如何在Python中找到两个一维数组之间的马氏距离？

Question

如何在Python中找到两个一维数组之间的马氏距离？

5

我有两个一维数组，需要找出它们之间的马氏距离。

数组1

-0.125510275,0.067021735,0.140631825,-0.014300184,-0.122152582,0.002372072,-0.050777748,-0.106606245,0.149123222,-0.159149423,0.210138127,0.031959131,-0.068411253,-0.038253143,-0.024590122,0.101361006,-0.160774037,-0.183688596,-0.07163775,-0.096662685,-0.000117288,0.14251323,-0.030461289,-0.006710192,-0.217195332,-0.338565469,-0.030219197,-0.100772612,0.144092739,-0.092911556,-0.008420993,0.042907588,-0.212668449,-0.009366207,-7.01E-05,0.134508118,-0.015715659,-0.050884761,0.18804647,0.04946585,-0.242626131,0.099951334,0.053660966,0.275807977,0.216019884,-0.009127878,0.019819722,-0.043750495,0.12940146,-0.259942383,0.061821692,0.107142501,0.098196507,0.022301452,0.079412982,-0.131031215,-0.049483716,0.126781181,-0.195536733,0.077051811,0.061049294,-0.039563753,0.02573989,0.025330214,0.204785526,0.099218346,-0.050533134,-0.109173119,0.205652237,-0.168003649,-0.062734045,0.100320764,-0.063513778,-0.120843001,-0.223983109,0.075016715,0.481291831,0.107607022,-0.141365036,0.075003348,-0.042418435,-0.041501854,0.096700639,0.083469011,-0.033227846,-0.050748199,-0.045331556,0.065955319,0.26927036,0.082820699,-0.014033476,0.176714703,0.042264186,-0.011814327,0.041769091,-0.00132945,-0.114337325,-0.013483777,-0.111367472,-0.051828772,-0.022199111,0.030011443,0.015529033,0.171916366,-0.172722578,0.214662731,-0.0219073,-0.067695767,0.040487193,0.04814541,0.003313571,-0.01360167,0.115932293,-0.235844463,0.185181856,0.130868644,0.010789306,0.171733275,0.059378762,0.003508842,0.039326921,0.024174646,-0.195897669,-0.088932432,0.025385177,-0.134177506,0.08158315,0.049005955

还有，数组 2

-0.120652862,0.030241199,0.146165773,-0.044423241,-0.138606027,-0.048646796,-0.00780057,-0.101798892,0.185339138,-0.210505784,0.1637595,0.015000292,-0.10359703,0.102251172,-0.043159217,0.183324724,-0.171825036,-0.173819616,-0.112194099,-0.161590934,-0.002507193,0.163269699,-0.037766434,0.041060638,-0.178659558,-0.268946916,-0.055348843,-0.11808344,0.113775767,-0.073903576,-0.039505914,0.032382272,-0.159118786,0.007761603,0.057116233,0.043675732,-0.057895001,-0.104836114,0.22844176,0.055832602,-0.245030299,0.006276659,0.140012532,0.21449241,0.159539059,-0.049584024,0.016899824,-0.074179329,0.119686954,-0.242336214,-0.001390997,0.097442642,0.059720818,0.109706804,0.073196828,-0.16272822,0.022305552,0.102650747,-0.192103565,0.104134969,0.099571452,-0.101140082,-0.038911857,0.071292967,0.202927336,0.12729995,-0.047885433,-0.165100336,0.220239595,-0.19612211,-0.075948663,0.096906625,-0.07410948,-0.108219706,-0.155030385,-0.042231761,0.484629512,0.093194947,-0.105109185,0.072906494,-0.056871444,-0.057923764,0.101847053,0.092042476,-0.061295755,-0.031595342,-0.01854251,0.074671492,0.266587347,0.052284949,0.003548023,0.171518356,0.053180017,-0.022400264,0.061757766,0.038441688,-0.139473096,-0.05759665,-0.101672307,-0.074863717,-0.02349415,-0.011674869,0.010008151,0.141401738,-0.190440938,0.216421023,-0.028323224,-0.078021556,-0.011468113,0.100600921,-0.019697987,-0.014288296,0.114862509,-0.162037179,0.171686187,0.149788797,-0.01235011,0.136169329,0.008751356,0.024811052,0.003802934,0.00500867,-0.1840965,-0.086204343,0.018549766,-0.110649876,0.068768717,0.03012047

我发现Scipy已经实现了该函数。然而，我对IV的值感到困惑。我尝试了以下操作

V = np.cov(np.array([array_1, array_2]))
IV = np.linalg.inv(V)
print(mahalanobis(array_1, array_2, IV))

然而，我遇到了以下错误：

文件 "C:\Users\XXXXXX\AppData\Local\Continuum\anaconda3\envs\face\lib\site-packages\scipy\spatial\distance.py" 的第 1043 行，mahalanobis 函数中， m = np.dot(np.dot(delta, VI), delta)

ValueError: 形状为 (128,) 和 (2,2) 的数组无法对齐：128 (dim 0) != 2 (dim 0)

编辑:

array_1 = [-0.10577646642923355, 0.09617947787046432, 0.029290344566106796, 0.02092641592025757, -0.021434104070067406, -0.13410840928554535, 0.028282659128308296, -0.12082239985466003, 0.21936850249767303, -0.06512433290481567, 0.16812698543071747, -0.03302834928035736, -0.18088334798812866, -0.04598559811711311, -0.014739632606506348, 0.06391328573226929, -0.15650317072868347, -0.13678401708602905, 0.01166679710149765, -0.13967938721179962, 0.14632365107536316, 0.025218486785888672, 0.046839646995067596, 0.09690812975168228, -0.13414686918258667, -0.2883925437927246, -0.1435326784849167, -0.17896348237991333, 0.10746842622756958, -0.09142691642045975, 0.04860316216945648, 0.031577128916978836, -0.17280976474285126, -0.059613555669784546, -0.05718057602643967, 0.0401446670293808, 0.026440180838108063, -0.017025159671902657, 0.22091664373874664, 0.024703698232769966, -0.15607595443725586, -0.0018572667613625526, -0.037675946950912476, 0.3210170865058899, 0.10884962230920792, 0.030370134860277176, 0.056784629821777344, -0.030112050473690033, 0.023124486207962036, -0.1449904441833496, 0.08885903656482697, 0.17527811229228973, 0.08804896473884583, 0.038310401141643524, -0.01704210229218006, -0.17355971038341522, -0.018237406387925148, 0.030551932752132416, -0.23085585236549377, 0.13475817441940308, 0.16338199377059937, -0.06968289613723755, -0.04330683499574661, 0.04434924200177193, 0.22637797892093658, 0.07463733851909637, -0.15070196986198425, -0.07500549405813217, 0.10863590240478516, -0.22288714349269867, 0.0010778247378766537, 0.057608842849731445, -0.12828609347343445, -0.17236559092998505, -0.23064571619033813, 0.09910193085670471, 0.46647992730140686, 0.0634111613035202, -0.13985536992549896, 0.052741192281246185, -0.1558966338634491, 0.022585246711969376, 0.10514408349990845, 0.11794176697731018, -0.06241249293088913, 0.06389056891202927, -0.14145469665527344, 0.060088545083999634, 0.09667345881462097, -0.004665130749344826, -0.07927791774272919, 0.21978208422660828, -0.0016187895089387894, 0.04876316711306572, 0.03137822449207306, 0.08962501585483551, -0.09108036011457443, -0.01795950159430504, -0.04094596579670906, 0.03533276170492172, 0.01394269522279501, -0.08244197070598602, -0.05095399543642998, 0.04305890575051308, -0.1195211187005043, 0.16731074452400208, 0.03894471749663353, -0.0222858227789402, -0.07944411784410477, 0.0614166259765625, -0.1481470763683319, -0.09113290905952454, 0.14758692681789398, -0.24051085114479065, 0.164126917719841, 0.1753545105457306, -0.003193420823663473, 0.20875433087348938, 0.03357946127653122, 0.1259773075580597, -0.00022807717323303223, -0.039092566817998886, -0.13582147657871246, -0.01937306858599186, 0.015938198193907738, 0.00787206832319498, 0.05792934447526932, 0.03294186294078827]
array_2 = [-0.1966051608324051, 0.0940953716635704, -0.0031937970779836178, -0.03691547363996506, -0.07240629941225052, -0.07114037871360779, -0.07133384048938751, -0.1283963918685913, 0.15377545356750488, -0.091400146484375, 0.10803385823965073, -0.09235749393701553, -0.1866973638534546, -0.021168243139982224, -0.09094691276550293, 0.07300164550542831, -0.20971564948558807, -0.1847742646932602, -0.009817334823310375, -0.05971141159534454, 0.09904412180185318, 0.0278592761605978, -0.012554554268717766, 0.09818517416715622, -0.1747943013906479, -0.31632938981056213, -0.0864541232585907, -0.13249783217906952, 0.002135572023689747, -0.04935726895928383, 0.010047778487205505, 0.04549024999141693, -0.26334646344184875, -0.05263081565499306, -0.013573898002505302, 0.2042253464460373, 0.06646320968866348, 0.08540669083595276, 0.12267164140939713, -0.018634958192706108, -0.19135263562202454, 0.01208433136343956, 0.09216200560331345, 0.2779296934604645, 0.1531585156917572, 0.10681629925966263, -0.021275708451867104, -0.059720948338508606, 0.06610126793384552, -0.21058350801467896, 0.005440462380647659, 0.18833838403224945, 0.08883830159902573, 0.025969548150897026, 0.0337764173746109, -0.1585341989994049, 0.02370697632431984, 0.10416869819164276, -0.19022507965564728, 0.11423652619123459, 0.09144753962755203, -0.08765758574008942, -0.0032832929864525795, -0.0051014479249715805, 0.19875964522361755, 0.07349056005477905, -0.1031823456287384, -0.10447365045547485, 0.11358538269996643, -0.24666038155555725, -0.05960353836417198, 0.07124857604503632, -0.039664581418037415, -0.20122921466827393, -0.31481748819351196, -0.006801256909966469, 0.41940364241600037, 0.1236235573887825, -0.12495145946741104, 0.12580059468746185, -0.02020396664738655, -0.03004150651395321, 0.11967054009437561, 0.09008713811635971, -0.07470540702342987, 0.09324200451374054, -0.13763070106506348, 0.07720538973808289, 0.19568027555942535, 0.036567769944667816, 0.030284458771348, 0.14119629561901093, -0.03820852190256119, 0.06232285499572754, 0.036639824509620667, 0.07704029232263565, -0.12276224792003632, -0.0035170004703104496, -0.13103705644607544, 0.027697769924998283, -0.01527332328259945, -0.04027168080210686, -0.03659897670149803, 0.03330300375819206, -0.12293602526187897, 0.09043421596288681, -0.019673841074109077, -0.07563626766204834, -0.13991905748844147, 0.014788001775741577, -0.07630413770675659, 0.00017269013915210962, 0.16345393657684326, -0.25710681080818176, 0.19869503378868103, 0.19393865764141083, -0.07422225922346115, 0.19553625583648682, 0.09189949929714203, 0.051557887345552444, -0.0008843056857585907, -0.006250975653529167, -0.1680600494146347, -0.10320111364126205, 0.03232177346944809, -0.08931156992912292, 0.11964476853609085, 0.00814182311296463]

以上数组的协方差矩阵是奇异矩阵，因此我无法求其逆矩阵。为什么会变成奇异矩阵？

编辑2：解决方法

由于这里的协方差矩阵是奇异矩阵，我必须使用np.linalg.pinv(V)进行伪逆操作。

- Parthapratim Neog

3

IV 应该是从采样向量的 128 维分布的协方差矩阵的逆矩阵。你可以使用 V = np.cov(np.array([array_1, array_2]).T) 来估算协方差矩阵（在 np.cov 中，行是变量，列是观测值），但这只能使用这两个样本。理想情况下，你应该使用更好的估计方法，例如整个数据集的协方差矩阵。 - jdehesa

@jdehesa 我没有从中提取128维数组的分布，这些值是从图像生成的。基本上是面部的编码。当我尝试使用你上面提到的.T时，我在尝试反转协方差矩阵时遇到了错误，即numpy.linalg.linalg.LinAlgError: Singular matrix。 - Parthapratim Neog

在计算协方差之前，您只需要对数组进行转置即可。而且@jdehesa是正确的，从两个观测值计算协方差是一个坏主意。首先，您应该使用整个图像计算cov。然后，您可以使用相同的协方差矩阵找到任意两行之间的马氏距离。 - tel

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- tel · Accepted Answer

从numpy.cov文档中可以看到，第一个参数应该是一个数组m，其中：

矩阵m的每一行代表一个变量，每一列代表所有这些变量的单个观测值。

因此，在调用cov之前，请将您的数组转置（使用.T）以修复您的代码：

V = np.cov(np.array([array_1, array_2]).T)
IV = np.linalg.inv(V)
print(mahalanobis(array_1, array_2, IV))

我刚刚在一些随机数据上进行了测试，可以确认它有效。

此外，仅从两个观测值计算协方差是一个不好的想法，也不太可能非常准确。如果您的数据来自图像，则在计算协方差矩阵时应使用整个图像 img（或至少感兴趣的整个区域），然后使用该矩阵来找到两个感兴趣向量之间的马氏距离：

V = np.cov(np.array(img))
IV = np.linalg.inv(V)
print(mahalanobis(array_1, array_2, IV))

根据您生成array_1和array_2的方式，您可能需要将img替换为img.T，也可能不需要。

如果您得到的是奇异协方差矩阵，那么您面临的是一个数学问题，而不是代码问题。显然，这是一个常见的问题，已经有人提出了问题“为什么我的协方差矩阵是奇异的？”并得到了解答。非常广泛地说，当足够多的数据点在某种意义上“太相似”时，它似乎会发生。我想使用只有两个数据点也会增加这种情况发生的可能性。