Question

我有一个列表列表，我想将列表与特定订单合并。见例：

id list  0         1           2
       [[0], [2, 6, 1, 4], [3, 7, 5]]

  Order          Resulting List
[1, 0, 2] = [2, 6, 1, 4, 0, 3, 7, 5]
[0, 2, 1] = [0, 3, 7, 5, 2, 6, 1, 4]
[2, 1, 0] = [3, 7, 5, 2, 6, 1, 4, 0]

有人可以建议下面提出的更优雅的算法吗？

    groups = [[0], [2, 6, 1, 4], [3, 7, 5]]
    orders = [[1, 0, 2], [0, 2, 1], [2, 1, 0]]

    for order in orders:
        LC = []
        for i in order:
            LC += groups[i]
    return LC

让我更好地解释一下我需要的东西：

groups = [[0], [2, 6, 1, 4], [3, 7, 5]] 
orders = [[0, 2, 1], [1, 0, 2], [2, 1, 0]] # Order of each group in LC
solutions = [] # I want to put the created LC here
for order in orders:
    LC = [] # I need this because a want LCs individualy and not one with all 
    for i in order: # for each order I pick de index (i) of the group
         LC += grupos[i] # and merge then according with index of group
        solutions.append([LC])
    print(solutions)

我想要这个（每个订单一个LC）：

[[0, 3, 7, 5, 2, 6, 1, 4], [2, 6, 1, 4, 0, 3, 7, 5], [3, 7, 5, 2, 6, 1, 4, 0]]

而不是这个：

[0, 3, 7, 5, 2, 6, 1, 4, 2, 6, 1, 4, 0, 3, 7, 5, 3, 7, 5, 2, 6, 1, 4, 0]

上面的算法有效，但需要另一个更优雅和高效。

输出的一些例子：

groups = [[0], [2, 1], [3, 7, 5], [4], [6]]

Order = [1, 0, 2, 3, 4]
LC = [2, 1, 0, 3, 7, 5, 4, 6]

    [2, 1, 0, 3, 4]
    [3, 7, 5, 2, 1, 0, 4, 6]

    [3, 1, 2, 0, 4]
    [4, 2, 1, 3, 7, 5, 0, 6]

    [4, 1, 2, 3, 0]
    [6, 2, 1, 3, 7, 5, 4, 0]

    [0, 2, 1, 3, 4]
    [0, 3, 7, 5, 2, 1, 4, 6]

    [0, 3, 2, 1, 4]
    [0, 4, 3, 7, 5, 2, 1, 6]

    [0, 4, 2, 3, 1]
    [0, 6, 3, 7, 5, 4, 2, 1]

    [0, 1, 3, 2, 4]
    [0, 2, 1, 4, 3, 7, 5, 6]

    [0, 1, 4, 3, 2]
    [0, 2, 1, 6, 4, 3, 7, 5]

    [0, 1, 2, 4, 3]
    [0, 2, 1, 3, 7, 5, 6, 4]

Answer 1

你可以使用其他一些技巧，比如理解。以下将返回一个单位列表：

return [part for order in orders for i in order for part in parts[i]]

以下将返回2D列表：

return [[part for i in order for part in parts[i]] for order in orders]

Answer 2

此解决方案与您提议的解决方案基本相同，但使用列表解析更多Python式。

>>> def merge_lists(desired_order):
...     merged_list = [element for i in desired_order for element in parts[i]]
...     return merged_list
... 
>>> desired_order = orders[0]
>>> merge_lists(desired_order)
[2, 6, 1, 4, 0, 3, 7, 5]

Answer 3

只需在索引上调用itertools.chain并与 operator.itemgetter 结合使用：

 frIn [9]: groups = [[0], [2, 6, 1, 4], [3, 7, 5]]

In [10]: orders = [[0, 2, 1], [1, 0, 2], [2, 1, 0]] # Ord

In [11]: from itertools import chain

In [12]: from operator import itemgetter


In [13]: [list(chain(*itemgetter(*o)(groups))) for o in orders]
    [[0, 3, 7, 5, 2, 6, 1, 4], [2, 6, 1, 4, 0, 3, 7, 5], [3, 7, 5, 2, 6, 1, 4, 0]]

在您自己的代码中，您只返回最后一个LC，因此无法正常工作：

for order in orders:
    LC = []  # overwritten each iteration so you only get the last sublists.
    for i in order:
        LC += parts[i]
return LC

Answer 4

inputs = [[1,2,3], [4,5,6], [7]]
orders = [[0,1,2], [2,1,0]]

result = [input_element for order in orders for order_element in order for input_element in inputs[order_element]]
print(result)

按特定顺序合并列表

4 个答案: