标签：covariance python numpy pca
我正在使用3维numpy数组,最终将在其上执行PCA.我首先将3-D数组展平为2-D,以便可以计算协方差(然后是特征值和特征向量).
在计算协方差矩阵时,我使用numpy.cov与numpy.dot得出了不同的结果.如果我的二维数组是(5,9),我想得到一个5×5(即NxN)协方差矩阵.这是我使用numpy.dot获得的.使用numpy.cov,我得到的协方差矩阵为9×9.这与我所需的形状不符,但说实话,我不知道哪一个是正确的.在我研究的示例中,我已经看到了两种用于计算协方差的方法.
如果我通过numpy.linalg.eig计算携带numpy.dot vs. numpy.cov,我显然会得到不同的答案(全部打印在示例输出中).因此,在这一点上,我对于哪种方法正确或我可能在哪里出错感到非常困惑.
这是带有输出的测试代码.谢谢你的帮助.
import numpy as np
a = np.random.random(((5,3,3))); # example of what real input will look like
# create 2D flattened version of 3D input array
d1,d2,d3 = a.shape
b = np.zeros([d1,d2*d3])
for i in range(len(a)):
b[i] = a[i].flatten()
print "shape of 3D array: ", a.shape
print "shape of flattened 2D array: ", b.shape, "\n"
print "flattened 2D array:\n", b, "\n"
# mean-center the flattened array
b -= np.mean(b, axis=0)
# calculate the covariance matrix of the flattened array
covar1 = np.cov(b, rowvar=0) # this makes a 9x9 array
covar2 = np.dot(b, b.T) # this makes a 5x5 array
print "covariance via numpy.cov:\n", covar1, "\n"
print "covariance via numpy.dot:\n", covar2, "\n"
# calculate eigenvalues and eigenvectors
eval1, evec1 = np.linalg.eig(covar1)
eval2, evec2 = np.linalg.eig(covar2)
print "eigenvalues via numpy.cov covariance matrix:\n", eval1, "\n"
print "eigenvectors via numpy.cov covariance matrix:\n", evec1, "\n"
print "eigenvalues via numpy.dot covariance matrix:\n", eval2, "\n"
print "eigenvectors via numpy.dot covariance matrix:\n", evec2, "\n"
======= Output =======
shape of 3D array: (5, 3, 3)
shape of flattened 2D array: (5, 9)
flattened 2D array:
[[ 0.94964127 0.71015973 0.80994774 0.49727821 0.38270025 0.89136202
0.19876615 0.72461047 0.43646456]
[ 0.00502329 0.70593521 0.44001479 0.97576486 0.37261107 0.6318449
0.86301405 0.21820704 0.91507706]
[ 0.75411747 0.98462782 0.65109776 0.1083943 0.12867679 0.63172813
0.85803498 0.89507165 0.62291308]
[ 0.88589874 0.02797773 0.6421045 0.17255432 0.5713524 0.28589519
0.55888288 0.7961657 0.4453764 ]
[ 0.85774793 0.19511453 0.92167001 0.27340606 0.41849435 0.98349776
0.19354437 0.2974041 0.52064868]]
covariance via numpy.cov():
[[ 0.15180806 -0.04977355 0.05733885 -0.11340765 0.00840097 0.01461576
-0.08596712 0.07512366 -0.07509614]
[-0.04977355 0.15853367 -0.02337953 0.0357429 -0.05604085 0.02600021
0.06158462 0.0229808 0.03506849]
[ 0.05733885 -0.02337953 0.0335786 -0.03485899 0.00294469 0.03209583
-0.05378417 0.00490397 -0.02751816]
[-0.11340765 0.0357429 -0.03485899 0.12340238 0.0052609 0.0144986
0.02494029 -0.07492008 0.05109007]
[ 0.00840097 -0.05604085 0.00294469 0.0052609 0.02529647 -0.01263607
-0.02327657 -0.01136774 -0.01037048]
[ 0.01461576 0.02600021 0.03209583 0.0144986 -0.01263607 0.07415853
-0.05387152 -0.0345835 -0.00342481]
[-0.08596712 0.06158462 -0.05378417 0.02494029 -0.02327657 -0.05387152
0.11053971 0.00903926 0.04727671]
[ 0.07512366 0.0229808 0.00490397 -0.07492008 -0.01136774 -0.0345835
0.00903926 0.09436665 -0.03526195]
[-0.07509614 0.03506849 -0.02751816 0.05109007 -0.01037048 -0.00342481
0.04727671 -0.03526195 0.03900974]]
covariance via numpy.dot():
[[ 0.3211555 -0.34304471 -0.01453859 -0.1071505 0.14357829]
[-0.34304471 1.24506647 -0.11174019 -0.43907983 -0.35120174]
[-0.01453859 -0.11174019 0.57018674 -0.10412646 -0.3397815 ]
[-0.1071505 -0.43907983 -0.10412646 0.60465919 0.0456976 ]
[ 0.14357829 -0.35120174 -0.3397815 0.0456976 0.50170735]]
eigenvalues via numpy.cov covariance matrix:
[ 3.34339027e-01 +0.00000000e+00j 1.98268985e-01 +0.00000000e+00j
5.71434551e-02 +0.00000000e+00j 1.13399310e-01 +0.00000000e+00j
3.38418299e-18 +1.46714498e-17j 3.38418299e-18 -1.46714498e-17j
1.20944017e-18 +0.00000000e+00j -8.89005842e-18 +0.00000000e+00j
-6.59244508e-18 +0.00000000e+00j]
eigenvectors via numpy.cov covariance matrix:
[[-0.33898927+0.j 0.01567746+0.j -0.32410513+0.j
0.01868249+0.j 0.03901578-0.09858459j 0.03901578+0.09858459j
-0.17596347+0.j 0.08294235+0.j 0.04883282+0.j ]
[ 0.03740184+0.j -0.01106985+0.j 0.11199662+0.j
-0.36257285+0.j 0.66513867+0.j 0.66513867+0.j
0.34810753+0.j -0.05174886+0.j -0.21147240+0.j ]
[ 0.42193056+0.j 0.10153367+0.j -0.52774125+0.j
-0.57292678+0.j -0.02584078-0.15425679j -0.02584078+0.15425679j
-0.02594397+0.j -0.23132722+0.j -0.33824532+0.j ]
[-0.08723679+0.j -0.17700647+0.j -0.04490487+0.j
0.14531440+0.j -0.08669754+0.21485879j -0.08669754-0.21485879j
-0.73208352+0.j 0.04474123+0.j -0.09159437+0.j ]
[-0.26991334+0.j 0.39182156+0.j 0.18023454+0.j
-0.14727224+0.j -0.21261400+0.1100362j -0.21261400-0.1100362j
0.15211635+0.j 0.54168898+0.j -0.36386803+0.j ]
[-0.39361702+0.j 0.48389127+0.j 0.12668909+0.j
0.07739853+0.j 0.31569702-0.34166187j 0.31569702+0.34166187j
0.11287735+0.j -0.74889136+0.j -0.42472067+0.j ]
[-0.29962418+0.j -0.01577641+0.j 0.35742257+0.j
-0.68969822+0.j -0.28182091+0.13998238j -0.28182091-0.13998238j
-0.40124817+0.j 0.06419507+0.j 0.47506061+0.j ]
[-0.57032501+0.j -0.60505095+0.j -0.30688172+0.j
-0.11823642+0.j 0.07618472-0.0915626j 0.07618472+0.0915626j
0.32272841+0.j -0.10872383+0.j -0.25867852+0.j ]
[-0.23498699+0.j 0.45164240+0.j -0.57569388+0.j
0.03856674+0.j -0.07478874+0.27512969j -0.07478874-0.27512969j
-0.10101603+0.j 0.25440413+0.j 0.47403650+0.j ]]
eigenvalues via numpy.dot covariance matrix:
[ 1.33735611e+00 7.93075942e-01 2.08276008e-16 4.53597239e-01
2.28573820e-01]
eigenvectors via numpy.dot covariance matrix:
[[ 0.1223889 -0.87441162 -0.4472136 -0.13172011 0.05545353]
[-0.54658696 0.08157704 -0.4472136 0.61361759 0.34360056]
[ 0.70163289 0.24699239 -0.4472136 0.41717057 -0.26958257]
[-0.41754523 0.17603863 -0.4472136 -0.33135976 -0.69632398]
[ 0.1401104 0.36980356 -0.4472136 -0.56770828 0.56685246]]
解决方法:
np.dot只是两个矩阵的矩阵乘积.那不是协方差.为什么使用rowvar = 0？如果只是执行np.cov(b),它将给出正确尺寸的矩阵.
标签：covariance,python,numpy,pca 来源： https://codeday.me/bug/20191031/1979214.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。