我有这样的身体字典:
[
{
"children": [
{
"children": [
{
"children": [
{
"label": "Something20",
"id": 11
},
{
"label": "Something19",
"id": 12
}
],
"label": "Something18",
"id": 5
},
{
"children": [
{
"label": "Something15",
"id": 13
}
],
"label": "Something14",
"id": 6
}
],
"label": "Something2",
"id": 2
},
{
"children": [
{
"children": [
{
"label": "Something10",
"id": 14
}
],
"label": "Something9",
"id": 7
},
{
"label": "Something8",
"id": 8
}
],
"label": "Somethin7",
"id": 3
},
{
"children": [
{
"label": "Something5",
"id": 9
},
{
"label": "Something4",
"id": 10
}
],
"label": "Something2",
"id": 4
}
],
"label": "Something1",
"id": 1
}
]
我如何递归地搜索该命令以生成Graphviz数据,如下所示:
Something1->Something2
Something1->Something7
例如称为edges = []
的列表,并将一些数据附加到列表中(仅标签):
[("Something1", "Something2"), ("Something1", "Something7")]
生成数据后,我可以简单地在GraphWiz(生成PNG图像)中选对象,以表示所提供字典的树结构。
答案 0 :(得分:2)
如果要递归执行此操作,则可以使用一个函数,该函数从传递给每个子节点的节点生成边,并为每个子节点调用自身。然后,您要做的就是为根列表中的每个节点调用它(并且您需要在列表中收集所有生成的边)。可以按以下方式实现(root
保存您的数据):
def get_edges(node):
if "children" not in node: # ensure the node has children
return []
label = node["label"]
children = node["children"]
result = [(label, c["label"]) for c in children] # create the edges
for c in children:
result.extend(get_edges(c)) # travers the tree recursively and collect all edges
return result
edges = sum(map(get_edges, root), []) # create and combine all the edge lists
使用您提供的数据运行此代码将产生以下结果:
[('Something1', 'Something2'), ('Something1', 'Somethin7'),
('Something1', 'Something2'), ('Something2', 'Something18'),
('Something2', 'Something14'), ('Something18', 'Something20'),
('Something18', 'Something19'), ('Something14', 'Something15'),
('Somethin7', 'Something9'), ('Somethin7', 'Something8'),
('Something9', 'Something10'), ('Something2', 'Something5'),
('Something2', 'Something4')]