beq) #Set up options if opts is not None: for k, v in opts.items(): solvers.options[k] = v #Run CVXOPT.SQP solver sol = solvers.qp(Q, q.T, A, b, Aeq, beq, None, x0) return sol def lsqnonneg(C, d, opts): ''' Solves nonnegative linear least-squares problem: min_x ||C*x - d||_2^2, where x >= 0 ''' return lsqlin(C, d, reg = 0, A = None, b = None, Aeq = None, \ beq = None, lb = 0, ub = None, x0 = None, opts = opts) if __name__ == '__main__': # simple Testing routines C = np.array(np.mat('''0.9501,0.7620,0.6153,0.4057; 0.2311,0.4564,0.7919,0.9354; 0.6068,0.0185,0.9218,0.9169; 0.4859,0.8214,0.7382,0.4102; 0.8912,0.4447,0.1762,0.8936''')) sC = sparse.coo_matrix(C) csC = scipy_sparse_to_spmatrix(sC) A = np.array(np.mat('''0.2027,0.2721,0.7467,0.4659; 0.1987,0.1988,0.4450,0.4186; 0.6037,0.0152,0.9318,0.8462''')) sA = sparse.coo_matrix(A) csA = scipy_sparse_to_spmatrix(sA) d = np.array([0.0578, 0.3528, 0.8131, 0.0098, 0.1388]) md = matrix(d) b = np.array([0.5251, 0.2026, 0.6721]) mb = matrix(b) lb = np.array([-0.1] * 4) mlb = matrix(lb) mmlb = -0.1 ub = np.array([2] * 4) mub = matrix(ub) mmub = 2 #solvers.options[show_progress'] = False opts = {'show_progress': False} for iC in [C, sC, csC]: for iA in [A, sA, csA]: for iD in [d, md]: for ilb in [lb, mlb, mmlb]: for iub in [ub, mub, mmub]: for ib in [b, mb]: ret = lsqlin(iC, iD, 0, iA, ib, None, None, ilb, iub, None, opts) print ret['x'].T print 'Should be [-1.00e-01 -1.00e-01 2.15e-01 3.50e-01]' #test lsqnonneg C = np.array([[0.0372, 0.2869], [0.6861, 0.7071], [0.6233, 0.6245], [0.6344, 0.6170]]); d = np.array([0.8587, 0.1781, 0.0747, 0.8405]); ret = lsqnonneg(C, d, {'show_progress': False}) print ret['x'].T print 'Should be [2.5e-07; 6.93e-01]'
7 comments:
Thanks for the post. One point that is potentially significant, depending upon your problem: if the matrix C is ill-conditioned, then you are worsening the condition number significantly by supplying C^TC to the quadratic programming routine of CVXOPT.
Thanks for the very helpful post. The original authors website is gone so this is the only record of this code.
I'm attempting to use this to fit a polynomial to some data with weights. I get the same answer using this as I do from Matlab if I do not use the bounds. I'm trying to set an upper limit on the quartic term only so my bounds are [0 inf inf inf inf]. This results in a domain error in cvxopt.compute_scaling.
Any ideas?
Ah, my bad. You are the original author. Good work!
I appear to have solved this problem by doing the following:
lbargs = isfinite(lb)
if sum(lbargs) > 0: # Modify 'A' and 'b' to add lb inequalities
if lb.size == 1:
lb = repeat(lb, nvars)
lb_A = -eye(nvars, nvars)
A = numpy_None_vstack(A, lb_A[lbargs, 0:nvars])
b = numpy_None_concatenate(b, -lb[lbargs])
ubargs = isfinite(ub)
if sum(ubargs) > 0: # Modify 'A' and 'b' to add ub inequalities
if ub.size == 1:
ub = repeat(ub, nvars)
ub_A = eye(nvars, nvars)
A = numpy_None_vstack(A, ub_A[ubargs, 0:nvars])
b = numpy_None_concatenate(b, ub[ubargs])
Thanks, very helpful!
The link to the script seems to be broken though. Also, if I want to get the test case to work, I have to take out the line:
csA = scipy_sparse_to_spmatrix(sA)
and further references to csA.
Hi, that should be r = - d'*C (minus sign) in the derivation
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