计算Python中字典键的唯一值的数量

Question

我已经使用从错误日志中提取的值创建了嵌套字典。

ErrorTypes是“消息”和“CMD”

dict1 = { 
    '1234': [ 
     ['2014-09-19 09:55:41', 1234, 'Msg', 'N/A', 'N/A'], 
     ['2014-09-19 10:22:22', 123, 'Msg', 123, 'N/A'], 
     ['2014-09-19 12:22:21', 1234, 'Cmd', 123, 'N/A'] 
    ] 
} 

什么是我怎样才能得到一个计数，所以我知道了每个错误类型的多为每个键存在的代码？

这是我到目前为止有：

from collections import Counter 

dict2 = {} 
for k in dict1: 
     for v in k: 
       dict2[str(k)].append(str(k),collection.Counter(v[k] for v in dict1)) 

请帮帮忙！谢谢。

Answer 1

您可以遍历列表并制作set错误类型，然后检查该集合的长度。根据定义，集合可能只包含独特的项目，因此您不会重复。

dict1 = {'1234': [['2014-09-19 09:55:41', 1234, 'Msg', 'N/A', 'N/A'], 
        ['2014-09-19 10:22:22', 123, 'Msg', 123, 'N/A'], 
        ['2014-09-19 12:22:21', 1234, 'Cmd', 123, 'N/A']]} 

>>> set(i[2] for i in dict1['1234']) 
set(['Msg', 'Cmd']) 

>>> len(set(i[2] for i in dict1['1234'])) 
2 

为了让您在您的评论

dict1 = {'1234': [['2014-09-19 09:55:41', 1234, 'Msg', 'N/A', 'N/A'], 
        ['2014-09-19 10:22:22', 123, 'Msg', 123, 'N/A'], 
        ['2014-09-19 12:22:21', 1234, 'Cmd', 123, 'N/A']], 
     '5678': [['2014-09-19 09:55:41', 1234, 'Foo', 'N/A', 'N/A'], 
        ['2014-09-19 10:22:22', 123, 'Bar', 123, 'N/A'], 
        ['2014-09-19 12:22:21', 1234, 'Bar', 123, 'N/A']]} 

from collections import Counter 

>>> {key : Counter([i[2] for i in dict1[key]]) for key in dict1.keys()} 
{'1234': Counter({'Msg': 2, 'Cmd': 1}), 
'5678': Counter({'Bar': 2, 'Foo': 1})}