有没有办法在python字典中保留重复的键 [英] Is there a way to preserve duplicate keys in python dictionary

问题描述

有没有办法在字典中存储重复的键?

我有一个特定的要求来形成请求和响应对.

从特定节点到另一个特定节点的请求形成相同的密钥.我需要存储这两个.

但是如果我尝试将它们添加到字典中，第一个将被第二个替换.有什么办法吗?

解决方案

假设您想继续使用字典，我可以想到两个简单的选项.

1. 您可以将键映射到项目列表.collections 模块中的 defaultdict 使这变得容易.

   <预><代码>>>>进口藏品>>>数据 = collections.defaultdict(list)>>>对于 k, v in (('a', 'b'), ('a', 'c'), ('b', 'c')):...数据[k].append(v)...>>>数据defaultdict(, {'a': ['b', 'c'], 'b': ['c']})

2. 您可以使用其他数据来消除键的歧义.这可以是时间戳、唯一 ID 号或其他内容.这样做的优点是保留键和值之间的一对一关系，缺点是查找更复杂，因为您总是必须指定 id.下面的例子展示了这可能是如何工作的；它是否对您有好处取决于问题域:

   <预><代码>>>>对于 k, v in (('a', 'b'), ('a', 'c'), ('b', 'c')):...我= 0... while (k, i) 在数据中:...我+= 1... 数据[(k, i)] = v...>>>数据{('a', 1): 'c', ('b', 0): 'c', ('a', 0): 'b'}

is there any way to store duplicate keys in a dictionary?

I have a specific requirement to form pairs of requests and responses.

Requests from a particular node to another particular node form same keys. I need to store both those.

But if I tried to add them to dictionary, first one is being replaced by second. Is there any way?

解决方案

I can think of two simple options, assuming you want to keep using a dictionary.

1. You could map keys to lists of items. A defaultdict from the collections module makes this easy.

   >>> import collections
   >>> data = collections.defaultdict(list)
   >>> for k, v in (('a', 'b'), ('a', 'c'), ('b', 'c')):
   ...     data[k].append(v)
   ... 
   >>> data
   defaultdict(<type 'list'>, {'a': ['b', 'c'], 'b': ['c']})
   

2. You could use additional data to disambiguate the keys. This could be a timestamp, a unique id number, or something else. This has the advantage of preserving a one-to-one relationship between keys and values, and the disadvantage of making lookup more complex, since you always have to specify an id. The example below shows how this might work; whether it's good for you depends on the problem domain:

   >>> for k, v in (('a', 'b'), ('a', 'c'), ('b', 'c')):
   ...     i = 0
   ...     while (k, i) in data:
   ...         i += 1
   ...     data[(k, i)] = v
   ... 
   >>> data
   {('a', 1): 'c', ('b', 0): 'c', ('a', 0): 'b'}
   

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