使用python的社交网络中的同质性

Question

我正在尝试确定以节点为键、颜色为值的数据集的同质性，然后是同质性的机会。

例子：

Node  Target   Colors 
A       N        1
N       A        0 
A       D        1
D       A        1
C       X        1
X       C        0
S       D        0
D       S        1
B                0
R       N        2
N       R        2

颜色与节点列相关联，范围从 0 到 2 (int)。 计算特征 z（在我的情况下为颜色）上的同质性概率的步骤如下所示：

c_list=df[['Node','Colors']].set_index('Node').T.to_dict('list')
print("\nChance of same color:", round(chance_homophily(c_list),2))

其中chance_homophily定义如下：

#  The function below takes a dictionary with characteristics as keys and the frequency of their occurrence as values.
# Then it computes the chance homophily for that characteristic (color)

def chance_homophily(dataset):
    freq_dict = Counter([tuple(x) for x in dataset.values()])
    df_freq_counter = freq_dict
    c_list = list(df_freq_counter.values())
    
    chance_homophily = 0
    for class_count in c_list:
        chance_homophily += (class_count/sum(c_list))**2
    return chance_homophily

那么同质性计算如下：

def homophily(G, chars, IDs):
    """
    Given a network G, a dict of characteristics chars for node IDs,
    and dict of node IDs for each node in the network,
    find the homophily of the network.
    """
    num_same_ties = 0
    num_ties = 0
    for n1, n2 in G.edges():
        if IDs[n1] in chars and IDs[n2] in chars:
            if G.has_edge(n1, n2):
                num_ties+=1
                if chars[IDs[n1]] == chars[IDs[n2]]:
                    num_same_ties+=1
    return (num_same_ties / num_ties) 

G 应该从我上面的数据集构建（因此同时考虑节点和目标列）。 我并不完全熟悉这个网络属性，但我认为我在实现中遗漏了一些东西（例如，它是否正确计算了网络中节点之间的关系？）。在网上找到的另一个示例（使用不同的数据集）

https://campus.datacamp.com/courses/using-python-for-research/case-study-6-social-network-analysis?ex=1

特征也是颜色（虽然它是一个字符串，而我有一个数字变量）。我不知道他们是否考虑到节点之间的关系来确定，可能使用邻接矩阵：这部分尚未在我的代码中实现，我正在使用

G = nx.from_pandas_edgelist(df, source='Node', target='Target')

Answer 1

您的代码运行良好。您唯一缺少的是 IDs 字典，它将您的节点名称映射到图 G 中的节点名称。通过从 pandas 边缘列表创建图，您已经在命名您的节点，因为它们在数据。

这使得“IDs”字典的使用变得不必要。查看下面的示例，一次使用 ID 字典，一次使用普通字典以使用原始函数：

import networkx as nx
import pandas as pd
from collections import Counter

df = pd.DataFrame({"Node":["A","N","A","D","C","X","S","D","B","R","N"],
                  "Target":["N","A","D","A","X","C","D","S","","N","R"],
                  "Colors":[1,0,1,1,1,0,0,1,0,2,2]})

c_list=df[['Node','Colors']].set_index('Node').T.to_dict('list')

G = nx.from_pandas_edgelist(df, source='Node', target='Target')

def homophily_without_ids(G, chars):
    """
    Given a network G, a dict of characteristics chars for node IDs,
    and dict of node IDs for each node in the network,
    find the homophily of the network.
    """
    num_same_ties = 0
    num_ties = 0
    for n1, n2 in G.edges():
        if n1 in chars and n2 in chars:
            if G.has_edge(n1, n2):
                num_ties+=1
                if chars[n1] == chars[n2]:
                    num_same_ties+=1
    return (num_same_ties / num_ties)

print(homophily_without_ids(G, c_list))


#create node ids map - trivial in this case
nodes_ids = {i:i for i in G.nodes()}

def homophily(G, chars, IDs):
    """
    Given a network G, a dict of characteristics chars for node IDs,
    and dict of node IDs for each node in the network,
    find the homophily of the network.
    """
    num_same_ties = 0
    num_ties = 0
    for n1, n2 in G.edges():
        if IDs[n1] in chars and IDs[n2] in chars:
            if G.has_edge(n1, n2):
                num_ties+=1
                if chars[IDs[n1]] == chars[IDs[n2]]:
                    num_same_ties+=1
    return (num_same_ties / num_ties) 

print(homophily(G, c_list, nodes_ids))