使用NetworkX扁平化父子层级并添加额外列的技术问题

我已经用NetworkX实现了父子层级的扁平化，输出了各层级节点的DataFrame，但不知道如何将原数据中的Description列对应添加到结果中。现有代码如下：

import pandas as pd

data = [
    ['Unit A', 'Department Q', 'Lorem'],
    ['Unit A', 'Department R', 'Ipsum'],
    ['Unit A', 'Department S', 'dolor'],
        ['Department S', 'Office 1', 'sit'],
        ['Department S', 'Office 2', 'amet'],
    ['Unit B', 'Department X', 'consetetur'],
    ['Unit B', 'Department Y', 'sadipscing'],
    ['Unit B', 'Department Z', 'elitr'],
        ['Department Z', 'Office 3', 'sed'],
        ['Department Z', 'Office 4', 'diam'],
            ['Office 4', 'Place K', 'nonumy'] ,
            ['Office 4', 'Place L', 'eirmod']   
]
  
df = pd.DataFrame(data, columns=['Parent', 'Child', 'Description'])

接着创建层级结构并转换为DataFrame：

import networkx as nx

G = nx.from_pandas_edgelist(df, source='Parent', target='Child',
                            create_using=nx.DiGraph, edge_attr=True,
                           )

roots = (v for v, d in G.in_degree() if d == 0)
leaves = [v for v, d in G.out_degree() if d == 0]

out = (pd.DataFrame(path for root in roots for path in
                    nx.all_simple_paths(G, root, leaves))
        .add_prefix('Node_')
      )

print(out)

当前输出结果：

Node_0        Node_1    Node_2   Node_3
0  Unit A  Department Q      None     None
1  Unit A  Department R      None     None
2  Unit A  Department S  Office 1     None
3  Unit A  Department S  Office 2     None
4  Unit B  Department X      None     None
5  Unit B  Department Y      None     None
6  Unit B  Department Z  Office 3     None
7  Unit B  Department Z  Office 4  Place K
8  Unit B  Department Z  Office 4  Place L

解决方案 #

原数据中的Description是每条父-子边的属性，我们可以在生成路径时，同时提取每条边对应的描述，再将这些描述作为额外列添加到结果DataFrame中。

修改后的完整代码：

import pandas as pd
import networkx as nx

data = [
    ['Unit A', 'Department Q', 'Lorem'],
    ['Unit A', 'Department R', 'Ipsum'],
    ['Unit A', 'Department S', 'dolor'],
        ['Department S', 'Office 1', 'sit'],
        ['Department S', 'Office 2', 'amet'],
    ['Unit B', 'Department X', 'consetetur'],
    ['Unit B', 'Department Y', 'sadipscing'],
    ['Unit B', 'Department Z', 'elitr'],
        ['Department Z', 'Office 3', 'sed'],
        ['Department Z', 'Office 4', 'diam'],
            ['Office 4', 'Place K', 'nonumy'] ,
            ['Office 4', 'Place L', 'eirmod']   
]
  
df = pd.DataFrame(data, columns=['Parent', 'Child', 'Description'])

G = nx.from_pandas_edgelist(df, source='Parent', target='Child',
                            create_using=nx.DiGraph, edge_attr=True,
                           )

roots = (v for v, d in G.in_degree() if d == 0)
leaves = [v for v, d in G.out_degree() if d == 0]

# 生成包含节点和对应边描述的字典列表
path_data = []
for root in roots:
    for path in nx.all_simple_paths(G, root, leaves):
        # 先构建节点列的键值对
        row = {'Node_' + str(i): node for i, node in enumerate(path)}
        # 提取每条边的Description，存入对应的Desc列
        for i in range(len(path)-1):
            parent_node = path[i]
            child_node = path[i+1]
            row[f'Desc_{i}'] = G[parent_node][child_node]['Description']
        path_data.append(row)

# 转换为DataFrame，缺失的列自动填充为None
out = pd.DataFrame(path_data)

print(out)

最终输出结果 #

Node_0        Node_1    Node_2   Node_3      Desc_0 Desc_1  Desc_2
0  Unit A  Department Q      None     None       Lorem   None    None
1  Unit A  Department R      None     None       Ipsum   None    None
2  Unit A  Department S  Office 1     None       dolor    sit    None
3  Unit A  Department S  Office 2     None       dolor   amet    None
4  Unit B  Department X      None     None  consetetur   None    None
5  Unit B  Department Y      None     None  sadipscing   None    None
6  Unit B  Department Z  Office 3     None       elitr    sed    None
7  Unit B  Department Z  Office 4  Place K       elitr   diam  nonumy
8  Unit B  Department Z  Office 4  Place L       elitr   diam  eirmod

关键说明 #

• 遍历每条路径时，先构建节点的键值对（如Node_0、Node_1）
• 对路径中每一对相邻节点，从图中提取对应的Description属性，存入Desc_0、Desc_1等列
• 用pd.DataFrame()转换时，缺失的列会自动填充为None，保证结果格式统一

内容的提问来源于stack exchange，提问作者Frank