Question

我创建了嵌套字典，其中包含从错误日志中提取的值。

ErrorTypes是“Msg”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']
    ]
}

我如何获取计数的代码是什么，所以我知道每个密钥存在多少个错误类型？

这是我到目前为止所做的：

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))

请帮忙！谢谢。

Answer 1

您可以遍历列表列表并生成set错误类型，然后只检查该集的长度。根据定义，集合可能只包含唯一的项目，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

制作评论中描述的词典

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})}

计算Python中字典键的唯一值的数量

1 个答案: