python中的自定义json格式答案

【问题标题】：custom json formatting in pythonpython中的自定义json格式
【发布时间】：2018-11-16 04:32:10
【问题描述】：

我有以下代码来生成列表列表的 json 表示。

Levels=[['L1','L1','L2'],
        ['L1','L1','L3'],
        ['L1','L2'],
        ['L2','L2','L3'],
        ['L2','L2','L1'],
        ['L3','L2'],
        ['L4','L2','L1'],
        ['L4','L2','L4']]

def append_path(root, paths):
    if paths:
        child = root.setdefault(paths[0], {})
        append_path(child, paths[1:])

for p in Levels:
    append_path(root, p)

def convert(d):
    return [{'name': k, 'children': convert(v) if v else [{}]} for k, v in d.items()]



# Print results
import json
print(json.dumps(convert(root),  indent=4))

输出：

[
"name": "L1",
      "children": [
        {
          "name": "L1",
           "children":[
              {
                "name":"L3",
                "children":[{}]
              },
              {
                "name":"L1",
                "children":[{}]
              }]
        },
        {
            "name":"L2",
            "children":[{}]
        }

      ]

关卡

Levels=[['L1','L1','L2'],
        ['L1','L1','L3'],
        ['L1','L2'],

我还需要编码每个级别的计数

例如，从L1 开始的路径有两个一级子级L1(2) 和L2(1)，后跟L2(1) 和L3(1) 用于下一级。

L1(3)-->L1(2)-->L2(1)
             -->L3(1)
     -->L2(1)

如何在我的 json 输出中编码这个计数。

我希望我的最终输出看起来像这样

"name": "L1(3)",
      "children": [
        {
          "name": "L1(2)",
           "children":[

【问题讨论】：

您想添加可以从已有数据中提取的值，其格式既不直观也不易于机器读取？
“count”是指叶子节点的个数还是节点的高度？

标签： python json python-3.x

【解决方案1】：

root={}
Levels=[['L1','L1','L2'],
        ['L1','L1','L3'],
        ['L1','L2'],
        ['L2','L2','L3'],
        ['L2','L2','L1'],
        ['L3','L2'],
        ['L4','L2','L1'],
        ['L4','L2','L4']]

def append_path(root, paths):
    if paths:
        child = root.setdefault(paths[0], {})
        append_path(child, paths[1:])

for p in Levels:
    append_path(root, p)

def convert(d):
    templist=[]
    noofchildren=0
    if(len(d.items())==0):
        return ([{}],1)
    for k,v in d.items():
        temp,children=convert(v)
        noofchildren+=children
        if(temp):
            templist.append({"name":k+"("+str(children)+")",'children':temp})
        else:
            templist.append({'name': k+"("+str(children)+")", 'children':[{}]})

    return (templist,noofchildren)    

# Print results
import json
print(json.dumps(convert(root)[0],  indent=2))

输出

[
  {
    "name": "L1(3)",
    "children": [
      {
        "name": "L1(2)",
        "children": [
          {
            "name": "L2(1)",
            "children": [
              {}
            ]
          },
          {
            "name": "L3(1)",
            "children": [
              {}
            ]
          }
        ]
      },
      {
        "name": "L2(1)",
        "children": [
          {}
        ]
      }
    ]
  },
  {
    "name": "L2(2)",
    "children": [
      {
        "name": "L2(2)",
        "children": [
          {
            "name": "L3(1)",
            "children": [
              {}
            ]
          },
          {
            "name": "L1(1)",
            "children": [
              {}
            ]
          }
        ]
      }
    ]
  },
  {
    "name": "L3(1)",
    "children": [
      {
        "name": "L2(1)",
        "children": [
          {}
        ]
      }
    ]
  },
  {
    "name": "L4(2)",
    "children": [
      {
        "name": "L2(2)",
        "children": [
          {
            "name": "L1(1)",
            "children": [
              {}
            ]
          },
          {
            "name": "L4(1)",
            "children": [
              {}
            ]
          }
        ]
      }
    ]
  }
]

【讨论】：