Python中利用itertools处理排列组合：生成与特定属性概率计算

import itertools

main_list = ['a', 'b', 'c', 'd']
all_permutations = []

# 遍历所有可能的长度 r，从 1 到 len(main_list)
for r in range(1, len(main_list) + 1):
    # 生成长度为 r 的所有排列
    perms_of_length_r = itertools.permutations(main_list, r)
    # 将这些排列添加到总列表中
    # 注意：itertools.permutations 返回的是元组，这里转换为列表以便后续处理
    all_permutations.extend([list(p) for p in perms_of_length_r])

# 另一种更简洁的写法，使用 itertools.chain
# all_permutations = [
#     list(p) for p in itertools.chain(*(itertools.permutations(main_list, r)
#                                        for r in range(1, len(main_list) + 1)))
# ]

print(f"生成的总排列数: {len(all_permutations)}")
# 打印部分排列示例
# for i, p in enumerate(all_permutations[:10]):
#     print(f"  {i+1}: {p}")

import itertools

main_list = ['a', 'b', 'c', 'd']
all_permutations = []

# 生成所有长度的排列
for r in range(1, len(main_list) + 1):
    perms_of_length_r = itertools.permutations(main_list, r)
    all_permutations.extend([list(p) for p in perms_of_length_r])

N = len(all_permutations) # 总排列数

probabilities = {}

# 计算每个排列的“元素构成概率”
for current_perm in all_permutations:
    # 获取当前排列的唯一元素集合
    current_perm_set = set(current_perm)

    # 计数与当前排列具有相同唯一元素集合的排列数量
    count_same_set_perms = 0
    for other_perm in all_permutations:
        if current_perm_set == set(other_perm):
            count_same_set_perms += 1

    # 计算“元素构成概率”
    # 注意：这里将结果存储为元组，因为列表不可哈希作为字典键
    probabilities[tuple(current_perm)] = (N - count_same_