Advanced structures - Dictionary of tuples in Python

Today, I want to show you how you can create a dictionary object with a dictionary as key and dictionary as value.

So, it is a Dict of Dict. The final object is stored as a tuple.

Remember the thumb rule that any key for a python dictionary must be hashable. So, we will use a feature called as frozendict in python.

Frozendict is hashable and can be used as a key in Python Dictionary.
You can use the below program effectively for searching for a key in Dictionary of Dictionaries in Python.

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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