Question

我的Python代码中有一个元组结构，它声明了以下内容：

match_entry = (util.frozendict(rule_match), priority, version)

当我打印match_entry时，我看到以下内容：

print match_entry
({'switch': 1, 'dstmac': 00:00:00:00:00:01, 'srcmac': 00:00:00:00:00:01}, 60000, 5)

我正在寻找这个元组的元组字母，让我们称之为dict_of_tuples; dict的相应输出如下。

print dict_of_tuples

{({'switch': 5, 'dstmac': '00:00:00:00:00:00', 'srcmac': '00:00:00:00:00:01'}, 59999, 7): [CountBucket 140271056467472, CountBucket 140271056411280], ({'switch': 5, 'dstmac': '00:00:00:00:00:00', 'srcmac': '00:00:00:00:00:01'}, 59999, 5): [CountBucket 140271056467472, CountBucket 140271056411280], ({'switch': 1, 'dstmac': '00:00:00:00:00:01', 'srcmac': '00:00:00:00:00:01'}, 60000, 5): [CountBucket 140271057099664, CountBucket 140271056501008]}

但是，当我检查匹配条目是否在元组中时：

if match_entry in dict_of_tuples:

我没有看到任何结果，即使match_entry明显在dict_of_tuple中。有这种情况的原因是什么？

Answer 1

字典不可删除，所以我认为你的结构可能不可能：

>>> {({1:2, 2:3}, 2, 3): [1,2,3]}
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'dict'

词典不允许在其键中存在dict。所以要确保它真的不像字符串那样。

修改：

作为评论中的mentioned，冻结字典（不可变字典）将是可散列的，因此可以解决此问题。但是你必须修改默认的dict类型。通常，您可以在运行时修改dict：

d = {} d["key"] = "val"

已在Object中使用__setattr__启用了该功能。当我们需要不可变类？基本上，不可变对象在内存上更有效。以下是a topic。

Answer 2

        import frozendict as util
        from collections import defaultdict

        ### Create the first entry using hashable frozendict
        match = util.frozendict({'switch': 1, 'dstmac': '00:00:00:00:00:01', 'srcmac': '00:00:00:00:00:01'})
        match_array = tuple[match,60000,5]
        count_bucket2 = dict(CountBucket1 = 140271057099664, CountBucket2 = 140271056501008)

        ### Create the second entry using hashable frozendict
        match_entry = util.frozendict(switch= 5, dstmac= '00:00:00:00:00:00', srcmac= '00:00:00:00:00:01')
        match_array1 = tuple([match_entry, 59999, 7])

        count_bucket1 = dict(CountBucket1 = 140271056467472, CountBucket2 = 140271056411280)

        # Initialize the dictionary of tuples
        dict_of_tuples = ({tuple(match_array) : count_bucket2},{tuple(match_array1) : count_bucket1})

        ####### Your match entry
        match_entry = [{'switch': 1, 'dstmac': '00:00:00:00:00:01', 'srcmac': '00:00:00:00:00:01'},60000,5]                

    #Treating the final structure as a tuple. Each element of the tuple is a #dictionary.
        k = 0
        while k < len(dict_of_tuples):
            key = dict_of_tuples[k].iterkeys()
            val = dict_of_tuples[k].itervalues()
            if key.next() == tuple(match_entry):
                print ('Has key','corresponding value',val.next())
            else:
                print "Key not present"
            k+= 1

在dict中搜索元组

2 个答案: