Networkx Graph plot node weights

Question

I would like to assign node weights to each node in an undirected graph. I use the following MWE:

import sys
import matplotlib.pyplot as plt
import networkx as nx
G = nx.Graph()
G.add_node(0)
G.add_node(1, weight=2)
G.add_node(2, weight=3)
nx.draw(G, with_labels=True)
plt.show()

Then I have a figure of the following form:

I would like to plot a graph with the weights given in a new color next to the nodes, such as:

What is the easiest way to implement this? On SO the materials are mostly for edge weights, or changing node sizes w.r.t. the node weights.

vurmux · Accepted Answer · 2019-05-24 14:52:21Z

3

You can use labels attribute with corresponding dict and node_color attribute with corresponding list. For this code:

G = nx.Graph()
G.add_node(0, weight=8)
G.add_node(1, weight=5)
G.add_node(2, weight=3)
labels = {n: G.nodes[n]['weight'] for n in G.nodes}
colors = [G.nodes[n]['weight'] for n in G.nodes]
nx.draw(G, with_labels=True, labels=labels, node_color=colors)

Networkx will draw:

If you want to draw both node ID and its weight, you can write something like this:

labels = {n: str(n) + '; ' + str(G.nodes[n]['weight']) for n in G.nodes}

If you have missing weight attributes in nodes and want to draw them, you can use this code:

labels = {
    n: str(n) + '\nweight=' + str(G.nodes[n]['weight']) if 'weight' in G.nodes[n] else str(n)
    for n in G.nodes
}

I think it is nearly impossible to draw weights near nodes with different color. It is the best I can suggest to you.

edited May 24, 2019 at 14:52

answered May 24, 2019 at 14:42

vurmux

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4 Comments

independentvariable Over a year ago

Hi, thanks for your reply! However, I don't want to change the colors, and only add the weights next to the original node index. Your last line labels = {n: str(n) + '; ' + str(G.nodes[n]['weight']) for n in G.nodes} does not work in my code. Is there any method to generate what I want?

vurmux Over a year ago

You should set weight attribute for each node for this code to work.

independentvariable Over a year ago

At least it works, but I hope there is an even more precise way :)

vurmux Over a year ago

Updated the answer.

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