python - 不同层的六边形点阵由网络x

Question

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 的六边形点阵图。 m=1 表示 1 个正六边形，边为 1，中心在原点，对于 m=2，在初始六边形周围添加 6 个正六边形，对于 m=3，添加第 3 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()

Answer 1

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()产生 m x n六边形的矩形网格。因此，任务是先绘制一个大网格，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，删除哪些节点。更加棘手，因为奇数和偶数 m行为略有不同。所以必须仔细计算中心坐标。

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=3，产生以下结果:

对于 m=4:

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