Question

我正在尝试使用以下代码在python中生成自定义JSON

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))

输出是

[
  {
    "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": [
              {}
            ]
          }
        ]
      }
    ]
  }
]

我的数据集有些变化

 Levels=[[['L1','L1','L2'],[10,20,30]],
        [[['L1','L1','L3'],[10,15,20]],
        [[['L1','L2'],[20,10]],
        [[['L2','L2','L3'],[20,20,30]],
        [[['L2','L2','L1'],[10,20,30]]
        [[['L3','L2'],[10,20]]
        [[['L4','L2','L1'],[10,20,10]]
        [[['L4','L2','L4'],[20,40,50]]]

我想要的输出是水平和计数的平均值

[
  {
    "name": "L1(3)#(13)", // taking avg of 10,10,20
    "children": [
      {
        "name": "L1(2)#(17)", // taking avg of 20,15
        "children": [
          {
            "name": "L2(1)#(30)",
            "children": [
              {}
            ]
          },
          {
            "name": "L3(1)#(20)",
            "children": [
              {}
            ]
          }
        ]
      },
      {
        "name": "L2(1)#10",
        "children": [
          {}
        ]
      }
    ]
  },
  {
    "name": "L2(2)#(15)", // avg of 20,10
    "children": [
      {
        "name": "L2(2)#(20)", // avg of 20,20
        "children": [
          {
            "name": "L3(1)#(30)",
            "children": [
              {}
            ]
          },
          {
            "name": "L1(1)#(30)",
            "children": [
              {}
            ]
          }
        ]
      }
    ]
  },
  {
    "name": "L3(1)#(10)",
    "children": [
      {
        "name": "L2(1)#(10)",
        "children": [
          {}
        ]
      }
    ]
  },
  {
    "name": "L4(2)#(15)",// avg of 10,20
    "children": [
      {
        "name": "L2(2)#(30)", // avg of 20,40
        "children": [
          {
            "name": "L1(1)# (10)",
            "children": [
              {}
            ]
          },
          {
            "name": "L4(1)#(50)",
            "children": [
              {}
            ]
          }
        ]
      }
    ]
  }
]

如何更改代码以添加此信息？

Answer 1

前言

在进入解决方案之前，我想提一些要点：

充分利用Python的面向对象编程功能！这使您自己和将来的读者都可以更清楚地了解数据结构。
使用自定义类还使我们更容易存储元数据（即节点的实例数及其总值），同时构造中间树结构，而不是同时转换。由于使用后一种方法，简单的朴素遍历算法将重复访问节点！
如果您希望输出（可靠地）保持插入路径的顺序，则应使用 OrderedDict （来自collections）普通dict（{}）。
为没有子节点的节点输出一个空列表比一个具有单个空字典的列表更合乎逻辑：
```
// Before
"children": [
  {}
]

// After
"children": []
```
原因是，任何稍后将解析此数据的软件都可以安全地假设所有对象都具有"name"和"children"字段，而空字典则没有。

Levels数组中的列表边界和元素似乎格式较差；你是说：

Levels = [
    [['L1','L1','L2'],[10,20,30]],
    [['L1','L1','L3'],[10,15,20]],
    [['L1','L2'],[20,10]],
    [['L2','L2','L3'],[20,20,30]],
    [['L2','L2','L1'],[10,20,30]],
    [['L3','L2'],[10,20]],
    [['L4','L2','L1'],[10,20,10]],
    [['L4','L2','L4'],[20,40,50]],
]

在数据主题上，由于节点和值遵循1对1映射（在每个路径内），因此使用元组列表会更合适而不是两个平行列表的列表：

Levels = [
    [('L1', 10), ('L1', 20), ('L2', 30)],
    [('L1', 10), ('L1', 15), ('L3', 20)],
    [('L1', 20), ('L2', 10)],
    [('L2', 20), ('L2', 20), ('L3', 30)],
    [('L2', 10), ('L2', 20), ('L1', 30)],
    [('L3', 10), ('L2', 20)],
    [('L4', 10), ('L2', 20), ('L1', 10)],
    [('L4', 20), ('L2', 40), ('L4', 50)]
]

您的预期输出似乎有误：

{
  "name": "L3(1)#(10)",
  "children": [
    {
      "name": "L2(1)#(10)", <--- this should be #(20)
      "children": [
        {}
      ]
    }
  ]
},

实施

对于您当前的数据格式（列表对）：

# A dictionary here corresponds to an array of nodes in JSON
# the "name" fields serve as the keys and "children" as the values
class data_node(OrderedDict):
    def __init__(self, **kwargs):
        super(data_node, self).__init__(**kwargs)
        self.instances = 0
        self.total = 0

    def insert(self, names, values):
        # Python lists are truthy, so no need for len(path) == 0
        if not (names or values):
            return

        # create the child node if it doesn't exist
        child = self.get(names[0], data_node())

        # add the value to the total
        # and increment the instance counter
        child.instances += 1
        child.total += values[0]

        # recursive call on the child 
        # with the rest of the path
        child.insert(names[1:], values[1:])
        self[names[0]] = child

    def convert(self):
        return [
            {
                "name": "{}({})#({})".format(
                    name,
                    child.instances,
                    child.total / child.instances # mean
                ),
                "children": child.convert()
            }
            for name, child in self.items()
        ]

tree = data_node()
for path in Levels:
    tree.insert(path[0], path[1])
print json.dumps(tree.convert(), indent=2)

对于我建议的数据格式（元组列表）：

# only the insertion method differs
# all other parts of the class are unchanged
def insert(self, path):
    if not path:
        return
    name, value = path[0]
    child = self.get(name, data_node())
    child.instances += 1
    child.total += value
    child.insert(path[1:])
    self[name] = child

...

for path in Levels:
    tree.insert(path) # simpler function signature

编辑：

如果出于某种原因，您希望叶节点格式为[{}]而不是[]，则只需进行简单的更改即可：

# in convert()
{
    "name": ..., # as before

    # again exploiting the truthy-ness property of arrays
    "children": child.convert() or [{}] 
}

输出

根据我在序言中的评论，这两种实现都提供正确的JSON输出：

[
  {
    "name": "L1(3)#(13)",
    "children": [
      {
        "name": "L1(2)#(17)",
        "children": [
          {
            "name": "L2(1)#(30)",
            "children": []
          },
          {
            "name": "L3(1)#(20)",
            "children": []
          }
        ]
      },
      {
        "name": "L2(1)#(10)",
        "children": []
      }
    ]
  },
  {
    "name": "L2(2)#(15)",
    "children": [
      {
        "name": "L2(2)#(20)",
        "children": [
          {
            "name": "L3(1)#(30)",
            "children": []
          },
          {
            "name": "L1(1)#(30)",
            "children": []
          }
        ]
      }
    ]
  },
  {
    "name": "L3(1)#(10)",
    "children": [
      {
        "name": "L2(1)#(20)",
        "children": []
      }
    ]
  },
  {
    "name": "L4(2)#(15)",
    "children": [
      {
        "name": "L2(2)#(30)",
        "children": [
          {
            "name": "L1(1)#(10)",
            "children": []
          },
          {
            "name": "L4(1)#(50)",
            "children": []
          }
        ]
      }
    ]
  }
]

在python中将节点添加到json

1 个答案:

前言

实施

输出