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我在本教程之后听说了 mahotas ，希望在 python 中找到 Zernike 多项式的良好实现。这再简单不过了。但是，我需要比较原始图像和从 Zernike 矩重建的图像之间的欧几里得差异。我问mahotas 的作者是否可以将重建功能添加到他的库中，但他没有时间构建它。

如何使用 mahotas 提供的 Zernike 矩在 OpenCV 中重建图像？

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2 回答 2

8

根据fireant在他的回答中提到的代码，我开发了以下代码进行重建。我还找到了研究论文 [ A. Khotanzad 和 YH Hong，“Zernike 矩的不变图像识别”</a>] 和 [ S.-K. 黄和 W.-Y。Kim，“一种快速计算 Zernike 矩的新方法”</a>] 非常有用。

函数_slow_zernike_poly构造二维 Zernike 基函数。在zernike_reconstruct函数中，我们将图像投影到_slow_zernike_poly返回的基函数上并计算矩。然后我们使用重建公式。

下面是使用此代码完成的示例重建：

输入图像

输入

输入喷射

使用 12 阶重建图像的实部

重建订单 12

'''
Copyright (c) 2015
Dhanushka Dangampola <dhanushkald@gmail.com>

Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in
all copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
THE SOFTWARE.
'''

import numpy as np
from math import atan2
from numpy import cos, sin, conjugate, sqrt

def _slow_zernike_poly(Y,X,n,l):
    def _polar(r,theta):
        x = r * cos(theta)
        y = r * sin(theta)
        return 1.*x+1.j*y

    def _factorial(n):
        if n == 0: return 1.
        return n * _factorial(n - 1)
    y,x = Y[0],X[0]
    vxy = np.zeros(Y.size, dtype=complex)
    index = 0
    for x,y in zip(X,Y):
        Vnl = 0.
        for m in range( int( (n-l)//2 ) + 1 ):
            Vnl += (-1.)**m * _factorial(n-m) /  \
                ( _factorial(m) * _factorial((n - 2*m + l) // 2) * _factorial((n - 2*m - l) // 2) ) * \
                ( sqrt(x*x + y*y)**(n - 2*m) * _polar(1.0, l*atan2(y,x)) )
        vxy[index] = Vnl
        index = index + 1

    return vxy

def zernike_reconstruct(img, radius, D, cof):

    idx = np.ones(img.shape)

    cofy,cofx = cof
    cofy = float(cofy)
    cofx = float(cofx)
    radius = float(radius)    

    Y,X = np.where(idx > 0)
    P = img[Y,X].ravel()
    Yn = ( (Y -cofy)/radius).ravel()
    Xn = ( (X -cofx)/radius).ravel()

    k = (np.sqrt(Xn**2 + Yn**2) <= 1.)
    frac_center = np.array(P[k], np.double)
    Yn = Yn[k]
    Xn = Xn[k]
    frac_center = frac_center.ravel()

    # in the discrete case, the normalization factor is not pi but the number of pixels within the unit disk
    npix = float(frac_center.size)

    reconstr = np.zeros(img.size, dtype=complex)
    accum = np.zeros(Yn.size, dtype=complex)

    for n in range(D+1):
        for l in range(n+1):
            if (n-l)%2 == 0:
                # get the zernike polynomial
                vxy = _slow_zernike_poly(Yn, Xn, float(n), float(l))
                # project the image onto the polynomial and calculate the moment
                a = sum(frac_center * conjugate(vxy)) * (n + 1)/npix
                # reconstruct
                accum += a * vxy
    reconstr[k] = accum
    return reconstr

if __name__ == '__main__':

    import cv2
    import pylab as pl
    from matplotlib import cm

    D = 12

    img = cv2.imread('fl.bmp', 0)

    rows, cols = img.shape
    radius = cols//2 if rows > cols else rows//2

    reconst = zernike_reconstruct(img, radius, D, (rows/2., cols/2.))

    reconst = reconst.reshape(img.shape)

    pl.figure(1)
    pl.imshow(img, cmap=cm.jet, origin = 'upper')
    pl.figure(2)    
    pl.imshow(reconst.real, cmap=cm.jet, origin = 'upper')
于 2015-10-26T05:56:40.217 回答
3

这并不难，我认为您可以自己编写代码。首先，请记住每个矩/矩阵的逆矩阵（也称为基础图像）是该矩阵的转置，因为它们是正交的。然后查看该库的作者用于测试该功能的代码。这比库中的代码更简单，因此您可以阅读并理解它的工作原理（当然也慢得多）。您需要为每个时刻获取那些作为基础图像的矩阵。您可以修改_slow_znl以获取在x,y主循环内计算的值，for x,y,p in zip(X,Y,P):并将其存储在与输入图像大小相同的矩阵中。将白色图像传递给_slow_zernike并使所有矩矩阵达到您想要的径向度。要使用系数重建图像，只需使用这些矩阵的转置，就像使用 Haar 变换一样。

于 2015-10-21T19:03:15.247 回答