如何提取字典列表中不属于出现频率最高城市的客户名单

以下实现时间复杂度为O(n)，仅需遍历两次客户列表，无冗余计算，是性能最优的写法：

from collections import Counter

# 输入数据
customers = [{'customer': 'Charles', 'city': 'Paris'}, {'customer': 'John', 'city': 'New York'}, {'customer': 'Jean', 'city': 'Paris'}]

# 统计城市频次、取最高频城市
city_counts = Counter(c['city'] for c in customers)
most_common_city = city_counts.most_common(1)[0][0]

# 列表推导式过滤不符合要求的客户
result = [c['customer'] for c in customers if c['city'] != most_common_city]

示例输入下运行后result值为['John']，和预期一致。

如果存在多个城市出现频次相同且均为最高的场景，可以使用如下兼容写法，避免只匹配第一个最高频城市的问题：

from collections import Counter

customers = [{'customer': 'Charles', 'city': 'Paris'}, {'customer': 'John', 'city': 'New York'}, {'customer': 'Jean', 'city': 'Paris'}, {'customer': 'Amy', 'city': 'New York'}]

city_counts = Counter(c['city'] for c in customers)
# 取最高频次
max_count = max(city_counts.values())
# 收集所有最高频的城市
most_common_cities = {city for city, cnt in city_counts.items() if cnt == max_count}

# 过滤
result = [c['customer'] for c in customers if c['city'] not in most_common_cities]

内容的提问来源于stack exchange，提问作者Synops