Python:从路径列表创建嵌套字典 [英] Python: Created nested dictionary from list of paths

问题描述

我有一个看起来与此类似的元组列表(这里简化了，有超过 14,000 个这些元组的路径比 Obj.part 更复杂)

I have a list of tuples the looks similar to this (simplified here, there are over 14,000 of these tuples with more complicated paths than Obj.part)

[ (Obj1.part1, {}), (Obj1.partN, {}), (ObjK.partN, {}) ]

Obj 从 1 到 1000，从 0 到 2000 的部分.这些键"都有一个与之关联的规范字典，作为检查另一个二进制文件的查找参考.specs dict 包含路径 ObjK.partN 指向的数据的位偏移、位大小和 C 类型等信息.

Where Obj goes from 1 - 1000, part from 0 - 2000. These "keys" all have a dictionary of specs associated with them which act as a lookup reference for inspecting another binary file. The specs dict contains information such as the bit offset, bit size, and C type of the data pointed to by the path ObjK.partN.

例如:Obj4.part500 可能有这个规范，{'size':32, 'offset':128, 'type':'int'} 这会让我知道在二进制文件中访问 Obj4.part500我必须从偏移 128 解压 32 位.

For example: Obj4.part500 might have this spec, {'size':32, 'offset':128, 'type':'int'} which would let me know that to access Obj4.part500 in the binary file I must unpack 32 bits from offset 128.

所以，现在我想获取我的字符串列表并创建一个嵌套字典，在简化的情况下它看起来像这样

So, now I want to take my list of strings and create a nested dictionary which in the simplified case will look like this

data = { 'Obj1' : {'part1':{spec}, 'partN':{spec} }, 
         'ObjK' : {'part1':{spec}, 'partN':{spec} }
       }

为此，我目前正在做两件事，1. 我正在使用 dotdict 类，以便能够使用点表示法来获取/设置字典.该类看起来像这样:

To do this I am currently doing two things, 1. I am using a dotdict class to be able to use dot notation for dictionary get / set. That class looks like this:

class dotdict(dict):
    def __getattr__(self, attr):
        return self.get(attr, None)
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

创建嵌套dotdict"的方法如下所示:

The method for creating the nested "dotdict"s looks like this:

def addPath(self, spec, parts, base):
    if len(parts) > 1:
        item = base.setdefault(parts[0], dotdict())
        self.addPath(spec, parts[1:], item)
    else:
        item = base.setdefault(parts[0], spec)
    return base

然后我就做类似的事情:

Then I just do something like:

for path, spec in paths:
    self.lookup = dotdict()
    self.addPath(spec, path.split("."), self.lookup)

所以，最后
self.lookup.Obj4.part500 指向规范.

So, in the end
self.lookup.Obj4.part500 points to the spec.

有没有更好(更pythonic)的方法来做到这一点?

Is there a better (more pythonic) way to do this?

推荐答案

除非您更喜欢使用点表示法访问规范，否则请尝试将它们直接放入字典中.在下面的代码中，名称 d 跟踪路径上访问的最里面的字典:

Unless you prefer to access the specs with dot notation, try putting them into the dictionary directly. In the below code, the name d tracks the innermost dictionary visited on the path:

specs = {}
for path, spec in paths:
    parts = path.split('.')
    d = specs
    for p in parts[:-1]:
        d = d.setdefault(p, {})
    d[parts[-1]] = spec

如果每个路径只有两个部分(比如 ObjN 和 partN)，你可以这样做:

If you have only two parts per path (ObjN and partN say), you could just do this:

specs = {}
for path, spec in paths:
    [obj, part] = path.split('.')
    specs.setdefault(obj, {})[part] = spec

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