在Python中存储字典路径 [英] Storing dictionary path in Python

问题描述

python的新手，假设我有一个字典:

Brand new to python, Let's say I have a dict:

kidshair = {'allkids':{'child1':{'hair':'blonde'},
                      'child2':{'hair':'black'},
                      'child3':{'hair':'red'},
                      'child4':{'hair':'brown'}}}

如果child3定期更改其头发颜色，我可能想编写一个应用程序以加快数据维护速度.在此示例中，我将使用:

If child3 changes their hair colour regularly, I might want to write an application to speed up the data maintenance. In this example i'd use:

kidshair['allkids']['child3']['hair'] = ...

是否有任何方法可以将此路径存储为变量，以便我可以随意访问它?显然

Is there any way to store this path as a variable so I can access it at my leasure? Obviously

mypath = kidshair['allkids']['child3']['hair']

导致mypath ='red'.有没有办法对路径本身进行硬编码，所以我可以使用:

results in mypath = 'red'. Is there any possible way to hard code the path itself so I could use:

mypath = 'blue' 

代表

kidshair['allkids']['child3']['hair'] = 'blue'

谢谢，ATfPT

推荐答案

根据您的需要，最简单的选择可能是使用元组作为字典键而不是嵌套词典:

Depending on what you need, the easiest option may be to use tuples as dictionary keys instead of nested dictionaries:

kidshair['allkids', 'child3', 'hair']
mypath = ('allkids', 'child3', 'hair')
kidshair[mypath]

唯一的问题是您无法获得字典的一部分，因此，例如，您无法(轻松/高效)访问与'child3'相关的所有内容.根据您的使用情况，这可能是不合适的解决方案.

The only issue with this is that you can't get a portion of the dictionary, so, for example, you can't (easily/efficiently) access everything to do with 'child3'. This may or may not be an appropriate solution for you depending on your usage.

当前结构的另一种选择是执行以下操作:

An alternative with your current structure is to do something like this:

>>> from functools import partial
>>> test = {"a": {"b": {"c": 1}}}
>>> def itemsetter(item):
...     def f(name, value):
...         item[name] = value
...     return f
...
>>> mypath = partial(itemsetter(test["a"]["b"]), "c")
>>> mypath(2)
>>> test
{'a': {'b': {'c': 2}}}

在这里，我们创建一个函数 itemsetter()，该函数(与 operator.itemgetter() )给我们提供了一个功能，该功能可以在给定字典中设置相关键.然后，我们使用 functools.partial 生成该功能的版本，其中包含我们要预填的键.也不是 mypath = blue ，但这还不错.

Here we make a function itemsetter(), which (in the vein of operator.itemgetter()) gives us a function that that sets the relevant key in the given dictionary. We then use functools.partial to generate a version of this function with the key we want pre-filled. It's not quite mypath = blue either, but it's not bad.

如果您不想为使与 operator 模块保持一致的内容而烦恼，只需执行以下操作:

If you don't want to bother with making something consistent to the operator module, you could simply do:

def dictsetter(item, name):
     def f(value):
         item[name] = value
     return f

mypath = dictsetter(test["a"]["b"], "c")

mypath(2)

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