如何从Python中的pandas数据帧中获得networkx图的分支作为列表?

您可以定义 Recursive Function 以及保存路径和打印路径:

df = pd.DataFrame({
          'From':['Node1','Node1', 'Node2', 'Node2', 'Node3', 'Node3', 'Node4', 'Node5', 'Node6', 'Node7'],
          'TO'  :['Node2','Node3', 'Node4', 'Node5', 'Node6', 'Node7', 'Node8', 'Node9', 'Node10', 'Node11']
        })

def prntPath(lst, node, df, lst_vst):
    for val in df.values:
        if val[0] == node:
            lst.append(val[1])
            prntPath(lst, val[1], df, lst_vst)
    
    if not lst[-1] in lst_vst:
        print('-'.join(lst))
    for l in lst: lst_vst.add(l)
    lst.pop()
    return
    
lst_vst = set()
prntPath(['Node1'],'Node1', df, lst_vst)

输出

Node1-Node2-Node4-Node8
Node1-Node2-Node5-Node9
Node1-Node3-Node6-Node10
Node1-Node3-Node7-Node11

您可以检查并用于其他图形,如下所示:

import networkx as nx
import matplotlib.pyplot as plt
import pandas as pd
from itertools import chain
from networkx.drawing.nx_pydot import graphviz_layout

def prntPath(lst, node, df, lst_vst):
    for val in df.values:
        if val[0] == node:
            lst.append(val[1])
            prntPath(lst, val[1], df, lst_vst)
    if not lst[-1] in lst_vst: print('-'.join(lst))
    for l in lst: lst_vst.add(l)
    lst.pop()
    return

df = pd.DataFrame({
          'From':['Node1','Node1', 'Node2', 'Node3', 'Node3', 'Node5', 'Node7'],
          'TO'  :['Node2','Node3', 'Node5', 'Node6', 'Node7', 'Node9', 'Node11']
        })

g = nx.DiGraph()
g.add_nodes_from(set(chain.from_iterable(df.values)))
for edg in df.values:
    g.add_edge(*edg)
pos = graphviz_layout(g, prog="dot")
nx.draw(g, pos,with_labels=True, node_shape='s')
plt.draw()
plt.show() 

lst_vst = set()
prntPath(['Node1'],'Node1', df, lst_vst)

输出

Node1-Node2-Node5-Node9
Node1-Node3-Node6
Node1-Node3-Node7-Node11

使用networkx包解决此问题的另一种方法。

import networkx as nx
G = nx.DiGraph()
G.add_edges_from((r.From, r.To) for r in df.itertuples())

roots = [n for (n, d) in G.in_degree if d == 0]
print(roots)

leafs = [n for (n, d) in G.out_degree if d == 0]
print(leafs)

df1 = pd.DataFrame(nx.algorithms.all_simple_paths(G, roots[0], leafs))

for index in df1:
    print (df1.loc[index].to_list())
    

将创建一个有向图,并从添加边 df 节点1是G中唯一的节点,其中in_degree等于0,即它是整个图的根。因此 roots 等于节点0。

leafs 表示out_degree等于0的有向图上的节点。即[“Node8”、“Node9”、“Node 10”、“节点11”]。

nx.algorithms.all_simple_paths(G, roots[0], leafs) 提供从节点1开始并在叶中的每个节点处结束的所有路径。

数据帧的每一行 df1 使用for循环语句打印为列表。

我认为@I'mahdi的第一段代码假设一个子节点不会有多个父节点,即图是一个二叉树。当图形不是二叉树时,可以将代码修改为:

import pandas as pd

def prntPath(lst, node, df, lst_vst):
    for val in df.values:
        if val[0] == node:
            lst.append(val[1])
            # recursely find child node first until no child nodes are found
            prntPath(lst, val[1], df, lst_vst)
            
            # recursion is over
    
    # when no child nodes are found
    if len(lst)>=2:
        if (not lst[-1] in lst_vst) or (not lst[-2] in lst_vst):
            print('-'.join(lst))
    for l in lst:
        lst_vst.add(l)
    lst.pop()

# Two links are added to the graph: 1) from node 4 to node 9. 2) from node 1 to node 11.
 
df = pd.DataFrame({
          'From':['Node1','Node1', 'Node2', 'Node2', 'Node3', 'Node3', 'Node4', 'Node4','Node5', 'Node6', 'Node7','Node1'],
          'TO'  :['Node2','Node3', 'Node4', 'Node5', 'Node6', 'Node7', 'Node8', 'Node9','Node9', 'Node10', 'Node11','Node12']
        })
lst_vst = set()
prntPath(['Node1'],'Node1', df, lst_vst)

结果是:

Node1-Node2-Node4-Node8
Node1-Node2-Node4-Node9
Node1-Node2-Node5-Node9
Node1-Node3-Node6-Node10
Node1-Node3-Node7-Node11
Node1-Node12