递归遍历列表(python) [英] Recursively going through a list (python)

问题描述

说我有一个列表x = [1、2、3、4]

Say I have a list x = [1, 2, 3, 4]

是否有一种递归方法可以在列表中查找值?

Is there a recursive method where i can go through the list to find the value?

我希望最终能够将列表(或嵌套列表)中的返回值与任意数字进行比较，以查看它是否匹配.

I want to ultimately be able to compare a returned value in the list, (or nested list) to an arbitrary number to see it it matches.

我可以考虑使用for循环执行此操作的方法，但是我很难想象使用递归方法来执行相同的操作.我知道我无法设置计数器来跟踪我在列表中的位置，因为递归调用函数只会每次都重置计数器.

I can think a way to do this using a for loop, but i have trouble imagining a recursive method to do the same thing. I know that I can't set a counter to keep track of my position in the list because calling the function recursively would just reset the counter every time.

我当时以为我可以将函数的基本情况设置为数字和len 1列表之间的比较.

I was thinking I could set my base case of the function as a comparison between the number and a list of len 1.

我只想要一些提示.

推荐答案

这不是在Python中执行操作的方法，但是可以肯定-您可以递归遍历列表列表:

This is not the way to do things in Python, but surely - you can traverse a list of lists recursively:

def findList(lst, ele):
    if not lst:         # base case: the list is empty
        return False
    elif lst[0] == ele: # check if current element is the one we're looking
        return True
    elif not isinstance(lst[0], list): # if current element is not a list
        return findList(lst[1:], ele)
    else:                              # if current element is a list
        return findList(lst[0], ele) or findList(lst[1:], ele)

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