如何统计字典中相似列表组合的出现次数并生成命名列表？

输入的字典集合：

{'HH1': ['x'], 'HH2': ['y', 'x'], 'HH3': ['x', 'z'], 'HH4': ['x'], 'HH5': ['x'], 'HH6': ['x'], 'HH7': ['x'], 'HH8': ['x', 'y', 'z'], 'HH9': ['x'], 'HH10': ['x', 'y'], 'HH11': ['x'], 'HH12': ['x'], 'HH13': ['x'], 'HH14': ['x'], 'HH15': ['x', 'y'], 'HH16': ['x', 'y'], 'HH17': ['x', 'y'], 'HH18': ['x']}

需求 #

• 统计每个完全相同的元素组合的出现次数i（元素顺序不影响，比如['x','y']和['y','x']视为同一组合）
• 生成每个组合对应的结果，命名格式为n=i

期望输出 #

n=11: ('x',)
n=5: ('x', 'y')
n=1: ('x', 'z')
n=1: ('x', 'y', 'z')

两次尝试的问题 #

尝试1：遗漏组合

代码：

from collections import Counter
from itertools import combinations

# Counting occurrences of each combination of fuel types
combination_count = Counter()
unique_combinations = set()
for fuel_list in hfuels.values():
    for r in range(1, len(fuel_list) + 1):
        for combination in combinations(fuel_list, r):
            unique_combinations.add(tuple(sorted(combination)))

# Creating new lists for each combination
renamed_lists = {}
for combination in unique_combinations:
    count = sum(1 for fuel_list in hfuels.values() if set(combination) == set(fuel_list))
    if count:
        renamed_lists[f"n={count}"] = list(combination)

# Printing the renamed lists
for name, fuel_list in renamed_lists.items():
    print(f"{name}: {fuel_list}")

执行结果：

n=1: ['z', 'y', 'x']
n=5: ['y', 'x']
n=11: ['x']

问题原因：

1. 错误生成了所有子组合（比如从['x','y','z']中拆分出('x','y')等子组合），而非统计原列表的完整组合。
2. 用字典存储结果时，相同次数的组合会被覆盖（比如('x','z')和('x','y','z')的次数都是1，后者会覆盖前者，导致前者丢失）。

尝试2：统计子组合次数而非完整组合次数

代码：

from collections import Counter
from itertools import combinations

# Counting occurrences of each combination of fuel types
combination_count = Counter()
unique_combinations = set()
for fuel_list in hfuels.values():
    for r in range(1, len(fuel_list) + 1):
        for combination in combinations(fuel_list, r):
            combination_count[tuple(sorted(combination))] += 1
            unique_combinations.add(tuple(sorted(combination)))

# Creating new lists for each combination
renamed_lists = {}
for combination in unique_combinations:
    count = combination_count[combination]
    if count:
        renamed_lists[f"n={count}"] = list(combination)

# Printing the renamed lists
for name, fuel_list in renamed_lists.items():
    print(f"{name}: {fuel_list}")

执行结果：

n=1: ['z', 'y', 'x']
n=2: ['z']
n=6: ['y']
n=18: ['x']

问题原因：
代码统计的是所有子组合的出现次数（比如每个包含x的列表都会贡献一次('x')），而非原列表完整组合的出现次数，完全偏离需求。

正确实现方法 #

核心思路：直接将每个原列表转换为排序后的元组（解决列表不可哈希、元素顺序干扰的问题），然后用Counter统计每个元组的出现次数即可。

代码：

from collections import Counter

hfuels = {'HH1': ['x'], 'HH2': ['y', 'x'], 'HH3': ['x', 'z'], 'HH4': ['x'], 'HH5': ['x'], 'HH6': ['x'], 'HH7': ['x'], 'HH8': ['x', 'y', 'z'], 'HH9': ['x'], 'HH10': ['x', 'y'], 'HH11': ['x'], 'HH12': ['x'], 'HH13': ['x'], 'HH14': ['x'], 'HH15': ['x', 'y'], 'HH16': ['x', 'y'], 'HH17': ['x', 'y'], 'HH18': ['x']}

# 将每个列表转为排序后的元组，确保顺序不影响组合判断
sorted_combinations = [tuple(sorted(lst)) for lst in hfuels.values()]
# 统计每个组合的出现次数
counts = Counter(sorted_combinations)

# 按需求格式输出
for combo, count in counts.items():
    print(f"n={count}: {combo}")

执行结果：

n=11: ('x',)
n=5: ('x', 'y')
n=1: ('x', 'z')
n=1: ('x', 'y', 'z')

内容的提问来源于stack exchange，提问作者Nike