如何将MDAnalysis与一组原子一起用于principal_axes和moment_of_inertia?

时间:2018-03-12 15:55:30

标签: python python-3.x physics chemistry mdanalysis

我正在尝试使用MDAnalysis(MDAnalysis.__version__ == 0.17.0)API函数principal_axes()moment_of_inertia()来计算  这些矩阵用于doc

中描述的一组选定原子
import MDAnalysis
from MDAnalysis.tests.datafiles import PSF, DCD
import numpy as np

u = MDAnalysis.Universe(PSF, DCD)

CA = u.select_atoms("protein and name CA")

I = np.matrix(CA.moment_of_inertia())
U = np.matrix(CA.principal_axes())
print("center of mass", CA.center_of_mass())
print("moment of inertia", I)
print("principal axes", U)
print("Lambda = U'IU", np.transpose(U)*I*U)

输出:

center of mass [ 0.06873595 -0.04605918 -0.24643682]
moment of inertia [[ 393842.2070687     -963.01376005   -6050.68541811]
 [   -963.01376005  474434.9289629    -3902.61617054]
 [  -6050.68541811   -3902.61617054  520207.91703069]]
principal axes [[-0.04680878 -0.08278738  0.99546732]
 [ 0.01813292 -0.9964659  -0.08201778]
 [-0.99873927 -0.01421157 -0.04814453]]
Lambda = U'IU [[ 519493.24344558   -4093.3268841    11620.96444297]
 [  -4093.3268841   473608.1536763     7491.56715845]
 [  11620.96444297    7491.56715845  395383.6559404 ]]

这看起来不对,其中一个原因是U'IU不是doc中提到的对角线: enter image description here

也许我需要将蛋白质与质心对齐以计算相对于惯性矩的惯性矩。

2 个答案:

答案 0 :(得分:1)

问题是他们在文档中说U'IU,但U是CA.principal_axes()结果的转置,请参阅source code

    # Sort
    indices = np.argsort(e_val)[::-1]
    # Return transposed in more logical form. See Issue 33.
    return e_vec[:, indices].T

Matlab确认:

>> I=[ 393842.2070687     -963.01376005   -6050.68541811 ;  -963.01376005  474434.9289629    -3902.61617054;  -6050.68541811   -3902.61617054  520207.91703069];
>> U=[-0.04680878 -0.08278738  0.99546732; 0.01813292 -0.9964659  -0.08201778;-0.99873927 -0.01421157 -0.04814453];
>> U*I*U'

ans =

   1.0e+05 *

    5.2082    0.0000   -0.0000
    0.0000    4.7413   -0.0000
   -0.0000   -0.0000    3.9354

答案 1 :(得分:1)

AtomGroup.principal_axes()教程中的文档原则上是正确的,但令人困惑的是AtomGroup.principal_axes()的返回值不是矩阵U,而是其转置,U.T

AtomGroup.principal_axes()方法返回一个数组[p1, p2, p3],其中主轴p1p2p3是长度为3的数组;为方便起见,选择了 row 向量的布局(以便可以使用p1, p2, p3 = ag.principal_axes()提取向量)。要形成矩阵U,其中主轴是列向量,就像在主轴的通常处理中一样,必须进行转置。例如:

import MDAnalysis
from MDAnalysis.tests.datafiles import PSF, DCD
import numpy as np

u = MDAnalysis.Universe(PSF, DCD)

CA = u.select_atoms("protein and name CA")

I = CA.moment_of_inertia()
UT = CA.principal_axes()

# transpose the row-vector layout UT = [p1, p2, p3]
U = UT.T

# test that U diagonalizes I
Lambda = U.T.dot(I.dot(U))

print(Lambda)

# check that it is diagonal (to machine precision)
print(np.allclose(Lambda - np.diag(np.diagonal(Lambda)), 0))

矩阵Lambda应为对角线(最后print应显示True):

[[ 5.20816990e+05 -6.56706349e-10 -2.83491351e-12]
[-6.62283524e-10  4.74131234e+05 -2.06979926e-11]
[-6.56687024e-12 -2.07159142e-11  3.93536829e+05]]
True

最后,如果你想“手工”计算:

values, evecs = np.linalg.eigh(I)
indices = np.argsort(values)
U = evecs[:, indices]

这使U的主轴为列向量。