Python

Question

我想为每个元素只能出现一次的数组创建所有可能的排列，并限制元素数组索引位置。

ID = ["A","B","C","D","E","F","G","H","I","J"]

我想创建 original_array 的所有可能排列，但是每个元素的位置都限制在以下索引位置：

ID = ["A","B","C","D","E","F","G","H","I","J"]

Index_Options=[]
for i in range(len(ID)):
    List1=[]
    distance=3
    value = i - distance
    for j in range((int(distance)*2)):
        if value < 0 or value > len(ID):
            print("Disregard") #Outside acceptable distance range
        else:
            List1.append(value)
        value=value+1
    Index_Options.append(List1)
    print(Index_Options)

#Index_Options gives the possible index positions for each element. ie "A" can occur in only index positions 0,1,2, "B" can occur in only index positions 0,1,2,3 ect.

我只是在苦苦思索如何使用这些信息来创建所有输出排列。

如有任何帮助，我们将不胜感激

Answer 1

您可以使用 itertools.permutations 创建所有可能的排列，然后创建新列表并检查所有字母是否都在正确的位置

permutations = [p for p in itertools.permutations(ID, len(ID)) if all(i in Index_Options[ID.index(x)] for i, x in enumerate(p))]

Answer 2

您可以使用递归生成器函数来构建组合。与其从 ID 生成所有可能的排列然后基于 Index_Options 进行过滤，不如通过直接遍历 Index_Options 来生成 ID 的笛卡尔积更高效：

ID = ["A","B","C","D","E","F","G","H","I","J"]
def combos(d, c = [], s = []):
  if not d:
     yield c
  else:
     for i in filter(lambda x:x not in s and x < len(ID), d[0]):
        yield from combos(d[1:], c=c+[ID[i]], s=s+[i])

print(list(combos(Index_Options)))

输出（产生的前十个组合）：

[['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'], ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'J', 'I'], ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'I', 'H', 'J'], ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'I', 'J', 'H'], ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'J', 'H', 'I'], ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'J', 'I', 'H'], ['A', 'B', 'C', 'D', 'E', 'F', 'H', 'G', 'I', 'J'], ['A', 'B', 'C', 'D', 'E', 'F', 'H', 'G', 'J', 'I'], ['A', 'B', 'C', 'D', 'E', 'F', 'H', 'I', 'G', 'J'], ['A', 'B', 'C', 'D', 'E', 'F', 'H', 'I', 'J', 'G']]

Python - 使用输出数组索引约束创建排列

Python - Creating permutations with output array index constraints

iteration

permutation

itertools