Given list of dictionaries, convert to ordered key dictionary with each key contained dictionary as its nested value.
Input : test_list = [{“Gfg” : 3, 4 : 9}, {“is”: 8, “Good” : 2}]
Output : {0: {‘Gfg’: 3, 4: 9}, 1: {‘is’: 8, ‘Good’: 2}}
Explanation : List converted to dictionary with index keys.Input : test_list = [{“is”: 8, “Good” : 2}]
Output : {1: {‘is’: 8, ‘Good’: 2}}
Explanation : List converted to dictionary with index keys, just one row.
Method #1 : Using loop + enumerate()
This is brute way in which this task can be performed. In this, we iterate through the index and value together using enumerate and create custom required dictionary.
Python3
# Python3 code to demonstrate working of # Convert Dictionaries List to Order Key Nested dictionaries# Using loop + enumerate() # initializing liststest_list = [{"Gfg" : 3, 4 : 9}, {"is": 8, "Good" : 2}, {"Best": 10, "CS" : 1}] # printing original listprint("The original list : " + str(test_list)) # using enumerate() to extract key to map with dict values res = dict()for idx, val in enumerate(test_list): res[idx] = val # printing result print("The constructed dictionary : " + str(res)) |
The original list : [{'Gfg': 3, 4: 9}, {'is': 8, 'Good': 2}, {'Best': 10, 'CS': 1}]
The constructed dictionary : {0: {'Gfg': 3, 4: 9}, 1: {'is': 8, 'Good': 2}, 2: {'Best': 10, 'CS': 1}}
Method #2 : Using dictionary comprehension + enumerate()
This is similar to above method, the only difference is that dictionary comprehension is used instead of loop to perform task of encapsulation.
Python3
# Python3 code to demonstrate working of # Convert Dictionaries List to Order Key Nested dictionaries# Using dictionary comprehension + enumerate() # initializing liststest_list = [{"Gfg" : 3, 4 : 9}, {"is": 8, "Good" : 2}, {"Best": 10, "CS" : 1}] # printing original listprint("The original list : " + str(test_list)) # dictionary comprehension encapsulating result as one linerres = {idx : val for idx, val in enumerate(test_list)} # printing result print("The constructed dictionary : " + str(res)) |
The original list : [{'Gfg': 3, 4: 9}, {'is': 8, 'Good': 2}, {'Best': 10, 'CS': 1}]
The constructed dictionary : {0: {'Gfg': 3, 4: 9}, 1: {'is': 8, 'Good': 2}, 2: {'Best': 10, 'CS': 1}}
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