.. _graphspectralclusteringcorrectionrst: ========================================== 1A.algo - Spectral Clustering - correction ========================================== .. only:: html **Links:** :download:`notebook `, :downloadlink:`html `, :download:`python `, :downloadlink:`slides `, :githublink:`GitHub|_doc/notebooks/td1a_algo/graph_spectral_clustering_correction.ipynb|*` On veut couper un graphe en deux en coupant le moindre d’arcs possible. C’est un algorithme de `clustering `__. Correction du notebook qui contient l’énoncé. .. code:: ipython3 from jyquickhelper import add_notebook_menu add_notebook_menu() .. contents:: :local: Un graphe --------- .. code:: ipython3 # tutoriel_graphe noeuds = {0: 'le', 1: 'silences', 2: 'quelques', 3: '\xe9crit', 4: 'non-dits.', 5: 'Et', 6: 'risque', 7: '\xe0', 8: "qu'elle,", 9: 'parfois', 10: 'aim\xe9', 11: 'lorsque', 12: 'que', 13: 'plus', 14: 'les', 15: 'Minelli,', 16: "n'oublierai", 17: 'je', 18: 'prises', 19: 'sa', 20: 'la', 21: 'jeune,', 22: "qu'elle,", 23: '\xe0', 24: 'ont', 25: "j'ai", 26: 'chemin', 27: '\xe9tranger', 28: 'lente', 29: 'de', 30: 'voir', 31: 'quand', 32: 'la', 33: 'recul,', 34: 'de', 35: 'trop', 36: 'ce', 37: 'Je', 38: 'Il', 39: "l'extr\xeame", 40: "J'ai", 41: 'silences,', 42: "qu'elle,", 43: 'le', 44: 'trace,', 45: 'avec', 46: 'seras', 47: 'dire,', 48: 'femme', 49: 'soit'} arcs = {(3, 15): None, (46, 47): None, (42, 33): None, (35, 45): None, (1, 14): None, (22, 26): None, (26, 28): None, (43, 29): None, (40, 41): None, (29, 44): None, (17, 3): None, (32, 37): None, (24, 19): None, (46, 34): None, (11, 19): None, (34, 49): None, (22, 2): None, (37, 48): None, (14, 12): None, (3, 10): None, (5, 18): None, (12, 24): None, (34, 32): None, (45, 39): None, (37, 26): None, (33, 45): None, (34, 47): None, (36, 31): None, (29, 47): None, (13, 11): None, (12, 21): None, (2, 16): None, (5, 4): None, (33, 35): None, (28, 49): None, (25, 49): None, (21, 0): None, (3, 13): None, (18, 24): None, (12, 7): None, (13, 15): None, (11, 1): None, (16, 23): None, (37, 45): None, (27, 32): None, (32, 41): None, (8, 24): None, (10, 1): None, (2, 24): None, (24, 11): None, (2, 14): None, (47, 36): None, (48, 39): None, (30, 25): None, (30, 43): None, (15, 14): None, (26, 27): None, (6, 8): None, (20, 10): None, (19, 17): None, (5, 7): None, (44, 25): None, (27, 38): None, (2, 0): None, (3, 18): None, (3, 9): None, (25, 33): None, (42, 48): None, (2, 15): None, (26, 48): None, (26, 38): None, (7, 8): None, (8, 4): None} .. code:: ipython3 from mlstatpy.graph.graphviz_helper import draw_graph_graphviz draw_graph_graphviz(noeuds, arcs, "image.png") from IPython.display import Image Image("image.png", width=400) .. image:: graph_spectral_clustering_correction_4_0.png :width: 400px Partie 1 : clustering en pratique --------------------------------- Q1 ~~ L’algorithme `Minimum Spanning Tree `__ supprime le plus d’arcs possible tout en gardant une seule composante connexe. Q2 ~~ Q3 ~~ Soit :math:`M=(m_{ij})` ce laplacien : .. math:: m_{ij} = \left \{ \begin{array}{ll} d_i & \text{ si } i = j \\ -1 & \text{ s'il existe un arc reliant } i \text{ et } j \\ 0 & \text{ sinon} \end{array} \right . .. code:: ipython3 import numpy def Laplacien (edges) : mat = {} for k in edges : i,j = k if i != j : mat [i,j] = -1 mat [j,i] = -1 if (i,i) not in mat : mat [i,i] = 0 if (j,j) not in mat : mat [j,j] = 0 mat [i,i] += 1 mat [j,j] += 1 maxi = max(max(_) for _ in mat) + 1 nmat = numpy.zeros((maxi, maxi)) for (i, j), v in mat.items(): nmat[i,j] = v return nmat mat = Laplacien(arcs) mat .. parsed-literal:: array([[ 2., 0., -1., ..., 0., 0., 0.], [ 0., 3., 0., ..., 0., 0., 0.], [-1., 0., 6., ..., 0., 0., 0.], ..., [ 0., 0., 0., ..., 4., 0., 0.], [ 0., 0., 0., ..., 0., 4., 0.], [ 0., 0., 0., ..., 0., 0., 3.]]) Q4 ~~ Par construction, le vecteur :math:`(1, ..., 1)` est le vecteur propre associé à la valeur propre 0. Q5 ~~ .. code:: ipython3 def eigen (mat, sort = True) : l, v = numpy.linalg.eig(mat) if sort : li = list (l) li = [ (_,i) for i,_ in enumerate (li) ] li.sort () pos = [ _[1] for _ in li ] l = numpy.array ( [ _[0] for _ in li ] ) mat = v.copy() for i in range (0, len (pos)) : mat [ :,i] = v [ :,pos[i] ] return l,mat else : return l,v val, vec = eigen(mat) val .. parsed-literal:: array([ 2.37288236e-15, 2.97516377e-02, 2.10328629e-01, 2.90790022e-01, 3.11774822e-01, 3.71706160e-01, 4.10657750e-01, 6.06987653e-01, 6.47893457e-01, 6.95288785e-01, 9.17873789e-01, 9.62159368e-01, 1.00171710e+00, 1.22716812e+00, 1.43721861e+00, 1.46689480e+00, 1.50739404e+00, 1.65758626e+00, 1.67080933e+00, 2.01271960e+00, 2.06246640e+00, 2.12023335e+00, 2.15823780e+00, 2.40007127e+00, 2.42330441e+00, 2.44200543e+00, 2.48046940e+00, 2.83472417e+00, 2.88760137e+00, 2.98652806e+00, 3.05062459e+00, 3.45716799e+00, 3.55952462e+00, 3.72589427e+00, 3.86383519e+00, 3.98776759e+00, 4.24048853e+00, 4.51749090e+00, 4.76260241e+00, 5.09890091e+00, 5.16973825e+00, 5.38422915e+00, 5.54406301e+00, 5.82908856e+00, 6.06195525e+00, 6.21987207e+00, 6.51026374e+00, 7.31385944e+00, 7.36832387e+00, 8.10194808e+00]) Q6 ~~ .. code:: ipython3 vec[:, 1] .. parsed-literal:: array([ 0.12706645, 0.1471807 , 0.11300729, 0.1497489 , 0.15689742, 0.15497599, 0.15887778, 0.15239956, 0.1541509 , 0.15434079, 0.15308929, 0.14778115, 0.14353765, 0.14642567, 0.13629278, 0.13739056, 0.1202737 , 0.15177162, 0.1510202 , 0.14927889, 0.15778361, 0.13734517, 0.00981551, 0.12396176, 0.14384261, -0.14991691, -0.0936683 , -0.10998677, -0.11870976, -0.16228832, -0.15880678, -0.17617328, -0.12965455, -0.13772134, -0.14785981, -0.13627598, -0.17093183, -0.1189228 , -0.10336518, -0.12629383, -0.14222274, -0.13799138, -0.12981884, -0.16297189, -0.15845984, -0.13077619, -0.15656133, -0.16060489, -0.11805402, -0.14021942]) Q7 ~~ On classe les noeuds en deux classes selon qu’ils sont associés à une valeur positive ou négative d’après ce vecteur propre. On peut maintenant déterminer quel noeud appartient à la première composante, quel noeud appartient à la seconde. Ecrire une fonction qui calcule le nombre d’arcs qui relient un noeud de la première composante à un noeud de la seconde. Quel est le résultat ? Partie 3 : un peu plus loin --------------------------- On applique cette méthode à un problème de clusterisation. .. code:: ipython3 from pyquickhelper.helpgen import NbImage NbImage("images/tutgraphcl.png") .. image:: graph_spectral_clustering_correction_18_0.png On note :math:`d(X_1,X_2)` la distance euclidienne entre deux points :math:`X_1` et :math:`X_2`. On construit le Laplacien suivant à partir d’un ensemble de points du plan :math:`(X_i)_i`. .. math:: m_{ij} = \left \{ \begin{array}{ll} -e^{-d(X_i,X_j)^2} & \text{ si } i \neq j \\ \sum_{i \neq j} e^{-d(X_i,X_j)^2} & \text{ si } i = j \end{array} \right . .. code:: ipython3 points = [(0.84737386691659533, 0.95848816613228727), (0.28893525107454354, 0.66073249195336492), (0.60382037086559148, 0.13747945088383384), (0.21951613156582261, 0.040905525433785228), (0.21613062123493632, 0.096875623632852625), (0.99787588721497178, 0.79337171783327132), (0.18576957348508683, 0.78396225027633837), (0.23875443625588322, 0.35497638429086975), (0.8713637939628045, 0.22983756618811024), (0.28301724069085921, 0.99408996134013161), (0.39792684083973429, 0.77105362865540716), (0.75452041353842147, 0.330325155167562), (0.24824845436118537, 0.95998690078041737), (0.92318434139996397, 0.38115765401571988), (0.54660304309415886, 0.62093667623480242), (0.58899996464290505, 0.9017292023892568), 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5.6377915319211773), (0.95656595366184183, -3.5482370903224183), (4.6552715153624238, -0.42419842122106877), (3.9138981541477369, 1.5211086418661788), (-5.7643908686171743, 3.3462875243179644), (4.4001664954474204, 1.8715548148469952), (3.7209034976257116, -4.3132712976844925), (2.0077653108424371, -3.8044349295045858), (-2.7004396541700451, 3.6313151291578776), (2.7805282578575432, -1.3496033840422226), (2.5149407509344646, -4.4491799573779538), (-3.4969549443875327, 0.59052341158001964), (2.5871839418980924, -2.8626995345211439), (4.530084220131168, 0.73947783901217035), (-4.2278934560638541, -1.4480933790189707), (-3.6638968948801822, -1.8603129450393652), (1.0034748779660814, 4.3783603559660618), (-0.24711046251746965, 5.0245225170472958), (-0.75233017871629115, -3.4003624728787472), (-5.3204808270534789, 0.8530050107548528), (-0.66555456366565435, -3.210607962975542), (4.4312598575388913, -1.8510534338146063), (-1.0579141292803367, -3.8599892658343156), (5.1580465239922022, -1.6376354853614972), (-2.6525127599513731, 2.9406618825179196), (3.3353268107001339, 4.5193520805659642), (4.9838132614191322, -4.5937246171656669)] Q1 ~~ .. code:: ipython3 import math def distance (p1, p2) : dx = p1[0] - p2[0] dy = p1[1] - p2[1] return math.exp (- (dx**2 + dy**2) ) def CreateProximityMatrix (points) : n = len (points) mat = [ [ 0.0 for i in range (0,n)] for j in range (0,n) ] for i in range (0, n) : for j in range (0, n) : if i != j : mat [i][j] = - distance (points [i], points [j]) for i in range (0, n) : mat [i][i] = -sum (mat [i]) return numpy.matrix(mat) mat = CreateProximityMatrix(points) mat .. parsed-literal:: matrix([[ 3.33607823e+01, -6.69976843e-01, -4.80285954e-01, ..., -9.41648335e-08, -6.38278275e-09, -1.51753802e-21], [ -6.69976843e-01, 3.90919719e+01, -6.88702158e-01, ..., -9.65910722e-07, -3.18671103e-11, -2.73374647e-22], [ -4.80285954e-01, -6.88702158e-01, 3.45881912e+01, ..., -9.59966914e-09, -2.63553433e-12, -8.85061645e-19], ..., [ -9.41648335e-08, -9.65910722e-07, -9.59966914e-09, ..., 5.08380506e+00, -2.21989770e-17, -1.05000993e-50], [ -6.38278275e-09, -3.18671103e-11, -2.63553433e-12, ..., -2.21989770e-17, 5.56991352e+00, -5.65510715e-38], [ -1.51753802e-21, -2.73374647e-22, -8.85061645e-19, ..., -1.05000993e-50, -5.65510715e-38, 2.18345495e+00]]) Q2 ~~ Implémentez la méthode suggérée et dessinez le résultat.