将字典值的位置解析到列表中 [英] Parse positions of dictionary values into a list

问题描述

假设您有一本具有以下数据类型的字典:

Let's say you have a dictionary with the following type of data:

{'abc':'AGCTAC', 'def': 'AGGTAC', 'ghi':'AGGTAG'}

我希望能够运行一个显示每个位置值的函数，例如

I want to be able to run a function that shows the values at each position, e.g.

(('A','A','A'),('G','G','G'),('C','G','G'))

然后可以运行一个计数器，如人们建议使用Collection Counter的计数器.

And then be able to run a counter such as people have suggested with collections Counter.

推荐答案

>>> d = {'abc':'AGCTAC', 'def': 'AGGTAC', 'ghi':'AGGTAG'}
>>> zip(*d.values())
[('A', 'A', 'A'), ('G', 'G', 'G'), ('C', 'G', 'G'), ('T', 'T', 'T'), ('A', 'A', 'A'), ('C', 'G', 'C')]

请记住，字典是无序的，因此元组中的元素可能以不同的顺序出现.但是，所有元组的顺序都相同

Remember that dicts are unordered, so the elements in the tuples may occur in a different order. However the order will be the same for all the tuples

在Python3中，zip返回一个"zip对象"，因此您需要在其周围包裹tuple()

In Python3, zip returns a "zip object" so you need to wrap tuple() around it

>>> tuple(zip(*d.values()))
(('A', 'A', 'A'), ('G', 'G', 'G'), ('C', 'G', 'G'), ('T', 'T', 'T'), ('A', 'A', 'A'), ('C', 'G', 'C'))

如果不需要中间元组，只需传递给Counter

If you don't need the intermediate tuple, just pass to Counter

>>> from collections import Counter
>>> Counter(zip(*d.values()))
Counter({('A', 'A', 'A'): 2, ('C', 'G', 'G'): 1, ('G', 'G', 'G'): 1, ('T', 'T', 'T'): 1, ('C', 'G', 'C'): 1})

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