基于键的唯一字典列表

Question

我有一个 dics 列表：

     data = {}
     data['key'] = pointer_key
     data['timestamp'] = timestamp
     data['action'] = action
     data['type'] = type
     data['id'] = id

     list = [data1, data2, data3, ... ]

如何确保对于列表中的每个数据项，每个“键”只存在一个这样的元素？如果有如下所示的两个键，则最近的时间戳将获胜：

    list = [{'key':1,'timestamp':1234567890,'action':'like','type':'photo',id:245},
            {'key':2,'timestamp':2345678901,'action':'like','type':'photo',id:252},
            {'key':1,'timestamp':3456789012,'action':'like','type':'photo',id:212}]

    unique(list)

    list = [{'key':2,'timestamp':2345678901,'action':'like','type':'photo',id:252},
            {'key':1,'timestamp':3456789012,'action':'like','type':'photo',id:212}]

谢谢。

Answer 1

这是我的解决方案：

def uniq(list_dicts):
    return [dict(p) for p in set(tuple(i.items()) 
        for i in list_dicts)]

希望它会帮助某人。

Answer 2

我需要这个，但不喜欢这里的任何答案。所以我做了这个简单且高性能的版本。

def list_of_seq_unique_by_key(seq, key):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if x[key] not in seen and not seen_add(x[key])]

# Usage
# If you want most recent timestamp to win, just sort by timestamp first
list = sorted(list, key=lambda k: k['timestamp'], reverse=True)
# Remove everything with a duplicate value for key 'key'
list = list_of_seq_unique_by_key(list, 'key')

Answer 3

你不需要。根据定义，一个字典对于给定的键只能有一个条目。