如何从点分隔的字符串列表构造 JSON？

Question

人们会认为这样做很容易。我发誓它是，只是......我似乎无法弄清楚。我如何转换：

terrible_way_to_describe_nested_json=['a.b.c','a.b.d','a.e','a.f','g.h']

进入

{
    "a": {
        "b": {
            "c": None,
            "d": None
        },
        "e": None,
        "f": None
    },
    "g": {
        "h": None
    }
}

如果您认为 'a.b.c' 是解构 JSON 负载的 path，那么我有大约 200 个未排序的路径（无论顺序如何，转换都应该起作用），所有这些路径的深度最多为 8 个点热切地希望成为他们原始结构的一部分。我尝试过使用递归来解决这个问题，pandas 从内部节点中排​​序叶节点（荒谬？），疯狂的列表字典列表列表谁知道，甚至 autovivification。

毁灭和绝望的领域

这是我编写/放弃的 6 个部分实现之一。它甚至在边缘节点之前剥离嵌套键层，然后我就失去了理智。我几乎建议忽略它。

def dot_to_json(dotted_paths):
    scope_map=[line.split('.') for line in dotted_paths] #Convert dots list to strings
    # Sort and group list of strings according to length of list. longest group is last
    b=[]
    for index in range(max([len(x) for x in scope_map])+1):
        a=[]
        for item in scope_map:
            if len(item)==index:
                a.append(item)
        b.append(a)
    sorted_nest=[x for x in b if x] # finally group according to list length
    #Point AA
    # group string list 'prefix' with key:value
    child_path=[]
    for item in sorted_nest[-1]:
        child_path.append([item[:-1],{item[-1]:None}])
    # peel back a layer
    new_child_path=[]
    for scope in scope_map[-2]:
        value=None # set value to None if fringe node
        for index, path in enumerate(child_path):
            if path[0]==scope: # else, save key + value as a value to the new prefix key
                value=path[1]
                child_path.pop(index) # 'move' this path off child_path list
        new_child_path.append([scope[:-1],{scope[-1]:value}])
    new_child_path+=child_path
    #Point BB...
    #Loop in some intelligent way between Point AA and Point BB
    return new_child_path
#%%
dotted_json=['a.b.c','a.b.d','a.e','a.f','g.h']

scope_map=dot_to_json(dotted_json)

Answer 1

给你：

In [5]: terrible_way_to_describe_nested_json=['a.b.c','a.b.d','a.e','a.f','g.h']

In [6]: terrible_way_to_describe_nested_json = [s.split('.') for s in terrible_way_to_describe_nested_json]

In [7]: data = {}

In [8]: for path in terrible_way_to_describe_nested_json:
   ....:     curr = data
   ....:     for i, node in enumerate(path):
   ....:         if i == len(path) - 1:
   ....:             curr[node] = None
   ....:         else:
   ....:             curr = curr.setdefault(node,{})
   ....:             

In [9]: data
Out[9]: {'a': {'b': {'c': None, 'd': None}, 'e': None, 'f': None}, 'g': {'h': None}}

现在使用 json 模块进行漂亮的打印：

{
    "a": {
        "f": null,
        "b": {
            "d": null,
            "c": null
        },
        "e": null
    },
    "g": {
        "h": null
    }
}

这两个应该是等价的，但是是乱序的，或者至少是乱序打印的。

Answer 2

试用dpath 模块，它可用于轻松添加/过滤/搜索字典。通过将点替换为“/”字符，您可以创建所需的 dict。