Code source de mlstatpy.ml.voronoi

# -*- coding: utf-8 -*-
"""
:githublink:%|py|6
[docs]class VoronoiEstimationError(Exception): """ Raised when the algorithm failed. :githublink:%|py|16 """ pass
[docs]def voronoi_estimation_from_lr(L, B, C=None, D=None, cl=0, qr=True, max_iter=None, verbose=False): """ Determines a Voronoi diagram close to a convex partition defined by a logistic regression in *n* classes. :math:M \\in \\mathbb{M}_{nd} a row matrix :math:(L_1, ..., L_n). Every border between two classes *i* and *j* is defined by: :math:\\scal{L_i}{X} + B = \\scal{L_j}{X} + B. The function looks for a set of points from which the Voronoi diagram can be inferred. It is done through a linear regression with norm *L1*. See :ref:l-lrvor-connection. :param L: matrix :param B: vector :param C: additional conditions (see below) :param D: addition condictions (see below) :param cl: class on which the additional conditions applies :param qr: use quantile regression :param max_iter: number of condition to remove until convergence :param verbose: display information while training :return: matrix :math:P \\in \\mathbb{M}_{nd} The function solves the linear system: .. math:: \\begin{array}{rcl} & \\Longrightarrow & \\left\\{\\begin{array}{l}\\scal{\\frac{L_i-L_j}{\\norm{L_i-L_j}}}{P_i + P_j} + 2 \\frac{B_i - B_j}{\\norm{L_i-L_j}} = 0 \\\\ \\scal{P_i- P_j}{u_{ij}} - \\scal{P_i - P_j}{\\frac{L_i-L_j}{\\norm{L_i-L_j}}} \\scal{\\frac{L_i-L_j}{\\norm{L_i-L_j}}}{u_{ij}}=0 \\end{array} \\right. \\end{array} If the number of dimension is big and the number of classes small, the system has multiple solution. Addition condition must be added such as :math:CP_i=D where *i=cl*, :math:P_i is the Voronoï point attached to class *cl*. Quantile regression <https://fr.wikipedia.org/wiki/R%C3%A9gression_quantile>_ is not implemented in :epkg:scikit-learn. We use QuantileLinearRegression <http://www.xavierdupre.fr/app/mlinsights/helpsphinx/mlinsights/mlmodel/quantile_regression.html>_. After the first iteration, the function determines the furthest pair of points and removes it from the list of equations. If *max_iter* is None, the system goes until the number of equations is equal to the number of points * 2, otherwise it stops after *max_iter* removals. This is not the optimal pair to remove as they could still be neighbors but it should be a good heuristic. :githublink:%|py|69 """ labels_inv = {} nb_constraints = numpy.zeros((L.shape,)) matL = [] matB = [] for i in range(0, L.shape): for j in range(i + 1, L.shape): li = L[i, :] lj = L[j, :] c = (li - lj) nc = (c.T @ c) ** 0.5 # first condition mat = numpy.zeros((L.shape)) mat[i, :] = c mat[j, :] = c d = -2 * (B[i] - B[j]) matB.append(d) matL.append(mat.ravel()) labels_inv[i, j, 'eq1'] = len(matL) - 1 nb_constraints[i] += 1 nb_constraints[j] += 1 # condition 2 - hides multiple equation # we pick one coor = 0 found = False while not found and coor < len(c): if c[coor] == 0: coor += 1 continue if c[coor] == nc: coor += 1 continue found = True if not found: raise ValueError( "Matrix L has two similar rows {0} and {1}. Problem cannot be solved.".format(i, j)) c /= nc c2 = c * c[coor] mat = numpy.zeros((L.shape)) mat[i, :] = -c2 mat[j, :] = c2 mat[i, coor] += 1 mat[j, coor] -= 1 matB.append(0) matL.append(mat.ravel()) labels_inv[i, j, 'eq2'] = len(matL) - 1 nb_constraints[i] += 1 nb_constraints[j] += 1 nbeq = (L.shape * (L.shape - 1)) // 2 matL = numpy.array(matL) matB = numpy.array(matB) if max_iter is None: max_iter = matL.shape - matL.shape if nbeq * 2 <= L.shape * L.shape: if C is None and D is None: warnings.warn( "[voronoi_estimation_from_lr] Additional condition are required.") if C is not None and D is not None: matL = numpy.vstack([matL, numpy.zeros((1, matL.shape))]) a = cl * L.shape b = a + L.shape matL[-1, a:b] = C if not isinstance(D, float): raise TypeError("D must be a float not {0}".format(type(D))) matB = numpy.hstack([matB, [D]]) elif C is None and D is None: pass else: raise ValueError( "C and D must be None together or not None together.") sample_weight = numpy.ones((matL.shape,)) tol = numpy.abs(matL.ravel()).max() * 1e-8 / matL.shape order_removed = [] removed = set() for it in range(0, max(max_iter, 1)): if qr: clr = QuantileLinearRegression( fit_intercept=False, max_iter=max(matL.shape)) else: clr = LinearRegression(fit_intercept=False) clr.fit(matL, matB, sample_weight=sample_weight) score = clr.score(matL, matB, sample_weight) res = clr.coef_ res = res.reshape(L.shape) # early stopping if score < tol: if verbose: print('[voronoi_estimation_from_lr] iter={0}/{1} score={2} tol={3}'.format( it + 1, max_iter, score, tol)) break # defines the best pair of points to remove dist2 = pairwise_distances(res, res) dist = [(d, n // dist2.shape, n % dist2.shape) for n, d in enumerate(dist2.ravel())] dist = [_ for _ in dist if _ < _] dist.sort(reverse=True) # test equal points if dist[-1] < tol: _, i, j = dist[-1] eq1 = labels_inv[i, j, 'eq1'] eq2 = labels_inv[i, j, 'eq2'] if sample_weight[eq1] == 0 and sample_weight[eq2] == 0: sample_weight[eq1] = 1 sample_weight[eq2] = 1 nb_constraints[i] += 1 nb_constraints[j] += 1 else: keep = (i, j) pos = len(order_removed) - 1 while pos >= 0: i, j = order_removed[pos] if i in keep or j in keep: eq1 = labels_inv[i, j, 'eq1'] eq2 = labels_inv[i, j, 'eq2'] if sample_weight[eq1] == 0 and sample_weight[eq2] == 0: sample_weight[eq1] = 1 sample_weight[eq2] = 1 nb_constraints[i] += 1 nb_constraints[j] += 1 break pos -= 1 if pos < 0: forma = 'Two classes have been merged in a single Voronoi point (dist={0} < {1}). max_iter should be lower than {2}' raise VoronoiEstimationError( forma.format(dist[-1], tol, it)) dmax, i, j = dist pos = 0 while (i, j) in removed or nb_constraints[i] == 0 or nb_constraints[j] == 0: pos += 1 if pos == len(dist): break dmax, i, j = dist[pos] if pos == len(dist): break removed.add((i, j)) order_removed.append((i, j)) eq1 = labels_inv[i, j, 'eq1'] eq2 = labels_inv[i, j, 'eq2'] sample_weight[eq1] = 0 sample_weight[eq2] = 0 nb_constraints[i] -= 1 nb_constraints[j] -= 1 if verbose: print('[voronoi_estimation_from_lr] iter={0}/{1} score={2:.3g} tol={3:.3g} del P{4},{5} d={6:.3g}'.format( it + 1, max_iter, score, tol, i, j, dmax)) return res