NetworkX最大连接组件共享属性

Question

我知道在NetworkX中存在用于计算图的连接组件大小的函数。您可以为节点添加属性。在阿克塞尔罗德的文化传播模型中，一个有趣的测量是最大连接组件的大小，其节点具有几个属性。在NetworkX中有办法做到这一点吗?例如，假设我们有一个通过网络表示的总体。每个节点都有头发颜色和皮肤颜色的属性。我怎样才能得到最大的节点分量的大小，使子图中的每个节点都有相同的头发和皮肤颜色?谢谢你

Answer 1

对于一般的数据分析，最好使用pandas。使用像networkx或graph-tool这样的图形库来确定连接的组件，然后将该信息加载到可以分析的DataFrame中。在这种情况下，熊猫groupby和nunique(唯一元素的数量)功能将是有用的。

下面是一个使用graph-tool(使用这个网络)的独立示例。您还可以通过networkx计算连接的组件。

import numpy as np
import pandas as pd
import graph_tool.all as gt
# Download an example graph
# https://networks.skewed.de/net/baseball
g = gt.collection.ns["baseball", 'user-provider']
# Extract the player names
names = g.vertex_properties['name'].get_2d_array([0])[0]
# Extract connected component ID for each node
cc, cc_sizes = gt.label_components(g)
# Load into a DataFrame
players = pd.DataFrame({
'id': np.arange(g.num_vertices()),
'name': names,
'cc': cc.a
})
# Create some random attributes
players['hair'] = np.random.choice(['purple', 'pink'], size=len(players))
players['skin'] = np.random.choice(['green', 'blue'], size=len(players))
# For the sake of this example, manipulate the data so
# that some groups are homogenous with respect to some attributes.
players.loc[players['cc'] == 2, 'hair'] = 'purple'
players.loc[players['cc'] == 2, 'skin'] = 'blue'
players.loc[players['cc'] == 4, 'hair'] = 'pink'
players.loc[players['cc'] == 4, 'skin'] = 'green'
# Now determine how many unique hair and skin colors we have in each group.
group_stats = players.groupby('cc').agg({
'hair': 'nunique',
'skin': ['nunique', 'size']
})
# Simplify the column names
group_stats.columns = ['hair_colors', 'skin_colors', 'player_count']
# Select homogenous groups, i.e. groups for which only 1 unique
# hair color is present and 1 unique skin color is present
homogenous = group_stats.query('hair_colors == 1 and skin_colors == 1')
# Sort from large groups to small groups
homogenous = homogenous.sort_values('player_count', ascending=False)
print(homogenous)

打印以下内容:

hair_colors  skin_colors  player_count
cc
4             1            1             4
2             1            1             3