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      1. 是否可以将列表转换为嵌套的键字典*无需*递归?

        时间:2023-10-19

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                • 本文介绍了是否可以将列表转换为嵌套的键字典*无需*递归?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  假设我有一个如下列表:

                  Supposing I had a list as follows:

                  mylist = ['a','b','c','d']
                  

                  是否可以从这个列表中不使用使用递归/递归函数创建以下字典?

                  Is it possible to create, from this list, the following dict without using recursion/a recursive function?

                  {
                    'a': {
                      'b': {
                        'c': {
                          'd': { }
                        }
                      }
                    }
                  }
                  

                  推荐答案

                  对于简单的情况,简单的迭代和构建,从头到尾都可以:

                  For the simple case, simply iterate and build, either from the end or the start:

                  result = {}
                  for name in reversed(mylist):
                      result = {name: result}
                  

                  result = current = {}
                  for name in mylist:
                      current[name] = {}
                      current = current[name]
                  

                  第一个解决方案也可以使用 reduce():

                  The first solution can also be expressed as a one-liner using reduce():

                  reduce(lambda res, name: {name: res}, reversed(mylist), {})
                  

                  这篇关于是否可以将列表转换为嵌套的键字典*无需*递归?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

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