按其他列表中项目的存在对列表进行排序 [英] Sort a list by presence of items in another list

问题描述

假设我有两个列表:

a = ['30', '10', '90', '1111', '17']
b = ['60', '1201', '30', '17', '900']

您如何最有效地对此进行排序，例如:

How would you sort this most efficiently, such that:

列表b相对于a进行了排序. b中的唯一元素应放在排序列表的末尾. a中的唯一元素可以忽略.

list b is sorted with respect to a. Unique elements in b should be placed at the end of the sorted list. Unique elements in a can be ignored.

示例输出:

c = ['30', '17', '60', '1201', '900']

对不起，这是一个简单的问题.我的尝试停留在采取交叉路口的位置.

Sorry, it's a simple question. My attempt is stuck at the point of taking the intersection.

intersection = sorted(set(a) & set(b), key = a.index)

推荐答案

这里没有必要进行实际排序.您希望a中的元素位于b中，并且顺序与a中的元素相同；其次是b中不在a中的元素，其顺序与在b中的顺序相同.

There is no need to actually sort here. You want the elements in a which are in b, in the same order as they were in a; followed by the elements in b which are not in a, in the same order as they were in b.

我们可以使用两个过滤器，使用用于快速成员资格测试的集合来做到这一点:

We can just do this with two filters, using the sets for fast membership tests:

>>> a = ['30', '10', '90', '1111', '17']
>>> b = ['60', '1201', '30', '17', '900']
>>> a_set = set(a)
>>> b_set = set(b)
>>> [*filter(lambda x: x in b_set, a), *filter(lambda x: x not in a_set, b)]
['30', '17', '60', '1201', '900']

或者，如果您更喜欢理解:

Or if you prefer comprehensions:

>>> [*(x for x in a if x in b_set), *(x for x in b if x not in a_set)]
['30', '17', '60', '1201', '900']

两者都需要线性时间，这比排序要好.

Both take linear time, which is better than sorting.

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