使用 Graphviz 可视化是/否树

Question

我有一份症状诊断问卷数据，格式如下（python）： 路径字典列表。这是一个症状诊断示例，第一个初始症状 (A) 和之后的 2 个问题。

 qa=  [OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 1), ('D', 1), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 1), ('D', 1), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 1), ('D', 0), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 1), ('D', 0), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 0), ('D', 1), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 0), ('D', 1), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 0), ('D', 0), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 1), ('C', 0), ('D', 0), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 1), ('D', 1), ('C', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 1), ('D', 1), ('C', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 1), ('D', 0), ('C', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 1), ('D', 0), ('C', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 0), ('D', 1), ('C', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 0), ('D', 1), ('C', 0)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 0), ('D', 0), ('C', 1)]),
 OrderedDict([('A', 1), ('B', 1), ('F', 0), ('E', 0), ('D', 0), ('C', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 1), ('D', 1), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 1), ('D', 1), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 1), ('D', 0), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 1), ('D', 0), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 0), ('D', 1), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 0), ('D', 1), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 0), ('D', 0), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 1), ('C', 0), ('D', 0), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 1), ('D', 1), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 1), ('D', 1), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 1), ('D', 0), ('E', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 1), ('D', 0), ('E', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 0), ('E', 1), ('D', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 0), ('E', 1), ('D', 0)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 0), ('E', 0), ('D', 1)]),
 OrderedDict([('A', 1), ('B', 0), ('F', 0), ('C', 0), ('E', 0), ('D', 0)])]
    

虽然 1= 是，0 = 否

我想以决策树格式绘制诊断图，其中每个节点都分成“是”/“否”边，这些边通向下一个节点，依此类推。

当两者都可用于同一问题时，我将“是”和“否”分组，因为它在节点上运行 使用graphviz：

u = Digraph(name, strict=True ,filename='blabla',format='png',node_attr={'color': 'mediumpurple1', 'style': 'filled'})
u.attr(size='16,16')
answer_map = ['No','Yes']
nodes = []
edges = []
for path in qa:
    questions = [f'{j}_{lev}' for lev,j in enumerate(path.keys(), 1)]
    questions = [w.replace(':', '_') for w in questions]
    answers = [answer_map[item] for item in path.values()] 
    for i in range(len(questions)-1):
        #u.edge(questions[i], questions[i+1],label=answers[i])
        nodes.append((questions[i],questions[i+1]))
        edges.append(answers[i])
d = {'nodes':nodes,'edges':edges}
df_graph = pd.DataFrame(d).drop_duplicates()
df_graph_joined = df_graph.groupby('nodes')['edges'].apply(','.join).reset_index()

for row in df_graph_joined.itertuples():
    u.edge(row.nodes[0],row.nodes[1],label=row.edges)
u.render()

但是，如您所见，无法区分诊断路径。我想在每个“是”/“否”连接处拆分树，所以通过查看树我可以看到每个诊断路径。我该怎么做？

我希望它看起来像这样：

Answer 1

要拆分每个答案，您需要将节点编辑为不同的名称。我建议通过完整路径更改节点名称。例如，对于这个 OrderedDict：

('A', 1), ('B', 1), ('F', 1), ('C', 1), ('D', 1), ('E', 1)

你可以这样使用：

root, root-A1, root-A1-B1, root-A1-B1-F1, root-A1-B1-F1-C1, root-A1-B1-F1-C1-D1

在这个例子中：

• root代表A
• root-A1代表B，路径是A-->1
• root-A1-B1代表F，路径是A --> 1 --> B --> 1

这是一个例子：

# New node names
qa_tree = []
for path in qa:
    prefix = 'root'
    path_tree = OrderedDict()
    for i, (key, value) in enumerate(path.items()):
        key_tree = '{}'.format(prefix)
        prefix += '-{}{}'.format(key, value)
        path_tree[key_tree] = {'value': value, 'name': key}
    qa_tree.append(path_tree)

name = 'test'
u = Digraph(name, strict=True ,filename='blabla',format='png',node_attr={'color': 'mediumpurple1', 'style': 'filled'})
u.attr(size='16,16')
answer_map = ['No','Yes']
nodes = []
edges = []

for path in qa_tree:
    questions = [key for i, key in enumerate(path.keys(), 1)]
    answers = [answer_map[item.get('value')] for item in path.values()] 
    names = [item.get('name') for item in path.values()] 
    for i in range(len(questions)-1):
        u.node(questions[i], label = names[i])
        u.node(questions[i+1], label = names[i+1] )
        u.edge(questions[i], questions[i+1], label=answers[i])
u.render()