为给定字典的值返回具有键的字典，反之亦然 [英] Return a dictionary with keys for a given dictionary's values and vice versa

问题描述

在一个函数中，我如何创建一个字典谁的键是一个给定的字典的值，反之亦然，给定的字典每个键有多个值？
例如，给定一个字典：

  d = {1：[a，b ]，2：[b，a，c]，3：[c，d]} 
  

我需要返回一个字典，如：

  d2 = { 一：[ 1， 2]中， b：[ 1， 2]， C：[ 2， 3]， D：[ 3 ]} 
  

我唯一的想法是列出所有值，然后手动检查每个键的每个键值，但是我不知道如何做到这一点，而不知道键和值的数量。任何帮助将非常感谢。

解决方案

尝试：

  {letter：[d键中的键，d [键]中的字母]）（d [键]中的字母中的键的字母）} 
  

说明： set（d [key]中d为字母的字母）是一组出现在原始dict中的所有字母。然后我们创建一个新的字母，其中每个条目都是 letter：[在d [key]]中键入d的键），这意味着其中一个字母，映射到原始dict中映射到它的数字列表。

How would I, in a function, create a dictionary who's keys are a given dictionary's values, and vice versa, where the given dictionary has multiple values per key? For example, given a dictionary:

d = {"1":["a","b"],"2":["b","a","c"],"3":["c","d"]}

I'd need to return a dictionary like:

d2 = {"a":["1","2"],"b":["1","2"],"c":["2","3"],"d":["3"]}

My only idea was to make a list of all values, then manually check each key for each value, but I couldn't figure out how to do that without knowing the number of keys and values. Any help would be vastly appreciated.

解决方案

Try:

{letter: [key for key in d if letter in d[key]] for letter in set(letter for key in d for letter in d[key])}

Explanation: set(letter for key in d for letter in d[key]) is a set of all letters that appears in the original dict. Then we make a new dict, in which every entry is letter: [key for key in d if letter in d[key]], which means one of the letters, mapping to a list of numbers that mapped to it in the original dict.

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