Python3：统计列表中重复字典的出现次数

我有一个由字典组成的列表，需要统计其中各唯一字典的出现次数。

示例输入：

[{'Basic': 100},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 200}]

期望输出格式：

{'Basic': 100} -> 3
{'Basic': 100, 'Food Allowance': 1000} -> 3
{'Basic': 200} -> 1

OR

[index_no] -> count

字典属于不可哈希类型，无法直接作为统计工具的键，因此需要先将每个字典转换为可哈希的结构，再进行次数统计。

方法1：转为排序后的键值对元组（推荐） #

将字典的键值对排序后转成元组，保证不同顺序的同内容字典会被判定为同一类，再用collections.Counter统计：

from collections import Counter

dict_list = [{'Basic': 100},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 200}]

# 转换为可哈希的排序元组
hashable_items = [tuple(sorted(d.items())) for d in dict_list]
count_result = Counter(hashable_items)

# 转换回字典格式输出
for key, count in count_result.items():
    print(f"{dict(key)} -> {count}")

运行输出：

{'Basic': 100} -> 3
{'Basic': 100, 'Food Allowance': 1000} -> 3
{'Basic': 200} -> 1

方法2：使用frozenset（不关心键顺序时可用） #

如果不介意字典键的顺序，也可以把键值对转成frozenset实现哈希：

from collections import Counter

dict_list = [{'Basic': 100},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 200}]

hashable_items = [frozenset(d.items()) for d in dict_list]
count_result = Counter(hashable_items)

for key, count in count_result.items():
    print(f"{dict(key)} -> {count}")

输出结果与方法1一致。

方法3：手动遍历统计（无依赖） #

不想用collections模块的话，可手动遍历实现统计：

dict_list = [{'Basic': 100},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 200}]

count_dict = {}
for d in dict_list:
    key = tuple(sorted(d.items()))
    count_dict[key] = count_dict.get(key, 0) + 1

for key, count in count_dict.items():
    print(f"{dict(key)} -> {count}")

输出[index_no] -> count格式 #

若需要按唯一字典的首次出现索引输出，可先收集唯一字典列表再统计：

dict_list = [{'Basic': 100},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 100},
             {'Basic': 100, 'Food Allowance': 1000},
             {'Basic': 200}]

unique_dicts = []
seen_keys = set()
for d in dict_list:
    key = tuple(sorted(d.items()))
    if key not in seen_keys:
        seen_keys.add(key)
        unique_dicts.append(d)

for idx, item in enumerate(unique_dicts):
    print(f"[{idx}] -> {dict_list.count(item)}")

运行输出：

[0] -> 3
[1] -> 3
[2] -> 1

内容的提问来源于stack exchange，提问作者Shahzad Nasir