将 0/1 矩阵转换为 python 中的二维网格图

Question

给定一个包含 0/1 矩阵的 .txt 文件，在 python.I 中实现网格图应该为该矩阵创建一个网格图，但问题是我应该在那些有 0 的部分创建网格和块中有 1s 的部分。所以我的 txt 文件包含这个：

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1
0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1
0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1
1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

我应该根据上面的矩阵创建波纹管网格但不知道如何:

Answer 1

我们可以用 nx.grid_2d_graph 定义一个网格图，并只保留矩阵中值为 0 的那些节点。然后我们可以使用网格图中的坐标来定位节点：

from matplotlib import pyplot as plt
import numpy as np
import networkx as nx

# lines to 2d array
with open('myfile.txt') as f:
    a = np.array([list(map(int,i.split())) for i in f.readlines()])

# define grid graph according to the shape of a
G = nx.grid_2d_graph(*a.shape)

# remove those nodes where the corresponding value is != 0
for val,node in zip(a.ravel(), sorted(G.nodes())):
    if val!=0:
        G.remove_node(node)

plt.figure(figsize=(9,9))
# coordinate rotation
pos = {(x,y):(y,-x) for x,y in G.nodes()}
nx.draw(G, pos=pos, 
        node_color='grey',
        width = 4,
        node_size=400)

将 0/1 矩阵转换为 python 中的二维网格图

convert 0/1 matrix to a 2D grid graph in python

python

grid

graph

matrix

networkx