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Python 从字典列表返回唯一值的计数_Python_List_Dictionary_List Comprehension - Fatal编程技术网
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Python 从字典列表返回唯一值的计数

python list dictionary

Python 从字典列表返回唯一值的计数,python,list,dictionary,list-comprehension,Python,List,Dictionary,List Comprehension,我有一份字典清单： event_results = [{'device': 'prod2261.example.com'}, {'device': 'npbodb253.example.com'}, {'device': 'db221.example.com'}, {'device': 'db219.example.com'}, {'device': 'db209.example.com'}, {'device': 'db243.example.com'}, {'device': 'prod27

我有一份字典清单：

event_results = [{'device': 'prod2261.example.com'}, {'device': 'npbodb253.example.com'}, {'device': 'db221.example.com'}, {'device': 'db219.example.com'}, {'device': 'db209.example.com'}, {'device': 'db243.example.com'}, {'device': 'prod277.example.com'}, {'device': 'prod2228.example.com'}, {'device': 'prod2252.example.com'}, {'device': 'prod2116.example.com'}, {'device': 'prod224.example.com'}, {'device': 'db223.example.com'}, {'device': 'db229.example.com'}, {'device': 'prod2116.example.com'}, {'device': 'db221.example.com'}, {'device': 'db239.example.com'}, {'device': 'npbodb249.example.com'}, {'device': 'db210.example.com'}, {'device': 'db219.example.com'}, {'device': 'db243.example.com'}, {'device': 'prod2210.example.com'}, {'device': 'prod224.example.com'}, {'device': 'npbodb253.example.com'}, {'device': 'npbovwa018.example.com'}, {'device': 'npbovwa018.example.com'}, {'device': 'db221.example.com'}, {'device': 'db243.example.com'}, {'device': 'prod2228.example.com'}]
我需要获得值，以及唯一值的计数

示例输出需要是：

prod2261.example.com, 2
prod2261.example.com, 5
....etc....
到目前为止，我可以得到一组所有唯一的值：

unique_values = set(e['device'] for e in event_results)
但是我正在努力迭代这两个列表，只返回unique_值集中每个项目的计数，而不创建一组临时列表，这样可以在其中存储值，然后在最后返回len（）

我认为解决方案与列表理解有关，但我就是想不起来。

只需使用
集合。计数器
而不是
集合
就可以了：）


哇，你说得对！即使在设置中使用itemgetter运行
timeit
，这也比字典查找慢得多（11s vs 8s）。我被彻底纠正了

import collections
unique_counts = collections.Counter(e['device'] for e in event_results)