如何在python中对存储xy坐标的二维数组进行排序？

Question

从一个包含 1 和 0 的数组中，我试图获取该数组的边界并绘制它。这是我用来获取边界的代码

import numpy as np
import math
import matplotlib.pyplot as plt


binI = np.array([[0,0,0,0,1,0,0,0,0,0,0],
                 [0,0,0,1,1,1,1,0,0,0,0],
                 [0,0,1,1,1,1,1,0,0,0,0],
                 [0,1,1,1,1,0,0,0,0,0,0],
                 [0,1,1,1,0,0,0,0,0,0,0],
                 [0,0,1,1,1,1,0,0,0,0,0],
                 [0,0,0,1,1,1,0,0,0,0,0],
                 [0,0,0,1,1,0,0,0,0,0,0]])


def boundary_Tracer(arr):
    indices_list = []
    for i in range (np.shape(arr)[0]):
        for j in range (np.shape(arr)[1]):
            if arr[i,j] == 1:
                if i == 0 and j == 0:
                    if (arr[i+1,j] == 0) or (arr[i,j+1] == 0):
                        indices_list.append([i,j])
                elif (i == 0) and (j == np.shape(arr)[1]-1):
                    if (arr[i,j-1] == 0) or (arr[i+1,j] == 0):
                        indices_list.append([i,j])
                elif (i == (np.shape(arr)[0]-1)) and j == 0:
                    if (arr[i-1,j] == 0) or (arr[i,j+1] == 0):
                        indices_list.append([i,j])
                elif (i == np.shape(arr)[0]-1) and (j == np.shape(arr)[1]-1):
                    if (arr[i-1,j] == 0) or (arr[i,j-1] == 0):
                        indices_list.append([i,j])
                elif (i in range (1,np.shape(arr)[0]-1)) and (j == 0):
                    if (arr[i-1,j] == 0) or (arr[i,j+1] == 0) or (arr[i+1,j] == 0):
                        indices_list.append([i,j])
                elif (i in range (1,np.shape(arr)[0]-1)) and (j == np.shape(arr)[1]-1):
                    if (arr[i-1,j] == 0) or (arr[i,j-1] == 0) or (arr[i+1,j] == 0):
                        indices_list.append([i,j])
                elif (i == 0) and (j in range (1,np.shape(arr)[1]-1)):
                    if (arr[i,j-1] == 0) or (arr[i,j+1] == 0) or (arr[i+1,j] == 0):
                        indices_list.append([i,j])          
                elif (i == np.shape(arr)[0]-1) and (j in range (1,np.shape(arr)[1]-1)):
                    if (arr[i-1,j] == 0) or (arr[i,j-1] == 0) or (arr[i,j+1] == 0):
                        indices_list.append([i,j])
                else:
                    if (arr[i-1,j] == 0) or (arr[i+1,j] == 0) or (arr[i,j-1] == 0) or (arr[i,j+1] == 0):
                        indices_list.append([i,j])
                
    indicies_array = np.array(indices_list)
    x_all = indicies_array[:,1]
    x_init_bw = np.min(np.where(x_all == np.min(x_all)))
    origin = np.reshape(indicies_array[x_init_bw,:],(1,2))[0]
    indicies_array = np.vstack((indicies_array,origin))
    
    return indicies_array, origin

bw = boundary_Tracer(binI)[0]
origin = boundary_Tracer(binI)[1]

plt.plot(bw[:,1],bw[:,0])
plt.gca().invert_yaxis()

我知道这很丑，我确信有更好的方法来做到这一点，但这是我最好的。无论如何，当我绘制这个时，情节在点之间曲折。我想让情节只是连接边界而不跨越标有 1 的区域的中间。

用边界的 xy 坐标（bw）重新排列数组的最佳方法是什么？

Answer 1

您可以将数据转换为原点，并使用复杂的角度对数据进行排序

bw = boundary_Tracer(binI)[0]
origin = boundary_Tracer(binI)[1]

xs = (bw - bw.mean(0))
x_sort = bw[np.angle((xs[:,0] + 1j*xs[:,1])).argsort()]

# Plot to test your trace
plt.imshow(binI, cmap='gray')
plt.plot(x_sort[:,1],x_sort[:,0])
plt.gca().invert_yaxis()
plt.axis('off');

输出

这个顺时针的解决方案不是想要的结果。不在凸包上的点被排错了位置。

一个可用的排序来证明问题不在您的数据中，例如

x_sort = bw[[18,  7,  4,  1,  0,  2,  3,  6,  5,  8, 10, 12, 13, 15, 17, 16, 14, 11,  9]]
plt.imshow(binI, cmap='gray')
plt.plot(x_sort[:,1],x_sort[:,0], 'ro--')
plt.gca().invert_yaxis()
plt.axis('off');

输出

有'star shaped' 非凸形状的算法，但no general unique solution 没有更多信息。