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我正在做一个项目,该项目需要找出新图像相对于旧图像移动和旋转了多少。我正在尝试使用 fft 来实现它。但是,它适用于某些情况,但适用于其他情况。我遵循的步骤是:

  1. 取两张图片,实现canny边缘检测
  2. 使用 fft 找出班次并删除班次
  3. 将图像转换为极域并找到偏移并将答案转换为弧度

在某些情况下,我得到了正确的答案,但在其他情况下,我得到了(0,0)的移位和 0 弧度的旋转。请提出可能发生这种情况的原因。

这是代码:

class Register:
    '''
        Class for registering images based on FFT. The usage is as follows:
        >>> im0 = imread('image0.jpg', flatten = True)
        >>> im1 = imread('image1.jpg', flatten = True)
        >>> reg = Register(im0, im1)
        >>> shift = reg.shift
        >>> rotation = reg.theta

        Note:
            1. This image registration technique is not very reliable and
               is valid only for small rotation
            2. The class is very slow, since it depends on canny edge detection
               module for finding edges.
    '''
    def __init__(self,imin0, imin1, PROCESSED = False):
        '''
            This method is used to execute all the routines required to get the
            shift and the rotation
        '''
        # find edges to remove low frequency signals and suppress information
        if PROCESSED:
            im0 = imin0
            im1 = imin1
        else:
            im0 = Canny(imin0, 0.85, 5).grad
            im1 = Canny(imin1, 0.85, 5).grad

        # A major drawback of this method is that it can operate only on square
        # images. Hence we will make square image of any input image

        im0 = self.createsquareim(self.clearBorder(im0))
        im1 = self.createsquareim(self.clearBorder(im1))

        self.shift = self.findShift(im0,im1)
        imtrans = shift(im1, self.shift)
        # Remove the shift in the image. This is mandatory before we find theta
        impolar0 = self.makePolar(im0)
        impolar1 = self.makePolar(imtrans)
        self.index = self.findShift(impolar0, impolar1)[1]
        self.theta = ((self.index*90.0)/impolar1.shape[0])

    def clearBorder(self,im,width = 50, color = 255):
        '''
            A little house keeping to clear any border noise
        '''
        im[:,:width] = color
        im[:,-width:] = color
        im[:width,:] = color
        im[-width:,:] = color

        return im

    def createsquareim(self, im):
        """
        function createsquareim
        input:numpy ndarray
        output:numpy ndarray

        The function takes in an image array and converts it into square
        image by creating empty columns and rows.
        """
        lenmax = max(im.shape[0],im.shape[1])
        imout = zeros((lenmax,lenmax))
        imout[:,:] = 255
        imout[:im.shape[0],:im.shape[1]] = im
        return imout

    def findShift(self, im0, im1):
        '''
            This method is based on fft method of registering images.
        '''
        IM0 = fft2(im0)
        IM1 = fft2(im1)

        numer = IM0*conj(IM1)
        denom = abs(IM0*IM1)

        pulse_im = ifft2(numer/denom)
        mag = abs(pulse_im)
        x, y = where(mag == mag.max())

        x = array(x.tolist())   # Issues with read only arrays
        y = array(y.tolist())

        X, Y = im0.shape

        if x > X/2:
            x -= X
        if y > Y/2:
            y -= Y

        return [x[0], y[0]]

    def makePolar(self, im):
        '''
            This method will convert the cartesian coordinates image
            to polar coordinates image. The relation between the two
            domains is
              F(r,theta) = f(r*cos(theta),r*sin(theta))
            To make the process fast, we are using map_coordinates function
        '''
        m, n = im.shape
        r_max = hypot(m, n)

        r_mat = zeros_like(im)
        t_mat = zeros_like(im)

        r_mat.T[:] = linspace(0, r_max, m)
        t_mat[:] = linspace(0, pi/2, n)

        x = r_mat*cos(t_mat)
        y = r_mat*sin(t_mat)

        imout = zeros_like(im)
        imout = map_coordinates(im, [x, y], cval = 255)

        return imout
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1 回答 1

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我已经使用 SIFT 和 RANSAC 算法编写了一个用于图像配准的 python 脚本,您可以在http://github.com/vishwa91/pyimreg找到代码。此外,可以在http://cyroforge.wordpress.com/2012/07/05/image-registration-using-python/找到相同的小介绍

于 2012-07-06T17:05:43.607 回答