如何为任意列表构建递归? [英] How to build a recursion for arbitrary lists?

问题描述

为了简化我的问题，我将尝试举一个简单的例子来使其更清楚，例如，如果我有这个由列表列表组成的树，列表的数量可以无限大，例如

To simplify my question I will try give a simple example to make it clearer,if I have this Tree made of lists of lists for instance, the number of lists can be infinitely large e.g.

[[[[], []], [[], []]], [[], [], []]]

我想剪掉它的叶子，没有外向边的节点(空列表)，所以用前面的输入我得到这个:

And I want to cut off its leaves, the nodes that do not have outgoing edges(empty list), so with the previous input I get this :

[[[], []], []]

如何将递归应用于此类情况?

How can I apply the recursion to this type of situations ?

推荐答案

这应该可以:

prune = lambda tree : [prune(branch) for branch in tree if branch != []]

l = [[[[], []], [[], []]], [[], [], []]]
prune(l)

避免使用 lambda(改进评论后的风格，另见 如何将这个 lambda 函数转换为 def 格式?) 函数可以定义为:

Avoiding the use of lambda (improving style following comments, and see also How do I transform this lambda function into def format?) the function can be defined as:

def prune(tree):
   return [prune(branch) for branch in tree if branch != []]

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