Python从项目列表创建字典键 [英] Python creating dictionary key from a list of items

问题描述

我希望使用Python字典来跟踪一些正在运行的任务。这些任务中的每一个具有许多属性，使其独特，所以我想使用这些属性的函数来生成字典键，以便我可以通过使用相同的属性再次在字典中找到它们;类似以下内容：

I wish to use a Python dictionary to keep track of some running tasks. Each of these tasks has a number of attributes which makes it unique, so I'd like to use a function of these attributes to generate the dictionary keys, so that I can find them in the dictionary again by using the same attributes; something like the following:

class Task(object):
    def __init__(self, a, b):
        pass

#Init task dictionary
d = {}

#Define some attributes
attrib_a = 1
attrib_b = 10

#Create a task with these attributes
t = Task(attrib_a, attrib_b)

#Store the task in the dictionary, using a function of the attributes as a key
d[[attrib_a, attrib_b]] = t

显然这不工作列表是可变的，因此不能用作键（不可更改类型：列表）） - 所以从多个已知属性生成唯一键的规范方法是什么？

Obviously this doesn't work (the list is mutable, and so can't be used as a key ("unhashable type: list")) - so what's the canonical way of generating a unique key from several known attributes?

推荐答案

使用元组代替列表。元组是不可变的，可以用作字典键：

Use a tuple in place of the list. Tuples are immutable and can be used as dictionary keys:

d[(attrib_a, attrib_b)] = t

括号可以省略：

d[attrib_a, attrib_b] = t

然而，有些人似乎不喜欢这种语法。

However, some people seem to dislike this syntax.

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