【问题标题】:image information along a polar coordinate system沿极坐标系的图像信息
【发布时间】:2011-04-17 10:25:06
【问题描述】:

我有一组 png 图像,我想用 Python 和相关工具进行处理。每个图像代表一个已知尺寸的物理对象。

在每幅图像中,在某个像素/物理位置处都有对象的特定特征。每张图片的位置都不同。

我想在给定图像上施加一个极坐标系,原点位于该特征的位置。

然后我希望能够获得以下信息: - 图像强度作为给定极角的径向位置的函数 - 当值在所有极角上取平均值时,图像强度作为径向位置的函数。

我在 Python 编程以及 NumPy 和 SciPy 中的许多函数的使用方面经验丰富,但在图像分析方面我完全是新手。

如果您能就解决此问题的可能方法向我提供任何建议,我将不胜感激。

谢谢。

【问题讨论】:

  • 从你的描述来看,这一切似乎都是数组数学而不是图像处理,你说你熟悉Python中的数组数学,那么你想回答什么?也就是说,你所描述的并不是一个真正的图像处理特定问题,就像你没有试图找到对象、边缘、模糊或锐化等。你想知道如何将图像放入 Numpy(这将是我的方法的第一步)?

标签: python image-processing numpy scipy


【解决方案1】:

您所描述的并不完全是传统意义上的图像处理,但使用 numpy 等很容易做到。

这是一个相当大的示例,执行您提到的一些事情以使您指向正确的方向...请注意,示例图像都显示图像中心原点的结果,但函数采用原点论据,因此您应该能够直接根据您的目的调整事物。

import numpy as np
import scipy as sp
import scipy.ndimage

import Image

import matplotlib.pyplot as plt

def main():
    im = Image.open('mri_demo.png')
    im = im.convert('RGB')
    data = np.array(im)

    plot_polar_image(data, origin=None)
    plot_directional_intensity(data, origin=None)

    plt.show()

def plot_directional_intensity(data, origin=None):
    """Makes a cicular histogram showing average intensity binned by direction
    from "origin" for each band in "data" (a 3D numpy array). "origin" defaults
    to the center of the image."""
    def intensity_rose(theta, band, color):
        theta, band = theta.flatten(), band.flatten()
        intensities, theta_bins = bin_by(band, theta)
        mean_intensity = map(np.mean, intensities)
        width = np.diff(theta_bins)[0]
        plt.bar(theta_bins, mean_intensity, width=width, color=color)
        plt.xlabel(color + ' Band')
        plt.yticks([])

    # Make cartesian coordinates for the pixel indicies
    # (The origin defaults to the center of the image)
    x, y = index_coords(data, origin)

    # Convert the pixel indices into polar coords.
    r, theta = cart2polar(x, y)

    # Unpack bands of the image
    red, green, blue = data.T

    # Plot...
    plt.figure()

    plt.subplot(2,2,1, projection='polar')
    intensity_rose(theta, red, 'Red')

    plt.subplot(2,2,2, projection='polar')
    intensity_rose(theta, green, 'Green')

    plt.subplot(2,1,2, projection='polar')
    intensity_rose(theta, blue, 'Blue')

    plt.suptitle('Average intensity as a function of direction')

def plot_polar_image(data, origin=None):
    """Plots an image reprojected into polar coordinages with the origin
    at "origin" (a tuple of (x0, y0), defaults to the center of the image)"""
    polar_grid, r, theta = reproject_image_into_polar(data, origin)
    plt.figure()
    plt.imshow(polar_grid, extent=(theta.min(), theta.max(), r.max(), r.min()))
    plt.axis('auto')
    plt.ylim(plt.ylim()[::-1])
    plt.xlabel('Theta Coordinate (radians)')
    plt.ylabel('R Coordinate (pixels)')
    plt.title('Image in Polar Coordinates')

def index_coords(data, origin=None):
    """Creates x & y coords for the indicies in a numpy array "data".
    "origin" defaults to the center of the image. Specify origin=(0,0)
    to set the origin to the lower left corner of the image."""
    ny, nx = data.shape[:2]
    if origin is None:
        origin_x, origin_y = nx // 2, ny // 2
    else:
        origin_x, origin_y = origin
    x, y = np.meshgrid(np.arange(nx), np.arange(ny))
    x -= origin_x
    y -= origin_y
    return x, y

def cart2polar(x, y):
    r = np.sqrt(x**2 + y**2)
    theta = np.arctan2(y, x)
    return r, theta

def polar2cart(r, theta):
    x = r * np.cos(theta)
    y = r * np.sin(theta)
    return x, y


def bin_by(x, y, nbins=30):
    """Bin x by y, given paired observations of x & y.
    Returns the binned "x" values and the left edges of the bins."""
    bins = np.linspace(y.min(), y.max(), nbins+1)
    # To avoid extra bin for the max value
    bins[-1] += 1 

    indicies = np.digitize(y, bins)

    output = []
    for i in xrange(1, len(bins)):
        output.append(x[indicies==i])

    # Just return the left edges of the bins
    bins = bins[:-1]

    return output, bins

def reproject_image_into_polar(data, origin=None):
    """Reprojects a 3D numpy array ("data") into a polar coordinate system.
    "origin" is a tuple of (x0, y0) and defaults to the center of the image."""
    ny, nx = data.shape[:2]
    if origin is None:
        origin = (nx//2, ny//2)

    # Determine that the min and max r and theta coords will be...
    x, y = index_coords(data, origin=origin)
    r, theta = cart2polar(x, y)

    # Make a regular (in polar space) grid based on the min and max r & theta
    r_i = np.linspace(r.min(), r.max(), nx)
    theta_i = np.linspace(theta.min(), theta.max(), ny)
    theta_grid, r_grid = np.meshgrid(theta_i, r_i)

    # Project the r and theta grid back into pixel coordinates
    xi, yi = polar2cart(r_grid, theta_grid)
    xi += origin[0] # We need to shift the origin back to 
    yi += origin[1] # back to the lower-left corner...
    xi, yi = xi.flatten(), yi.flatten()
    coords = np.vstack((xi, yi)) # (map_coordinates requires a 2xn array)

    # Reproject each band individually and the restack
    # (uses less memory than reprojection the 3-dimensional array in one step)
    bands = []
    for band in data.T:
        zi = sp.ndimage.map_coordinates(band, coords, order=1)
        bands.append(zi.reshape((nx, ny)))
    output = np.dstack(bands)
    return output, r_i, theta_i

if __name__ == '__main__':
    main()

原图:

投影到极坐标:

强度作为图像中心方向的函数:

【讨论】:

  • 谢谢!这很有帮助。
  • @user458738,您应该接受他的回答,然后单击其左侧的复选标记。
  • 不错的答案。不过,看起来 x 轴标签是向后的。颈部在极坐标图像中显示为大约 pi/2 弧度,但在原始图像中显示为 -pi/2 弧度。
  • @JustinPeel 我知道这个答案很老了.. 但我有几个问题.. 如何从像第二张图像这样的变换图像计算从极坐标到笛卡尔坐标的逆投影,到底是什么# Make a regular (in polar space) grid based on the min and max r & theta 的使用
  • 非常感谢 Joe Kinghorn 的出色回答,为我节省了大量时间。如果有人觉得它有用,这个 reproject_image_into_polar 函数的简单“灰度”版本现在被合并到 PyAbel 包中:github.com/PyAbel/PyAbel/blob/master/abel/tools/polar.py
【解决方案2】:

这是我使用 scipy 的 geometric_transform 方法的结果:

topolar.py

import numpy as np
from scipy.ndimage.interpolation import geometric_transform

def topolar(img, order=1):
    """
    Transform img to its polar coordinate representation.

    order: int, default 1
        Specify the spline interpolation order. 
        High orders may be slow for large images.
    """
    # max_radius is the length of the diagonal 
    # from a corner to the mid-point of img.
    max_radius = 0.5*np.linalg.norm( img.shape )

    def transform(coords):
        # Put coord[1] in the interval, [-pi, pi]
        theta = 2*np.pi*coords[1] / (img.shape[1] - 1.)

        # Then map it to the interval [0, max_radius].
        #radius = float(img.shape[0]-coords[0]) / img.shape[0] * max_radius
        radius = max_radius * coords[0] / img.shape[0]

        i = 0.5*img.shape[0] - radius*np.sin(theta)
        j = radius*np.cos(theta) + 0.5*img.shape[1]
        return i,j

    polar = geometric_transform(img, transform, order=order)

    rads = max_radius * np.linspace(0,1,img.shape[0])
    angs = np.linspace(0, 2*np.pi, img.shape[1])

    return polar, (rads, angs)

下面是一些测试用法:

testpolar.py

from topolar import topolar
from skimage.data import chelsea

import matplotlib.pyplot as plt

img = chelsea()[...,0] / 255.
pol, (rads,angs) = topolar(img)

fig,ax = plt.subplots(2,1,figsize=(6,8))

ax[0].imshow(img, cmap=plt.cm.gray, interpolation='bicubic')

ax[1].imshow(pol, cmap=plt.cm.gray, interpolation='bicubic')

ax[1].set_ylabel("Radius in pixels")
ax[1].set_yticks(range(0, img.shape[0]+1, 50))
ax[1].set_yticklabels(rads[::50].round().astype(int))

ax[1].set_xlabel("Angle in degrees")
ax[1].set_xticks(range(0, img.shape[1]+1, 50))
ax[1].set_xticklabels((angs[::50]*180/3.14159).round().astype(int))

plt.show()

...和输出:

【讨论】:

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