基于Zernike矩的mahotas和opencv图像重建

Question

我听说过本教程后的mahotas ,希望在python中找到一个很好的Zernike多项式实现.这可不容易.但是,我需要比较原始图像和Zernike矩重建的图像之间的欧几里德差异.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时刻在OpenCV中重建图像？

Answer 1

基于他在答案中提到的代码,我开发了以下代码用于重建.我还发现了研究论文[ A. Khotanzad和YH Hong,"Zernike时刻的不变图像识别" ]和[ S.-K.黄和W.-Y. 金,"一种快速计算泽尼克时刻的新方法"非常有用.

函数_slow_zernike_poly构造了2-D 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

使用顺序12的重建图像的实部

'''
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')