使用组合复制列表中的项目

Question

Python 中有两个列表：

i = [0, 1, 2, 3, 4]
j = [1, 4]

我正在尝试复制 j 中的值并获取所有唯一组合。 i可以是任何（可散列的）值的列表。 它永远不会包含重复项，但值的顺序很重要。 j始终是i的子集。 整数仅用作示例，实际值可以是字符串。

output 应该是：

[
[0, 1, 2, 3, 4], # i is always included as a combination - this can be added independently if required
[0, 1, 1, 2, 3, 4],
[0, 1, 1, 2, 3, 4, 4],
[0, 1, 2, 3, 4, 4]
]

为清楚起见，另一个示例：

i = ["B", "E", "C", "A", "D"]
j = ["C", "A", "D"]

# desired output - the order of i should be preserved
# but the order of lists within the outer list is unimportant
[
["B", "E", "C", "A", "D"], # original list
["B", "E", "C", "C", "A", "D"], # duplicated C
["B", "E", "C", "C", "A", "A", "D"], # duplicated C, A
["B", "E", "C", "C", "A", "D", "D"], # duplicated C, D
["B", "E", "C", "C", "A", "A", "D", "D"], # duplicated C, A, D
["B", "E", "C", "A", "A", "D"], # duplicated A
["B", "E", "E", "C", "A", "A", "D", "D"], # duplicated A, D
["B", "E", "C", "A", "D", "D"], # duplicated D
]

我尝试了各种列表理解方法，以及一些 itertools 函数，如产品和组合，但还没有找到令人满意的解决方案。

半工半程的尝试如下，虽然这是一种丑陋的方法：

i = [0, 1, 2, 3, 4]
j = [1, 4]

all_lists = [i]
for x in j:
    new_list = i.copy()
    new_list.insert(i.index(x), x)
    all_lists.append(new_list)

print(all_lists)
# [[0, 1, 2, 3, 4], [0, 1, 1, 2, 3, 4], [0, 1, 2, 3, 4, 4]]
# still missing the [0, 1, 1, 2, 3, 4, 4] case

问题的背景 - i是 networkx 网络中的节点列表， j是具有自环的节点列表。 结果是节点之间所有非简单路径的组合。

Answer 1

您尝试使用itertools是对的。 function combinations就是您想要的。 只需遍历所有可能的长度（从 0 到 len(j)）：

import itertools

i = ["B", "E", "C", "A", "D"]
j = ["C", "A", "D"]

result = []
for l in range(len(j) + 1):
    for j_subset in itertools.combinations(j, l):
        result.append(sorted(i + list(j_subset), key=lambda x: i.index(x)))


print(result)

您说列表可以包含任何类型的元素。 这是使用自定义 class Node执行相同操作的示例：

import itertools

# here is my custom Node class
class Node():
    def __init__(self, name):
        self.name = name

    def __eq__(self, other):
        return self.name == other.name

# the names of the nodes
names_i = ["B", "E", "C", "A", "D"]
names_j = ["C", "A", "D"]


# the actual data we are working with - lists of distinct nodes
i = [Node(name) for name in names_i]
j = [Node(name) for name in names_j]

result = []

for l in range(len(j) + 1):
    for j_subset in itertools.combinations(j, l):
        result.append(sorted(i + list(j_subset), key=lambda x: i.index([node for node in i if node == x][0])))


# print the raw result
print(result)

# print the result but just showing the node names
print([[node.name for node in result_element] for result_element in result])

使用类的主要区别在于，使用相同名称实例化的两个节点类不是同一个 object，因此sorted键中的i.index(x)将不起作用，因为i中没有j的元素（即使它们有同名，并且“相等”）。 如果您愿意，我可以对此进行更多解释。

Answer 2

更改以反映j是要复制的值的列表

for k in range(len(j) + 1):
    for n in itertools.combinations(j, k):
        lst = i[:]
        for el in n:
            ix = lst.index(el)
            lst.insert(ix, lst[ix])
        print(lst)

使用insert进行编辑比我以前的效率更高

Answer 3

这是你要找的吗？

In [1]: %paste                                                                                                                                                                                                                                                                       
from itertools import combinations

i = [0, 1, 2, 3, 4]
j = [1, 4]

res = []
for r in range(len(j) + 1):
    for comb in combinations(j, r):
        res.append(sorted(i + list(comb)))
print(res)

## -- End pasted text --
[[0, 1, 2, 3, 4], [0, 1, 1, 2, 3, 4], [0, 1, 2, 3, 4, 4], [0, 1, 1, 2, 3, 4, 4]]

In [2]: