Question

在NumPy / SciPy或相关库中是否存在类似Matlab的procrustes函数的内容？

供参考。 Procrustes分析旨在通过去除刻度，平移和旋转扭曲分量来对齐2组点（换句话说，2个形状）以最小化它们之间的方形距离。

Matlab中的示例：

X = [0 1; 2 3; 4 5; 6 7; 8 9];   % first shape
R = [1 2; 2 1];                  % rotation matrix
t = [3 5];                       % translation vector
Y = X * R + repmat(t, 5, 1);     % warped shape, no scale and no distortion
[d Z] = procrustes(X, Y);        % Z is Y aligned back to X
Z

Z =

  0.0000    1.0000
  2.0000    3.0000
  4.0000    5.0000
  6.0000    7.0000
  8.0000    9.0000

NumPy中的相同任务：

X = arange(10).reshape((5, 2))
R = array([[1, 2], [2, 1]])
t = array([3, 5])
Y = dot(X, R) + t
Z = ???

注意：我只对对齐的形状感兴趣，因为方形错误（Matlab代码中的变量d）很容易从2个形状计算出来。

Answer 1

我不知道Python中预先存在的任何实现，但很容易使用edit procrustes.m查看MATLAB代码并将其移植到Numpy：

def procrustes(X, Y, scaling=True, reflection='best'):
    """
    A port of MATLAB's `procrustes` function to Numpy.

    Procrustes analysis determines a linear transformation (translation,
    reflection, orthogonal rotation and scaling) of the points in Y to best
    conform them to the points in matrix X, using the sum of squared errors
    as the goodness of fit criterion.

        d, Z, [tform] = procrustes(X, Y)

    Inputs:
    ------------
    X, Y    
        matrices of target and input coordinates. they must have equal
        numbers of  points (rows), but Y may have fewer dimensions
        (columns) than X.

    scaling 
        if False, the scaling component of the transformation is forced
        to 1

    reflection
        if 'best' (default), the transformation solution may or may not
        include a reflection component, depending on which fits the data
        best. setting reflection to True or False forces a solution with
        reflection or no reflection respectively.

    Outputs
    ------------
    d       
        the residual sum of squared errors, normalized according to a
        measure of the scale of X, ((X - X.mean(0))**2).sum()

    Z
        the matrix of transformed Y-values

    tform   
        a dict specifying the rotation, translation and scaling that
        maps X --> Y

    """

    n,m = X.shape
    ny,my = Y.shape

    muX = X.mean(0)
    muY = Y.mean(0)

    X0 = X - muX
    Y0 = Y - muY

    ssX = (X0**2.).sum()
    ssY = (Y0**2.).sum()

    # centred Frobenius norm
    normX = np.sqrt(ssX)
    normY = np.sqrt(ssY)

    # scale to equal (unit) norm
    X0 /= normX
    Y0 /= normY

    if my < m:
        Y0 = np.concatenate((Y0, np.zeros(n, m-my)),0)

    # optimum rotation matrix of Y
    A = np.dot(X0.T, Y0)
    U,s,Vt = np.linalg.svd(A,full_matrices=False)
    V = Vt.T
    T = np.dot(V, U.T)

    if reflection is not 'best':

        # does the current solution use a reflection?
        have_reflection = np.linalg.det(T) < 0

        # if that's not what was specified, force another reflection
        if reflection != have_reflection:
            V[:,-1] *= -1
            s[-1] *= -1
            T = np.dot(V, U.T)

    traceTA = s.sum()

    if scaling:

        # optimum scaling of Y
        b = traceTA * normX / normY

        # standarised distance between X and b*Y*T + c
        d = 1 - traceTA**2

        # transformed coords
        Z = normX*traceTA*np.dot(Y0, T) + muX

    else:
        b = 1
        d = 1 + ssY/ssX - 2 * traceTA * normY / normX
        Z = normY*np.dot(Y0, T) + muX

    # transformation matrix
    if my < m:
        T = T[:my,:]
    c = muX - b*np.dot(muY, T)

    #transformation values 
    tform = {'rotation':T, 'scale':b, 'translation':c}

    return d, Z, tform

Answer 2

它有一个Scipy函数：scipy.spatial.procrustes

我只是在这里发布它的例子：

>>> import numpy as np
>>> from scipy.spatial import procrustes

>>> a = np.array([[1, 3], [1, 2], [1, 1], [2, 1]], 'd')
>>> b = np.array([[4, -2], [4, -4], [4, -6], [2, -6]], 'd')
>>> mtx1, mtx2, disparity = procrustes(a, b)
>>> round(disparity)
0.0

Answer 3

您可以在 void display(void) { glColor3f(1, 0, 1); glMatrixMode(GL_PROJECTION); glLoadIdentity(); gluOrtho2D(0.0, 640.0, 0.0, 480.0); for (int k = 0; k <= i; k++) { glBegin(GL_LINES); glVertex2f(x[k], y[k]); glVertex2f(x[k + 1], y[k + 1]); glEnd(); } glFlush(); // flushes the frame buffer to the screen } int main(int argc, char** argv) { glutInit(&argc, argv); glutInitWindowSize(640, 480); glutInitWindowPosition(10, 10); glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB); glutCreateWindow("Line Drawing"); glClearColor(1, 1, 1, 0); glClear(GL_COLOR_BUFFER_BIT); glutDisplayFunc(display); glutMouseFunc(mouse); glutMainLoop(); } 中使用普通 Procrustes 分析和广义 Procrustes 分析，如下所示：

python

出于测试目的，每个算法的输出可以可视化如下：

import numpy as np
        
def opa(a, b):
    aT = a.mean(0)
    bT = b.mean(0)
    A = a - aT 
    B = b - bT
    aS = np.sum(A * A)**.5
    bS = np.sum(B * B)**.5
    A /= aS
    B /= bS
    U, _, V = np.linalg.svd(np.dot(B.T, A))
    aR = np.dot(U, V)
    if np.linalg.det(aR) < 0:
        V[1] *= -1
        aR = np.dot(U, V)
    aS = aS / bS
    aT-= (bT.dot(aR) * aS)
    aD = (np.sum((A - B.dot(aR))**2) / len(a))**.5
    return aR, aS, aT, aD 
        
def gpa(v, n=-1):
    if n < 0:
        p = avg(v)
    else:
        p = v[n]
    l = len(v)
    r, s, t, d = np.ndarray((4, l), object)
    for i in range(l):
        r[i], s[i], t[i], d[i] = opa(p, v[i]) 
    return r, s, t, d

def avg(v):
    v_= np.copy(v)
    l = len(v_) 
    R, S, T = [list(np.zeros(l)) for _ in range(3)]
    for i, j in np.ndindex(l, l):
        r, s, t, _ = opa(v_[i], v_[j]) 
        R[j] += np.arccos(min(1, max(-1, np.trace(r[:1])))) * np.sign(r[1][0]) 
        S[j] += s 
        T[j] += t 
    for i in range(l):
        a = R[i] / l
        r = [np.cos(a), -np.sin(a)], [np.sin(a), np.cos(a)]
        v_[i] = v_[i].dot(r) * (S[i] / l) + (T[i] / l) 
    return v_.mean(0)

Answer 4

可能您想尝试使用不同 Procrustes 方法的各种口味的这个包，https://github.com/theochem/procrustes。

使用NumPy进行Procrustes分析？

4 个答案: