基于键的字典的唯一列表 [英] Unique list of dicts based on keys

问题描述

我有一个dic列表：

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

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

如何我确保对于列表中的每个数据项，每个键仅存在一个这样的元素？如果有两个键（如下所示），则最新的时间戳将获胜：

How can I ensure that for each data item in the list, only one such element exists for each "key"? If there are two keys as seen below, the most recent timestamp would win:

    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。

推荐答案

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)]

希望对别人有帮助。

这篇关于基于键的字典的唯一列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！