在python中使用列表进行递归 [英] recursion with a list in python

问题描述

我刚刚开始学习python，有些递归问题似乎无法解决.最烦人的是:我需要构建一个函数 ind(e，L)，其中 e 是一个int值，而 L 是一个列表.

I just started learning python and there are some recursion questions that I can't seem to figure out. The most annoying one is this: I need to build a function ind(e,L) where e is an int and L is a list.

通过输入 e (如果它在列表中)，需要将输出作为其索引例如:

By entering e if it is in the list the output needs to be its index For example:

ind(42,[0,14,52,42,15]) -> 3

这是我到目前为止编写的代码，但是我得到的索引始终为0.有人可以向我解释我做错了什么吗?

This is the code I wrote this far but the index I get is always 0. Can someone please explain to me what I am doing wrong?

def location(e,L):
    if L == []:
        return False
    elif e == L[0]:
        A = L[:-1].index(e)
        return A
    else:
        return location(e,L[1:])

print(location(14,[1,2,14,1]))

谢谢:)

推荐答案

仅当 e 位于索引0时返回(可以跳过 L [:-1] ...项，它始终为0)，并且传播该值不变.而不是返回无意义的索引，而是返回递归数.最简单的方法是在函数递归时加1.

You only return if e is at index 0 (you can skip the L[:-1]... term, it is always 0) and propagate that unchanged. Instead of returning the meaningless index, return the number of recursions. The simplest way is to add 1 whenever the function recurses.

def location(element, sequence):
    if not sequence:
        # e is not in the list at all
        # it is not meaningful to return an index
        raise IndexError
    elif element == sequence[0]:
        # we already know where e is
        # since we checked it explicitly
        return 0
    else:
        # keep searching in the remainder,
        # but increment recursion level by 1
        return 1 + location(element, sequence[1:])

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