Python-给定三个不同的列表，以递归方式创建任意嵌套的dict [英] Python - Recursively create arbitrarily nested dict given three different lists

问题描述

假设给我三个列表:根列表，子列表(可以有子列表)和节点列表.在这些列表中的任何一项上，我都可以调用.children属性来获取其子项.

Suppose I am given three lists: a list of roots, a list of children (which can have children), and a list of nodes. On any of the items in these lists, I can call use an attribute .children to get its children.

目标是构建一个看起来像这样的任意嵌套的字典:

The goal is to construct an arbitrarily nested dictionary that could look like this:

{root_item: {child_item: {child_child_item: [node1, node2]}}}

到目前为止，我的尝试-在一种方法中，我称为递归方法:

My attempt so far -- in one method, I call a recursive method:

structure = {}
for root in list_of_roots:
    structure[root] = self._build_structure(root)

def _build_structure(self, item)
    if item in list_of_nodes:
        return item
    else:
        for child in item.children:
            self._build_structure(child)

因此，问题在于递归方法不返回任何中间内容，仅返回节点(没有子项的子项).我缺少了一些东西，需要一些帮助来弄清楚如何解决这个问题.

So, the problem with this is that the recursive method doesn't return any of the intermediate stuff, just the nodes (children without children). I am missing something and need some help figuring out how to go about this.

请记住，我是编程新手.

Please keep in mind that I am new to programming.

推荐答案

def _build_structure(self, item)
    if item in list_of_nodes:
        return item
    else:
        return [self._build_structure(child) for child in item.children]

您需要返回您的递归调用...如果您需要更多帮助，则需要发布一个list_of_nodes .....

you need to return your recursive call ... if you need more help you will need to post a minimal example of list_of_nodes ....

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