75

导入 fromJSON可以获得非常复杂和嵌套的结构。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'}

行走上述复杂结构的推荐方法是什么？

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

4

9 回答 9

66

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, pre + [key]):
                    yield d
            elif isinstance(value, list) or isinstance(value, tuple):
                for v in value:
                    for d in dict_generator(v, pre + [key]):
                        yield d
            else:
                yield pre + [key, value]
    else:
        yield pre + [indict]

它返回

[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']

更新：修复了注释中提到的从[key] + pre到的键列表。pre + [key]

于 2012-09-20T07:04:40.437 回答
51

如果您只需要遍历字典，我建议使用递归walk函数，该函数接受字典，然后递归遍历其元素。像这样的东西：

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，或查询多个通过特定条件的元素，请查看jsonpath模块。

于 2012-09-20T06:58:40.767 回答
8

json您可以根据任务从标准库模块扩展编码器和解码器，而不是编写自己的解析器。

我推荐这个，特别是如果您需要将属于自定义类的对象编码到 json 中。如果您必须执行一些也可以在 json 的字符串表示上完成的操作，请考虑迭代 JSONEncoder().iterencode

对于这两个参考是http://docs.python.org/2/library/json.html#encoders-and-decoders

于 2013-01-23T09:12:38.670 回答
7

如果您知道数据的含义，您可能希望创建一个parse函数来将嵌套容器转换为自定义类型的对象树。然后，您将使用这些自定义对象的方法来执行您需要对数据执行的任何操作。

对于您的示例数据结构，您可以创建Program、VariableDeclaration、VariableDeclarator、和类Identifier，然后在解析器中使用类似这样的内容：LiteralBinaryExpression

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.")
于 2012-09-20T07:53:10.367 回答
6

也许可以帮助：

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)

将产生如下输出：

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
于 2016-04-10T00:49:06.120 回答
4

如果接受的答案对您有用，但您还想要一个完整的、有序的路径，其中包含嵌套数组的数字索引，那么这种轻微的变化将起作用：

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,  pre + [key]):
                    yield d
            elif isinstance(value, list) or isinstance(value, tuple):
                for k,v in enumerate(value):
                    for d in dict_generator(v, pre + [key] + [k]):
                        yield d
            else:
                yield pre + [key, value]
    else:
        yield indict
于 2019-12-05T01:39:49.257 回答
2

对上述解决方案的一些补充（处理 json 包括列表）

#!/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)
于 2019-01-02T03:49:32.527 回答
1

我更灵活的版本如下：

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,)]
于 2021-04-23T07:20:16.313 回答
1

要遍历/映射整个 JSON 结构，可以使用以下代码：

def walk(node, key):
    if type(node) is dict:
        return {k: walk(v, k) for k, v in node.items()}
    elif type(node) is list:
        return [walk(x, key) for x in node]
    else:
        return YourFunction(node, key)

def YourFunction(node, key):
    if key == "yourTargetField":   # for example, you want to modify yourTargetField
        return "Modified Value"
    return node # return existing value

这将遍历您的整个 json 结构并通过您的函数运行每个叶子（端点键值对）。Yourfunction 返回修改后的值。整个 walk 函数将按顺序为您提供一个新的、已处理的对象。

于 2021-12-26T22:12:27.830 回答