如何扩展此代码以包含不同数量的六边形层? 我需要一个给定层数 m 的六方晶格图。 m=1 表示 1 个边长为 1、中心位于原点的正六边形,m=2 时在第一个正六边形周围添加 6 个正六边形,m=3 时添加第三层六边形,依此类推。
import networkx as nx
import matplotlib.pyplot as plt
G = nx.hexagonal_lattice_graph(m=2,n=3, periodic=False, with_positions=True,
create_using=None)
pos = nx.get_node_attributes(G, 'pos')
nx.draw(G, pos=pos, with_labels=True)
plt.show()
最佳答案
一个有趣的问题!我花了比我预期更长的时间。基本上,函数hexagonal_lattice_graph()
生成一个m x n
矩形 六边形网格。因此,任务是首先绘制一个大网格,然后删除最外层之外的节点。
我使用距离来决定保留哪些节点以及删除哪些节点。这更加棘手,因为奇数和偶数 m
的行为略有不同。所以中心坐标必须仔细计算。
import networkx as nx
import matplotlib.pyplot as plt
def node_dist(x,y, cx, cy):
"""Distance of each node from the center of the innermost layer"""
return abs(cx-x) + abs(cy-y)
def remove_unwanted_nodes(G, m):
"""Remove all the nodes that don't belong to an m-layer hexagonal ring."""
#Compute center of all the hexagonal rings as cx, cy
cx, cy = m-0.5, 2*m -(m%2) #odd is 2m-1, even is 2m
#in essence, we are converting from a rectangular grid to a hexagonal ring... based on distance.
unwanted = []
for n in G.nodes:
x,y = n
#keep short distance nodes, add far away nodes to the list called unwanted
if node_dist(x,y, cx, cy) > 2*m:
unwanted.append(n)
#now we are removing the nodes from the Graph
for n in unwanted:
G.remove_node(n)
return G
##################
m = 4 #change m here. 1 = 1 layer, single hexagon.
G = nx.hexagonal_lattice_graph(2*m-1,2*m-1, periodic=False,
with_positions=True,
create_using=None)
pos = nx.get_node_attributes(G, 'pos')
G = remove_unwanted_nodes(G, m)
#render the result
plt.figure(figsize=(9,9))
nx.draw(G, pos=pos, with_labels=True)
plt.axis('scaled')
plt.show()
对于 m=4:
欢迎来到SO!希望上述解决方案是清晰的并帮助您继续前进。
关于python - Networkx 不同层的六方晶格,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64222103/