以给定的合并顺序合并两个或多个列表 [英] Merge two or more lists with given order of merging

问题描述

一开始我有 2 个列表和 1 个列表，说明我应该以什么顺序合并这两个列表.例如，我的第一个列表等于 [a, b, c]，第二个列表等于 [d, e]，'merging' 列表等于 [0, 1, 0, 0, 1].

On start I have 2 lists and 1 list that says in what order I should merge those two lists. For example I have first list equal to [a, b, c] and second list equal to [d, e] and 'merging' list equal to [0, 1, 0, 0, 1].

这意味着:要首先制作合并列表，我需要从第一个列表中获取元素，然后是第二个，然后是第一个，然后是第一个，然后是第二个......最后我得到 [a, d, b, c, e].为了解决这个问题，我只使用了 for 循环和两个指针"，但我想知道我是否可以更 Python 地完成这个任务......我试图找到一些可以帮助我的函数，但没有真正的结果.

That means: to make merged list first I need to take element from first list, then second, then first, then first, then second... And I end up with [a, d, b, c, e]. To solve this I just used for loop and two "pointers", but I was wondering if I can do this task more pythonic... I tried to find some functions that could help me, but no real result.

推荐答案

您可以从这些列表创建迭代器，遍历排序列表，并调用 next 在其中一个迭代器上:

You could create iterators from those lists, loop through the ordering list, and call next on one of the iterators:

i1 = iter(['a', 'b', 'c'])
i2 = iter(['d', 'e'])
# Select the iterator to advance: `i2` if `x` == 1, `i1` otherwise
print([next(i2 if x else i1) for x in [0, 1, 0, 0, 1]]) # ['a', 'd', 'b', 'c', 'e']

可以将此解决方案推广到任意数量的列表，如下所示

It's possible to generalize this solution to any number of lists as shown below

def ordered_merge(lists, selector):
    its = [iter(l) for l in lists]
    for i in selector:
        yield next(its[i])

In [4]: list(ordered_merge([[3, 4], [1, 5], [2, 6]], [1, 2, 0, 0, 1, 2]))
Out[4]: [1, 2, 3, 4, 5, 6]

如果排序列表包含字符串、浮点数或任何其他不能用作列表索引的对象，请使用字典:

If the ordering list contains strings, floats, or any other objects that can't be used as list indexes, use a dictionary:

def ordered_merge(mapping, selector):
    its = {k: iter(v) for k, v in mapping.items()}
    for i in selector:
        yield next(its[i])

In [6]: mapping = {'A': [3, 4], 'B': [1, 5], 'C': [2, 6]}

In [7]: list(ordered_merge(mapping, ['B', 'C', 'A', 'A', 'B', 'C']))
Out[7]: [1, 2, 3, 4, 5, 6]

当然，您也可以使用整数作为字典键.

Of course, you can use integers as dictionary keys as well.

或者，您可以从每个原始列表的左侧逐个删除元素，然后将它们添加到结果列表中.快速示例:

Alternatively, you could remove elements from the left side of each of the original lists one by one and add them to the resulting list. Quick example:

In [8]: A = ['a', 'b', 'c']
   ...: B = ['d', 'e']
   ...: selector = [0, 1, 0, 0, 1]
   ...: 

In [9]: [B.pop(0) if x else A.pop(0) for x in selector]
Out[9]: ['a', 'd', 'b', 'c', 'e']

我希望第一种方法更有效(list.pop(0) 慢).

I would expect the first approach to be more efficient (list.pop(0) is slow).

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