我的print G.nodes(data=True)的输出是:
[('Bytes:\n620', {}), ('dIP:\n178.237.19.228', {}), ('sPort:\n2049', {}), ('sPort:\n60179', {}), ('sIP:\n16.37.97.29', {}), (153, {}), ('dPort:\n443', {}), ('dPort:\n80', {}), ('Packets:\n2', {}), ('Packets:\n1', {}), ('sPort:\n44492', {}), ('Bytes:\n100', {}), ('sIP:\n16.37.93.196', {}), ('dIP:\n178.237.17.97', {}), (188, {}), ('dIP:\n16.37.157.74', {}), ('sIP:\n16.37.97.222', {}), ('dIP:\n178.237.17.61', {}), ('sIP:\n16.37.97.17', {}), ('Bytes:\n46', {}), (224, {}), (227, {}), ('dPort:\n691', {}), ('dIP:\n104.131.44.62', {}), ('sPort:\n55177', {}), ('Protocol:\n6', {}), (120, {}), ('sPort:\n56326', {})]如何使用pos变量手动设置节点的位置(坐标)?
先谢谢你,
问候:)
发布于 2018-10-30 10:11:30
我认为你问的问题实际上不会解决你遇到的问题,但这仍然是一个相当常见的问题,所以它值得解释。
pos只是一个字典,它的键是图中的节点,值是节点的二维位置。
下面是你如何做到这一点。
import networkx as nx
import matplotlib.pyplot as plt
G = nx.Graph()
G.add_edge(0,1)
G.add_node(2)
pos = {}
pos[0] = (0,0)
pos[1] = (1,0)
pos[2] = (0.5, 1)
nx.draw_networkx(G, pos)
plt.show()

https://stackoverflow.com/questions/53052112
复制相似问题