递归地在python中构建分层的JSON树 [英] recursively build hierarchical JSON tree in python

问题描述

我有一个父子连接的数据库。数据看起来像下面的，但可以以任何你想要的方式（字典，列表列表，JSON等）。

  links =（（Tom，Dick），（Dick，Harry）， ，Larry），（Bob，Leroy），（Bob，Earl））
  

b $ b

我需要的输出是一个分层的JSON树，将使用d3呈现。在数据中有离散子树，我将附加到根节点。所以我需要递归地去经历链接，并建立树结构。我可以得到的最远的是遍历所有的人，并追加他们的孩子，但我不能想出做更高阶的链接（例如如何附加一个有孩子的人到别人的孩子）。这类似于。

I have a database of parent-child connections. The data look like the following but could be presented in whichever way you want (dictionaries, list of lists, JSON, etc).

links=(("Tom","Dick"),("Dick","Harry"),("Tom","Larry"),("Bob","Leroy"),("Bob","Earl"))

The output that I need is a hierarchical JSON tree, which will be rendered with d3. There are discrete sub-trees in the data, which I will attach to a root node. So I need to recursively go though the links, and build up the tree structure. The furthest I can get is to iterate through all the people and append their children, but I can't figure out to do the higher order links (e.g. how to append a person with children to the child of someone else). This is similar to another question here, but I have no way to know the root nodes in advance, so I can't implement the accepted solution.

I am going for the following tree structure from my example data.

{
"name":"Root",
"children":[
    {
     "name":"Tom",
     "children":[
         {
         "name":"Dick",
         "children":[
             {"name":"Harry"}
         ]
         },
         {
          "name":"Larry"}
     ]
    },
    {
    "name":"Bob",
    "children":[
        {
        "name":"Leroy"
        },
        {
        "name":"Earl"
        }
    ]
    }
]
}

This structure renders like this in my d3 layout.

解决方案

To identify the root notes you can unzip links and look for parents who are not children:

parents, children = zip(*links)
root_nodes = {x for x in parents if x not in children}

Then you can apply the recursive method:

import json

links = [("Tom","Dick"),("Dick","Harry"),("Tom","Larry"),("Bob","Leroy"),("Bob","Earl")]
parents, children = zip(*links)
root_nodes = {x for x in parents if x not in children}
for node in root_nodes:
    links.append(('Root', node))

def get_nodes(node):
    d = {}
    d['name'] = node
    children = get_children(node)
    if children:
        d['children'] = [get_nodes(child) for child in children]
    return d

def get_children(node):
    return [x[1] for x in links if x[0] == node]

tree = get_nodes('Root')
print json.dumps(tree, indent=4)

I used a set to get the root nodes, but if order is important you can use a list and remove the duplicates.

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