python 获取 json 键作为完整路径

我想解析一个 JSON 文件并获取包含访问密钥所需的所有路径的完整列表。如果我们使用keys方法，我们会得到单个键的列表，但不会得到访问数据所需的分层键的完整列表。

所以如果给定数据这样

data = {
    "glossary": {
        "title": "example glossary",
        "GlossDiv": {
            "title": "S",
            "GlossList": {
                "GlossEntry": {
                    "ID": "SGML",
                    "SortAs": "SGML",
                    "GlossTerm": "Standard Generalized Markup Language",
                    "Acronym": "SGML",
                    "Abbrev": "ISO 8879:1986",
                    "GlossDef": {
                        "para": "A meta-markup language, used to create markup languages such as DocBook.",
                        "GlossSeeAlso": ["GML", "XML"]
                    },
                    "GlossSee": "markup"
                }
            }
        }
    }
}

我可以返回一个如下所示的列表，其中包含密钥的所有完整路径。

[['glossary']['title'],['glossary']['GlossDiv']...]

读取和访问元素很好。为了达到结果，我尝试使用这个SO答案通过键列表访问嵌套字典项 https://stackoverflow.com/a/14692747

我真的不明白这是如何工作的，它只返回“词汇表”这个词。

这是我的代码。我使用 ChainMap 因为它可以更轻松地将 json 转换为字典并轻松访问密钥。

import json
from collections import ChainMap
from functools import reduce
import operator

myDataChained = ChainMap(data)

def getFromDict(data):
    return reduce(operator.getitem, data)

Json_Paths = getFromDict(myDataChained)
print(Json_Paths)

您不能使用与链接答案中相同的技术来执行相反的操作 - 您没有预先遍历的路径信息functools.reduce()/operator.getitem()组合 - 您正在尝试获取该信息，即标准化/扁平化您的字典结构。

为此，您必须迭代整个结构并收集all数据中可能的路径，例如：

import collections

def get_paths(source):
    paths = []
    if isinstance(source, collections.MutableMapping):  # found a dict-like structure...
        for k, v in source.items():  # iterate over it; Python 2.x: source.iteritems()
            paths.append([k])  # add the current child path
            paths += [[k] + x for x in get_paths(v)]  # get sub-paths, extend with the current
    # else, check if a list-like structure, remove if you don't want list paths included
    elif isinstance(source, collections.Sequence) and not isinstance(source, str):
        #                          Python 2.x: use basestring instead of str ^
        for i, v in enumerate(source):
            paths.append([i])
            paths += [[i] + x for x in get_paths(v)]  # get sub-paths, extend with the current
    return paths

现在如果你运行你的data通过它：

data = {
    "glossary": {
        "title": "example glossary",
        "GlossDiv": {
            "title": "S",
            "GlossList": {
                "GlossEntry": {
                    "ID": "SGML",
                    "SortAs": "SGML",
                    "GlossTerm": "Standard Generalized Markup Language",
                    "Acronym": "SGML",
                    "Abbrev": "ISO 8879:1986",
                    "GlossDef": {
                        "para": "A meta-markup language, used to create markup languages...",
                        "GlossSeeAlso": ["GML", "XML"]
                    },
                    "GlossSee": "markup"
                }
            }
        }
    }
}

paths = get_paths(data)

你会得到paths包含：

[['glossary'],
 ['glossary', 'title'],
 ['glossary', 'GlossDiv'],
 ['glossary', 'GlossDiv', 'title'],
 ['glossary', 'GlossDiv', 'GlossList'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'ID'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'SortAs'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossTerm'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Acronym'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Abbrev'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'para'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso', 0],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso', 1],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossSee']]  

你可以将其中任何一个输入其中functools.reduce()/operator.getitem()组合以获得目标值。