I need to generate a regular graph (also known as lattice network) which has 100x100
nodes. I started off with drawing a 10x10
graph with the following code:
import numpy
from numpy import *
import networkx as nx
from networkx import *
import matplotlib.pyplot as plt
G=nx.grid_2d_graph(10,10)
nx.draw(G)
plt.axis('off')
plt.show()
but what I get is this:
Is there any way of getting rid of this sort of rotation effect the output has? My final network must look like a chess table, just like this (please ignore the lables):
Also, I need to give each node its ID, ranging from 0 to 9999 (in the case of the 100x100 network). Any idea will be much appreciated!
By default, networkx.draw
uses a spring layout. Instead, you can provide your own positions with parameter pos
. This is actually really simple, since the labels of nodes given networkx.grid_2d_graph
actually are a (row, column) tuple:
>>> G=nx.grid_2d_graph(2,2)
[(0, 1), (1, 0), (0, 0), (1, 1)]
Thus you can use a node's name as its position. So you just need to create a dictionary mapping nodes to themselves, and pass that as the position.
pos = dict( (n, n) for n in G.nodes() )
However, since you also want to add node labels, you should use networkx.draw_networkx
, which takes a dictionary of custom labels as an optional parameter. You'll need a dictionary mapping nodes to their new labels. Since NetworkX gives each node the label (row, column) by default, we can just label each node with row * 10 + column:
labels = dict( ((i, j), i * 10 + j) for i, j in G.nodes() )
Putting it all together, you get the following code which yields the graph below:
import networkx as nx
import matplotlib.pyplot as plt
N = 10
G=nx.grid_2d_graph(N,N)
pos = dict( (n, n) for n in G.nodes() )
labels = dict( ((i, j), i * 10 + j) for i, j in G.nodes() )
nx.draw_networkx(G, pos=pos, labels=labels)
plt.axis('off')
plt.show()
EDIT
Using the suggestion from @AbdallahSobehy, we can label the nodes from left to right and top to bottom.
labels = dict( ((i, j), i + (N-1-j) * 10 ) for i, j in G.nodes() )