检查列表是否是子列表 [英] Checking if list is a sublist

问题描述

我需要检查 list1 是否是 list2 的子列表(真；如果 list2 中与 list1 相同的每个整数都与 list1 中的索引顺序相同)

I need to check if list1 is a sublist of list2 (True; if every integer in list2 that is common with list1 is in the same order of indexes as in list1)

def sublist(lst1,lst2):
    for i in range(len(lst1)):
        if lst1[i] not in lst2:
            return False
        for j in range(len(lst2)):
            if (lst1[j] in lst2) and (lst2.index(lst1[i+1]) > lst2.index(lst1[i])):
                return True

谁能帮帮我...为什么这不起作用?

Can anybody help me... why isn't this working?

推荐答案

我需要检查 list1 是否是 list2 的子列表(真；如果 list2 中与 list1 共有的每个整数都与 list1 中的索引顺序相同)

i need to check if list1 is a sublist to list2 (True; if every integer in list2 that is common with list1 is in the same order of indexes as in list1)

您的代码不起作用，因为只要 ls1 中的列表元素没有出现在 ls2 中，它就会立即返回 False.

Your code isn't working because as soon as a list element in ls1 doesn't occur in ls2 it will return False immediately.

这将创建两个仅包含公共元素(但按其原始顺序)的列表，然后在它们相同时返回 True:

This creates two lists that contain only the common elements (but in their original order) and then returns True when they are the same:

def sublist(lst1, lst2):
   ls1 = [element for element in lst1 if element in lst2]
   ls2 = [element for element in lst2 if element in lst1]
   return ls1 == ls2

内存高效的变体:

def sublist(ls1, ls2):
    '''
    >>> sublist([], [1,2,3])
    True
    >>> sublist([1,2,3,4], [2,5,3])
    True
    >>> sublist([1,2,3,4], [0,3,2])
    False
    >>> sublist([1,2,3,4], [1,2,5,6,7,8,5,76,4,3])
    False
    '''
    def get_all_in(one, another):
        for element in one:
            if element in another:
                yield element

    for x1, x2 in zip(get_all_in(ls1, ls2), get_all_in(ls2, ls1)):
        if x1 != x2:
            return False

    return True

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