使用k-means聚类基于单个变量进行聚类

Question

我只是试图让我的头围绕群集。

我有一系列数据点 - y - 具有与它们相关的噪声函数（高斯）

有两类值0和> 0（显然有噪音）。我正试图找到该组的中心点，即> 0。

我用简单的移动平均线绘制了点，以便能够观察数据。

移动平均线图：

如何根据y值对数据进行聚类？

我想要有两个星团 - 一个覆盖左右两点（大约＆lt; 120和＆gt; 260的外观），另一个覆盖中间点（x = 120到260）

如果我尝试使用两个群集，我会得到这个：

k表示情节 - k = 2：

我应该如何修改我的代码才能实现这个目标？

x = range(315)
y= [-0.0019438692324050865, 0.0028994208839327852, 0.0051483573976274649, -0.0033242993359676809, -0.007205517954705391, 0.0023493638544448323, 0.0021109981155292179, 0.0035990200904119076, -0.0039516797159245328, 0.0046512034107712786, -0.0019248189368846083, 0.0036744109953683823, 0.0007898612768152954, 0.0050059088808496474, -0.0021084425769681558, 0.0014692258570182986, -0.0030711206115484175, -0.0026614801222815628, 0.0022816301256991535, 0.00019923934682088178, -0.0013181161659271139, -0.0021956355547661358, 0.0012941895041076283, 0.00337197586896105, -0.0019792508536746402, -0.002020497762984554, 0.0014495021773240431, 0.0011887337096206894, 0.0016667792145975404, -0.0010119590445208419, -0.0024506337087077676, 0.0072264471843846339, -0.0014126073097276062, -0.00065673498034648755, -0.0011355352304356647, -0.00042657980930307281, -0.0032875547481258042, -0.002351265010099495, -0.00073344218847348742, -0.0031555991687002589, 0.0026170287799315104, 0.0019289080666337198, -0.0021804765064623076, 0.0026221290350876979, 0.0019831827145683828, -0.005422907223254632, -0.0014107046201467732, -0.0049438583709020423, 0.00081884635937855494, 0.0054783747880986361, -0.0011282600170147909, -0.00436581779762948, 0.0024421851848953177, -0.0018564229613786095, -0.0052492274840120123, 0.0051775747035086306, 0.0052413417491534494, 0.0030817295096650732, -0.0014106391941506153, 0.00074380887788818206, -0.0041507550699856439, -0.00074928547462217287, -9.3938667619130614e-05, -0.00060592968804004362, 0.0064913597798387348, 0.0018098075166183621, 0.00099550852535854441, 0.0037322288350247917, 0.0027039351321340869, 0.0060238021513650541, -0.006567405116575234, 0.0020858553839503175, -0.0040329574871009084, -0.0029337227854833213, 0.0020743996957790969, 0.0041249738085716511, -0.0016678673351373336, -0.00081387164524554967, -0.0028411340446090278, 0.00013572776045231967, -0.00025350369023925548, 0.00071609777542998309, -0.0018427036825796074, -0.0015513575887011904, -0.0016357115978466398, 0.0038235991426514866, 0.0017693050063256977, -0.00029816429542494152, -0.0016071303644783605, -0.0031883070092131086, -0.0010340123778528594, -0.0049194467790889653, 0.0012109237666701397, 0.0024532524488299246, 0.0069307209537693721, 0.0009573350812806618, -6.0022322637651027e-05, -0.00050143013334696311, 0.0023415017810229548, 0.0033053845403900849, -0.0061156769150035222, 0.00022216114877491691, 0.0017257349557975464, 4.6919738262423826e-05, -0.0035257466102171162, -0.0043673831041441185, -0.0016592116617178102, -0.003298933045964781, -0.001667158964114637, 0.0011283739877531254, -0.0055098513985193534, 0.0023564462221116358, 0.0041971132878626258, 0.0061727231077443314, 0.0047583822927202779, 0.0022475414486232245, 0.0048682822792560521, 0.0022415648209199016, 0.00044859963858686957, -0.0018519391698513449, 0.0031460918774998763, 0.0038614233082916809, -0.0043409564348247066, -0.0055560805453666326, -0.00025133196059449212, 0.012436346397552794, 0.01136022093203152, 0.011244278807602391, 0.01470018209739289, 0.0075560289478025277, 0.012568781764361209, 0.0076068752709663838, 0.011022209533236597, 0.010545997929846045, 0.01084340614623565, 0.011728388118710915, 0.0075043238708055885, 0.012860298948366296, 0.0097297636410632864, 0.0098800557729756874, 0.011536517297700085, 0.0082316420968713416, 0.012612386004592427, 0.016617154743589352, 0.0091391582296167315, 0.014952150276251052, 0.011675391002362373, 0.01568297072839233, 0.01537664322062633, 0.01622711654371662, 0.010708828344561546, 0.016625354383482532, 0.010757807468539406, 0.016867909081979202, 0.010354635736138377, 0.014345365677006765, 0.011114328315579219, 0.010034249196973242, 0.015846180181371881, 0.014303841146954242, 0.011608682896746103, 0.0086826955459553216, 0.0088576104599897426, 0.011250553207393772, 0.005522552439745569, 0.011185993425936373, 0.010241377537878162, 0.0079206732150164348, 0.0052965651546758108, 0.011104715912291204, 0.010506408714857187, 0.010153282642128673, 0.010286986015082572, 0.01187330766677645, 0.014541420264499783, 0.013092204890199896, 0.012979246400649271, 0.012595814351669916, 0.014714607377710237, 0.011727516021525658, 0.011035077266739704, 0.0089698030032708698, 0.0087245475140550147, 0.011139467365240661, 0.0094505568595650603, 0.014430361388952871, 0.0089241578716030695, 0.014616210804585136, 0.013295072783119581, 0.014430633057603408, 0.01200577022494694, 0.011315388654675421, 0.013359877656434442, 0.017704146495248471, 0.0089900858719559155, 0.014731590728415532, 0.0053244009632545759, 0.011199377929150522, 0.0098899254166580439, 0.012220397221188688, 0.015315682643295272, 0.0042842773538990919, 0.0098560854848898077, 0.0088592602102698509, 0.011682575531316278, 0.0098450268165344631, 0.015508017179782136, 0.0083959771972897564, 0.0057504382506886418, 0.010149849298310511, 0.011467172305959087, 0.019354427705224483, 0.013200207481702888, 0.0084555200083286791, 0.011458643458455485, 0.0067582116806278788, 0.01083616691886825, 0.013189184991857963, 0.011774794518724967, 0.014419252448288828, 0.011252283438046358, 0.013346699363583018, 0.0070752340082163006, 0.013215300343131422, 0.0083841320189162287, 0.0067600805611729283, 0.014043517055899181, 0.0098241497159076551, 0.011466675085574904, 0.01155354571355972, 0.012051701509217881, 0.010150596813866767, 0.0093930906430917619, 0.003368481869910186, 0.0048359029438027378, 0.0072083852964288445, 0.010112266453748613, 0.014009345326404186, 0.0050187514558796657, 0.0076315122645601551, 0.0098572381625301152, 0.0114902035403828, 0.018390212262653569, 0.020552166087412803, 0.010428735773226807, 0.011717974670325962, 0.011586303572796604, 0.0092978832913345726, 0.0040060048273946845, 0.012302496528511328, 0.0076707934776137684, 0.014700766223305586, 0.013491092168119941, 0.016244916923257174, 0.010387716692694397, 0.0072564046806323553, 0.0089420045528720883, 0.012125390630607462, 0.013274623392811291, 0.012783388635585766, 0.013859113028817658, 0.0080975189401925642, 0.01379241865445455, 0.012648552766643405, 0.011380280655911323, 0.010109646424218717, 0.0098577688652478051, 0.0064661895943772208, 0.010848835432253455, -0.0010986941731458047, -0.00052875821639583262, 0.0020423603076171414, 0.0035710440970171805, 0.001652886517437206, 0.0023512717524485573, -0.002695275440737862, 0.002253880812688683, -0.0080855104018828141, -0.0020090808966136161, -0.0029794078852333791, 0.00047537441103425869, -0.0010168825525621432, 0.0028683012479151873, -0.0014733214239664142, 0.0019432702158397569, -0.0012411849653504801, -0.00034507088510895141, -0.0023587874349834145, 0.0018156591123708393, 0.0040923006067568324, 0.0043522232127477072, -0.0055992642684123371, -0.0019368557792245147, 0.0026257395447205848, 0.0025594329536029635, 0.00053681548609292378, 0.0032186216144045742, -0.003338121135450386, 0.00065996843114729585, 0.006711173245189642, 0.0032877327776177517, 0.0039528629317296367, 0.0063732674764248719, -0.0026207617244284023, 0.0061381482567009048, -0.003024741769256066, -0.0023891419421980839, -0.004011235930513047, 0.0018372067754070733, -0.0045928077859572689, -0.0021420171112169601, 0.001665179522797816, 0.0074356736689407859, 0.0065680163280897891, -0.0038116640825467678]

data = np.column_stack([x,y])
from sklearn.cluster import KMeans
kmeans = KMeans(n_clusters=2)
kmeans.fit(data)
y_kmeans = kmeans.predict(data)
plt.scatter(data[:, 0], data[:, 1], c=y_kmeans, s=5, cmap='viridis')

centers = kmeans.cluster_centers_
plt.scatter(centers[:, 0], centers[:, 1], c='black', s=200, alpha=0.5);
plt.grid()

我还希望能够返回每个群集中值的最大值，最小值和平均值 - 这可能吗？

Answer 1

您可以将数据重新整形为n x 1阵列。

但如果您想考虑时间，我建议您调查时间序列中的更改检测。它可以检测到平均值的变化。

Answer 2

使用您的代码，获得所需内容的最简单方法是更改​​：

kmeans.fit(data)
y_kmeans = kmeans.predict(data)

到

kmeans.fit(data[:,1].reshape(-1,1))
y_kmeans = kmeans.predict(data[:,1].reshape(-1,1))

您可以使用索引获取max，min，mean等，例如：

np.max(data[:,1][y_kmeans == 1])

Answer 3

关于你的问题的一些想法。

k-means实际上是一种多变量方法，因此在您的情况下它可能不是一个好的选择。您可以通过查找y数据的核密度估计的最小值来利用数据的1维度。密度估计图将显示双峰密度函数，其中两个模式除以最小值，即您想要划分两个聚类的y值。 看看http://scikit-learn.org/stable/modules/density.html#kernel-density

要获得除以的x值，可以使用已计算的移动平均值。

但是，可能有更适合您的数据类型的方法。您可能想在https://stats.stackexchange.com/提出您的问题，因为这不是一个编程问题，而是一个关于适当方法的问题。