python – 从numpy数组中计算Ellipse中的像素

我想我可能在我的一个涉及劳厄衍射模式(材料科学)的研究项目中解决了与你类似的问题.我的任务是在给定每个峰的中心坐标的情况下找到衍射图中每个峰的长度,宽度和倾斜角.我的解决方案是在峰值和阈值,滤波器等周围选择感兴趣的区域,使得子图像中只有一个峰值.然后我构建了一个函数来将这些参数拟合到生成的椭圆中：

from xml.etree.cElementTree import parse
import numpy as np
from os import listdir,getcwd
from scipy import ndimage
import h5py
import multiprocessing
import time
#from dicttoxml import dicttoxml
#from xml.dom.minidom import parseString
import sys
import cPickle as pickle
from threading import Thread
from skimage.measure import moments

def fitEllipse(data):
    '''
    Returns the length of the long and short axis and the angle measure
    of the long axis to the horizontal of the best fit ellipsebased on
    image moments.

    usage: longAxis,shortAxis,angle = fitEllipse(N_by_M_image_as_array)
    '''
    # source:
    #     Kieran F. Mulchrone,Kingshuk Roy Choudhury,# Fitting an ellipse to an arbitrary shape:
    # implications for strain analysis,Journal of
    # Structural Geology,Volume 26,Issue 1,# January 2004,Pages 143-153,ISSN 0191-8141,# <http://dx.doi.org/10.1016/S0191-8141(03)00093-2.>
    #     Lourena Rocha,Luiz Velho,Paulo Cezar P. Carvalho
    # Image Moments-Based Structuring and Tracking of
    # Objects,IMPA-Instituto Nacional de Matematica Pura
    # e Aplicada. Estrada Dona Castorina,110,22460
    # Rio de Janeiro,RJ,Brasil,# <http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1167130>

    m = moments(data,2) # super fast compated to anything in pure python
    xc = m[1,0] / m[0,0]
    yc = m[0,1] / m[0,0]
    a = (m[2,0]) - (xc**2)
    b = 2 * ((m[1,0]) - (xc * yc))
    c = (m[0,2] / m[0,0]) - (yc**2)
    theta = .5 * (np.arctan2(b,(a - c)))
    w = np.sqrt(6 * (a + c - np.sqrt(b**2 + (a-c)**2)))
    l = np.sqrt(6 * (a + c + np.sqrt(b**2 + (a-c)**2)))
    return l,w,theta

我只是把它扔在一起,以防它与你想要的类似.如果您需要更多解释,请随时发表评论.我使用的来源(数学)在评论中.