如何从列表中删除每次出现的子列表

问题描述

我有两个列表:

big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]

我要删除big_list中所有出现的sub_list.

I want to remove all sub_list occurrences in big_list.

结果应为[2, 3, 4]

对于字符串，您可以使用以下代码:

For strings you could use this:

'2123124'.replace('12', '')

但是AFAIK不适用于列表.

But AFAIK this does not work for lists.

这不是从列表中删除子列表的重复项，因为我想从大列表中删除所有子列表.在另一个问题中，结果应为[5,6,7,1,2,3,4].

This is not a duplicate of Removing a sublist from a list since I want to remove all sub-lists from the big-list. In the other question the result should be [5,6,7,1,2,3,4].

更新:为简单起见，在此示例中，我采用了整数.但是列表项可以是任意对象.

Update: For simplicity I took integers in this example. But list items could be arbitrary objects.

Update2:

如果big_list = [1, 2, 1, 2, 1]和sub_list = [1, 2, 1]，我希望结果为[2, 1](如'12121'.replace('121', ''))

if big_list = [1, 2, 1, 2, 1] and sub_list = [1, 2, 1],I want the result to be [2, 1] (like '12121'.replace('121', ''))

Update3:

我不喜欢将StackOverflow中的源代码复制粘贴到我的代码中.这就是为什么我在软件建议中提出了第二个问题:

I don't like copy+pasting source code from StackOverflow into my code. That's why I created second question at software-recommendations: https://softwarerecs.stackexchange.com/questions/51273/library-to-remove-every-occurrence-of-sub-list-from-list-python

Update4:如果您知道要进行此方法调用的库，请将其编写为答案，因为这是我的首选解决方案.

Update4: if you know a library to make this one method call, please write it as answer, since this is my preferred solution.

测试应通过此测试:

def test_remove_sub_list(self):
    self.assertEqual([1, 2, 3], remove_sub_list([1, 2, 3], []))
    self.assertEqual([1, 2, 3], remove_sub_list([1, 2, 3], [4]))
    self.assertEqual([1, 3], remove_sub_list([1, 2, 3], [2]))
    self.assertEqual([1, 2], remove_sub_list([1, 1, 2, 2], [1, 2]))
    self.assertEquals([2, 1], remove_sub_list([1, 2, 1, 2, 1], [1, 2, 1]))
    self.assertEqual([], remove_sub_list([1, 2, 1, 2, 1, 2], [1, 2]))

推荐答案

您必须自己实现它.这是基本思想:

You'd have to implement it yourself. Here is the basic idea:

def remove_sublist(lst, sub):
    i = 0
    out = []
    while i < len(lst):
        if lst[i:i+len(sub)] == sub:
            i += len(sub)
        else:
            out.append(lst[i])
            i += 1
    return out

这将遍历原始列表的每个元素，并将其添加到输出列表(如果它不是子集的成员).这个版本的效率不是很高，但它的工作方式类似于您提供的字符串示例，因为它创建了一个不包含您的子集的新列表.只要它们支持==，它也适用于任意元素类型.从[1,1,1,1]中删除[1,1,1]会正确生成[1]，就像字符串一样.

This steps along every element of the original list and adds it to an output list if it isn't a member of the subset. This version is not very efficient, but it works like the string example you provided, in the sense that it creates a new list not containing your subset. It also works for arbitrary element types as long as they support ==. Removing [1,1,1] from [1,1,1,1] will correctly result in [1], as for a string.

这是一个 IDEOne链接，展示了

>>> remove_sublist([1, 'a', int, 3, float, 'a', int, 5], ['a', int])
[1, 3, <class 'float'>, 5]

这篇关于如何从列表中删除每次出现的子列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！