Question

从JSON导入可以获得非常复杂的嵌套结构。例如：

{u'body': [{u'declarations': [{u'id': {u'name': u'i',
                                       u'type': u'Identifier'},
                               u'init': {u'type': u'Literal', u'value': 2},
                               u'type': u'VariableDeclarator'}],
            u'kind': u'var',
            u'type': u'VariableDeclaration'},
           {u'declarations': [{u'id': {u'name': u'j',
                                       u'type': u'Identifier'},
                               u'init': {u'type': u'Literal', u'value': 4},
                               u'type': u'VariableDeclarator'}],
            u'kind': u'var',
            u'type': u'VariableDeclaration'},
           {u'declarations': [{u'id': {u'name': u'answer',
                                       u'type': u'Identifier'},
                               u'init': {u'left': {u'name': u'i',
                                                   u'type': u'Identifier'},
                                         u'operator': u'*',
                                         u'right': {u'name': u'j',
                                                    u'type': u'Identifier'},
                                         u'type': u'BinaryExpression'},
                               u'type': u'VariableDeclarator'}],
            u'kind': u'var',
            u'type': u'VariableDeclaration'}],
 u'type': u'Program'}

如上所述，推荐复杂结构的推荐方法是什么？

除了一些列表，主要是字典，结构可以变得更加重叠，所以我需要一个通用的解决方案。

Answer 1

您可以使用递归生成器将字典转换为平面列表。

def dict_generator(indict, pre=None):
    pre = pre[:] if pre else []
    if isinstance(indict, dict):
        for key, value in indict.items():
            if isinstance(value, dict):
                for d in dict_generator(value, [key] + pre):
                    yield d
            elif isinstance(value, list) or isinstance(value, tuple):
                for v in value:
                    for d in dict_generator(v, [key] + pre):
                        yield d
            else:
                yield pre + [key, value]
    else:
        yield indict

返回

[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'type', u'Literal']
[u'init', u'declarations', u'body', u'value', 2]
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'i']
[u'body', u'type', u'VariableDeclaration']
[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'type', u'Literal']
[u'init', u'declarations', u'body', u'value', 4]
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'j']
[u'body', u'type', u'VariableDeclaration']
[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'operator', u'*']
[u'right', u'init', u'declarations', u'body', u'type', u'Identifier']
[u'right', u'init', u'declarations', u'body', u'name', u'j']
[u'init', u'declarations', u'body', u'type', u'BinaryExpression']
[u'left', u'init', u'declarations', u'body', u'type', u'Identifier']
[u'left', u'init', u'declarations', u'body', u'name', u'i']
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'answer']
[u'body', u'type', u'VariableDeclaration']
[u'type', u'Program']

Answer 2

如果你只需要走字典，我建议使用一个递归的walk函数，它接受一个字典，然后递归遍历它的元素。像这样：

def walk(node):
    for key, item in node.items():
        if item is a collection:
            walk(item)
        else:
            It is a leaf, do your thing

如果您还想搜索元素或查询通过某些条件的多个元素，请查看jsonpath模块。

Answer 3

您可以从标准库json模块扩展编码器和解码器，而不是编写自己的解析器，具体取决于任务。

我推荐这个，特别是如果你需要将属于自定义类的对象编码到json中。如果你必须在json的字符串表示上做一些操作，也可以考虑迭代JSONEncoder（）。iterencode

两者的引用均为http://docs.python.org/2/library/json.html#encoders-and-decoders

Answer 4

如果您知道数据的含义，则可能需要创建parse函数以将嵌套容器转换为自定义类型对象树。然后，您可以使用这些自定义对象的方法来执行您需要处理的任何数据。

对于您的示例数据结构，您可以创建Program，VariableDeclaration，VariableDeclarator，Identifier，Literal和BinaryExpression类，然后对你的解析器使用这样的东西：

def parse(d):
    t = d[u"type"]

    if t == u"Program":
        body = [parse(block) for block in d[u"body"]]
        return Program(body)

    else if t == u"VariableDeclaration":
        kind = d[u"kind"]
        declarations = [parse(declaration) for declaration in d[u"declarations"]]
        return VariableDeclaration(kind, declarations)

    else if t == u"VariableDeclarator":
        id = parse(d[u"id"])
        init = parse(d[u"init"])
        return VariableDeclarator(id, init)

    else if t == u"Identifier":
        return Identifier(d[u"name"])

    else if t == u"Literal":
        return Literal(d[u"value"])

    else if t == u"BinaryExpression":
        operator = d[u"operator"]
        left = parse(d[u"left"])
        right = parse(d[u"right"])
        return BinaryExpression(operator, left, right)

    else:
        raise ValueError("Invalid data structure.")

Answer 5

也许可以提供帮助：

def walk(d):
    global path
      for k,v in d.items():
          if isinstance(v, str) or isinstance(v, int) or isinstance(v, float):
            path.append(k)
            print "{}={}".format(".".join(path), v)
            path.pop()
          elif v is None:
            path.append(k)
            ## do something special
            path.pop()
          elif isinstance(v, dict):
            path.append(k)
            walk(v)
            path.pop()
          else:
            print "###Type {} not recognized: {}.{}={}".format(type(v), ".".join(path),k, v)

mydict = {'Other': {'Stuff': {'Here': {'Key': 'Value'}}}, 'root1': {'address': {'country': 'Brazil', 'city': 'Sao', 'x': 'Pinheiros'}, 'surname': 'Fabiano', 'name': 'Silos', 'height': 1.9}, 'root2': {'address': {'country': 'Brazil', 'detail': {'neighbourhood': 'Central'}, 'city': 'Recife'}, 'surname': 'My', 'name': 'Friend', 'height': 1.78}}

path = []
walk(mydict)

会产生这样的输出：

Other.Stuff.Here.Key=Value 
root1.height=1.9 
root1.surname=Fabiano 
root1.name=Silos 
root1.address.country=Brazil 
root1.address.x=Pinheiros 
root1.address.city=Sao 
root2.height=1.78 
root2.surname=My 
root2.name=Friend 
root2.address.country=Brazil 
root2.address.detail.neighbourhood=Central 
root2.address.city=Recife

Answer 6

如果接受的答案对您有用，但是您还想要一条完整的，有序的路径，其中包括嵌套数组的数字索引，那么这种细微的变化将起作用：

TCP_DEFER_ACCEPT

Answer 7

上述解决方案的一些补充（用于处理包含列表的json）

#!/usr/bin/env python

import json

def walk(d):
   global path
   for k,v in d.items():
      if isinstance(v, str) or isinstance(v, int) or isinstance(v, float):
         path.append(k)
         print("{}={}".format(".".join(path), v)) 
         path.pop()
      elif v is None:
         path.append(k)
         # do something special
         path.pop()
      elif isinstance(v, list):
         path.append(k)
         for v_int in v:
            walk(v_int)
         path.pop()
      elif isinstance(v, dict):
         path.append(k)
         walk(v)
         path.pop()
      else:
         print("###Type {} not recognized: {}.{}={}".format(type(v), ".".join(path),k, v))

with open('abc.json') as f:
   myjson = json.load(f)

path = []
walk(myjson)

Answer 8

我比较灵活的版本如下：

def walktree(tree, at=lambda node: not isinstance(node, dict), prefix=(), 
                flattennode=lambda node:isinstance(node, (list, tuple, set))):
    """
    Traverse a tree, and return a iterator of the paths from the root nodes to the leaf nodes.
    tree: like '{'a':{'b':1,'c':2}}'
    at: a bool function or a int indicates levels num to go down. walktree(tree, at=1) equivalent to tree.items()
    flattennode: a bool function to decide whether to iterate at node value
    """
    if isinstance(at, int):
        isleaf_ = at == 0
        isleaf = lambda v: isleaf_
        at = at - 1
    else:
        isleaf = at
    if isleaf(tree):
        if not flattennode(tree):
            yield (*prefix, tree)
        else:
            for v in tree:
                yield from walktree(v, at, prefix, flattennode=flattennode)
    else:
        for k,v in tree.items():
            yield from walktree(v, at, (*prefix, k), flattennode=flattennode)

用法：

> list(walktree({'a':{'b':[0,1],'c':2}, 'd':3}))
[('a', 'b', 0), ('a', 'b', 1), ('a', 'c', 2), ('d', 3)]
> list(walktree({'a':{'b':[0,1],'c':2}, 'd':3}, flattennode=lambda e:False))
[('a', 'b', [0, 1]), ('a', 'c', 2), ('d', 3)]
> list(walktree({'a':{'b':[0,1],'c':2}, 'd':3}, at=1))
[('a', {'b': [0, 1], 'c': 2}), ('d', 3)]
> list(walktree({'a':{'b':[0,{1:9,9:1}],'c':2}, 'd':3}))
[('a', 'b', 0), ('a', 'b', 1, 9), ('a', 'b', 9, 1), ('a', 'c', 2), ('d', 3)]
> list(walktree([1,2,3,[4,5]]))
[(1,), (2,), (3,), (4,), (5,)]

如何完整遍历未知深度的复杂字典？

8 个答案: