Python K表示集群 [英] Python K means clustering

问题描述

我正在尝试在此网站上实施代码，以估计我应将哪些K值用于我的K均值聚类.

I am trying to implement the code on this website to estimate what value of K I should use for my K means clustering.

https://datasciencelab.wordpress.com/2014/01/21/selection-of-k-in-k-means-clustering-reloaded/

但是我没有获得任何成功-特别是我试图获得f(k)与簇数k的关系图，我可以用它来获取要使用的k的理想值.

However I am not getting any success - in particular I am trying to get the f(k) vs the number of clusters k graph which I can use to procure the ideal value of k to use.

我的数据格式如下:

每个坐标都有5个维度/变量，即它们是生活在5维空间中的数据点. 坐标列表在下面，例如，第一个数据点的坐标为( 35.38361202590826,-24.022420305129415, 0.9608968122051765, -11.700331772145386, -9.4393980963685).

Each of the coordinates have 5 dimensions/variables i.e. they are data points that live in a five-dimensional space. The list of the coordinates are below, where for example the first data point has coordinates ( 35.38361202590826,-24.022420305129415, 0.9608968122051765, -11.700331772145386, -9.4393980963685).

Variable1 = [35.38361202590826, 3.0, 10.0, 10.04987562112089, 5.385164807134505, 24.35159132377184, 10.77032961426901, 10.816653826391967, 18.384776310850235, 14.317821063276353, 24.18677324489565, 3.0, 24.33105012119288, 8.94427190999916, 2.82842712474619, 4.123105625617661, 4.47213595499958, 13.453624047073712, 12.529964086141668, 19.4164878389476, 5.385164807134505, 5.0, 24.041630560342618, 30.083217912982647, 15.132745950421555, 1.414213562373095, 21.470910553583888, 12.649110640673516, 9.0, 9.055385138137416, 16.124515496597102, 18.027756377319946, 7.615773105863908, 4.47213595499958, 5.0, 16.124515496597102, 8.246211251235321, 3.0, 23.02172886644268, 2.23606797749979, 10.0, 13.416407864998737, 14.7648230602334, 12.649110640673516, 2.82842712474619, 9.899494936611665, 12.806248474865697, 13.0, 10.19803902718557, 10.440306508910549]
Variable2 = [-24.022420305129415, -40.0, -21.0, -36.020346285601605, -14.298541039632994, -10.225204451297113, -7.242118188905023, -10.816653826391967, -16.263455967290593, -0.9079593845004517, -5.70559779110359, -1.0, -17.426292654367874, -0.4472135954999579, -12.727922061357855, -38.32062875574061, -15.205262246998569, -13.89960053482201, -6.943355894868313, -18.43793805396085, -14.298541039632994, -8.0, -9.899494936611665, -10.537436550735357, -9.251460406371256, -1.414213562373095, -0.23287321641631115, -4.743416490252569, -10.0, -25.951408627588936, -5.457528321925173, -11.648704120729812, -15.231546211727816, -9.838699100999074, -2.2, 4.713319914389921, -3.395498750508662, -32.0, -16.59301967354925, -4.47213595499958, -3.4, -13.416407864998737, 4.944183868793753, -3.478505426185217, -21.213203435596423, -18.384776310850235, -6.871645523098667, -21.0, -5.491251783869154, -8.620436566990362]
Variable3 = [0.9608968122051765, 22.0, 21.0, 18.507691737905798, 15.412713068695306, -8.08982038917884, -0.7427813527082074, -7.211102550927978, -14.849242404917499, -0.4190581774617469, -10.170848236315095, -7.0, 1.150792911137501, -5.366563145999495, -12.727922061357855, 4.85071250072666, 9.838699100999074, -8.473553267217696, 6.065460321953928, -10.249021432229634, 4.642383454426297, -9.0, 9.899494936611665, 4.354587344310195, -8.854969246098202, -8.48528137423857, -10.292996165600954, -11.067971810589327, -30.0, -10.932721081409808, -14.6360986815266, -22.188007849009164, 0.0, -7.155417527999327, -5.4, -12.279438724331637, 19.40285000290664, -7.0, 18.938629784469825, 8.94427190999916, 3.8, -8.94427190999916, -43.549455173073746, -8.538149682454623, -11.31370849898476, 1.4142135623730951, -10.619815808425212, 12.0, 7.060180864974626, -7.854175538813441]
Variable4 = [-11.700331772145386, -8.0, -5.0, -2.9851115706299676, -10.398938937914904, -8.459406092237773, -7.242118188905023, -10.539303728279352, -21.920310216782973, -8.03194840135015, -10.791021909261136, -10.0, -9.69954025101608, -2.6832815729997477, -23.33452377915607, -7.761140001162655, -17.44133022449836, -4.980070779856015, -2.7134954071899156, -6.48933015307002, -12.441587657862476, -5.2, -18.384776310850235, -10.603918800266811, -14.604091070057484, -4.949747468305833, -1.3506646552146047, -7.905694150420948, -14.0, -29.706080514133717, -2.4806946917841692, -23.574758339572238, -3.2826608214930637, -5.813776741499453, -13.4, -4.9613893835683385, -11.884245626780316, -19.0, -5.473090258814675, -2.23606797749979, -2.0, -2.6832815729997477, -6.163297699455227, -12.01665510863984, -12.727922061357855, -12.020815280171307, -8.589556903873333, -18.53846153846154, -5.491251783869154, -4.789131426105757]
Variable5 = [-9.4393980963685, -4.0, -2.0, -0.29851115706299675, -9.84185292338375, 6.118696639531204, -6.127946159842712, -2.218800784900916, 10.606601717798213, 0.6984302957695782, 0.7442084075352507, -0.0, 3.452378733412503, 1.3416407864998738, -6.363961030678928, 6.305926250944657, -5.813776741499453, -0.4459764877482998, -0.7980868844676221, 7.673890419106611, -1.4855627054164149, 1.4, -2.8284271247461903, -2.925218979383948, 3.9649116027305387, 0.7071067811865475, 0.4191717895493601, 1.5811388300841895, -4.0, 4.748555621218401, 4.341215710622296, 4.714951667914447, -5.120950881529179, 4.919349550499537, 6.2, 0.6201736729460423, -6.305926250944657, -9.0, -6.168085847235585, 0.0, -1.0, 1.3416407864998738, 3.3186987612451224, 4.427188724235731, 4.242640687119285, 4.949747468305833, 5.9346029517670305, 2.3076923076923075, -3.1378581622109447, 1.436739427831727]

我能够使用scikit-learn创建具有这些坐标的聚类，但是我有兴趣寻找要使用的最佳k值-但是scikit-learn不具有可以以此估计K的最佳值的功能技术(或据我所知的任何技术).

I am able to use scikit-learn to create clusters with these coordinates however I am interested in finding the optimal k value to use - however scikit-learn does not have a feature where I can estimate the optimal value of K with this technique (or any technique as far as I am aware).

推荐答案

您可以尝试代码. 这是一个简化的版本:

You can try the code in the last comment by Monte Shaffer. Here's a simplified version:

import numpy as np
import random
from numpy import zeros

class KMeansFK():
    def __init__(self, K, X):
        self.K = K
        self.X = X
        self.N = len(X)
        self.mu = None
        self.clusters = None
        self.method = None

    def _cluster_points(self):
        mu = self.mu
        clusters  = {}
        for x in self.X:
            bestmukey = min([(i[0], np.linalg.norm(x-mu[i[0]])) \
                             for i in enumerate(mu)], key=lambda t:t[1])[0]
            try:
                clusters[bestmukey].append(x)
            except KeyError:
                clusters[bestmukey] = [x]
        self.clusters = clusters

    def _reevaluate_centers(self):
        clusters = self.clusters
        newmu = []
        keys = sorted(self.clusters.keys())
        for k in keys:
            newmu.append(np.mean(clusters[k], axis = 0))
        self.mu = newmu

    def _has_converged(self):
        K = len(self.oldmu)
        return(set([tuple(a) for a in self.mu]) == \
               set([tuple(a) for a in self.oldmu])\
               and len(set([tuple(a) for a in self.mu])) == K)

    def find_centers(self, K, method='random'):
        self.method = method
        X = self.X
        K = self.K
        # https://stackoverflow.com/questions/44372231/population-must-be-a-sequence-or-set-for-dicts-use-listd
        self.oldmu = random.sample(list(X), K)
        if method != '++':
            # Initialize to K random centers
            self.mu = random.sample(list(X), K)
        while not self._has_converged():
            self.oldmu = self.mu
            # Assign all points in X to clusters
            self._cluster_points()
            # Reevaluate centers
            self._reevaluate_centers()

    def _dist_from_centers(self):
        cent = self.mu
        X = self.X
        D2 = np.array([min([np.linalg.norm(x-c)**2 for c in cent]) for x in X])
        self.D2 = D2

    def _choose_next_center(self):
        self.probs = self.D2/self.D2.sum()
        self.cumprobs = self.probs.cumsum()
        r = random.random()
        ind = np.where(self.cumprobs >= r)[0][0]
        return(self.X[ind])

    def init_centers(self,K):
        self.K = K
        #self.mu = random.sample(self.X, 1)
        self.mu = random.sample(list(self.X), 1)
        while len(self.mu) < self.K:
            self._dist_from_centers()
            self.mu.append(self._choose_next_center())

    def get_ak(self,k, Nd):
        if k == 2:
            return( 1 - 3.0 / (4.0 * Nd ) )
        else:
            previous_a = self.get_ak(k-1, Nd)
            return ( previous_a + (1.0-previous_a)/6.0 )

    def fK(self, thisk, Skm1=0):
        X = self.X
        Nd = len(X[0])

        self.find_centers(thisk, method='++')
        mu, clusters = self.mu, self.clusters
        Sk = sum([np.linalg.norm(mu[i]-c)**2 \
                 for i in range(thisk) for c in clusters[i]])
        if thisk == 1:
            fs = 1
        elif Skm1 == 0:
            fs = 1
        else:
            fs = Sk/(self.get_ak(thisk,Nd)*Skm1)
        return fs, Sk

    def run(self, maxk):
        ks = range(1,maxk)
        fs = zeros(len(ks))
        Wks,Wkbs,sks = zeros(len(ks)+1),zeros(len(ks)+1),zeros(len(ks)+1)
        # Special case K=1
        self.init_centers(1)
        fs[0], Sk = self.fK(1)
        # Rest of Ks
        for k in ks[1:]:
            self.init_centers(k)
            fs[k-1], Sk = self.fK(k, Skm1=Sk)
        self.fs = fs

然后在您的数据上运行它:

And then run it on your data:

X = np.array([Variable1, Variable2, Variable3, Variable4, Variable5])
km = kmeans.KMeansFK(2, X)
km.run(5)

现在km.clusters得到结果.

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