While working with Python, we can have a problem in which we need to initialize a list of a particular size with custom dictionaries. This task has it’s utility in web development to store records. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using {dict} + "*" operator
This task can be performed using the “*” operator. We can create a list containing single custom dictionary and then multiply it by Number that is size of list. The drawback is that similar reference dictionaries will be made which will point to similar memory location.
# Python3 code to demonstrate working of # Custom dictionary initialization in list # using {dict} + "*" operator # Initialize dict test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3} # Custom dictionary initialization in list # using {dict} + "*" operator res = [test_dict] * 6 print("The list of custom dictionaries is : " + str(res)) |
The list of custom dictionaries is : [{'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}]
Method #2 : Using {dict} + list comprehension
This is perhaps the better and correct way to perform this task. We initialize the each index of list with dictionary, this way, we have independently referring dictionaries and don’t point to single reference.
# Python3 code to demonstrate working of # Custom dictionary initialization in list # using {dict} + list comprehension # Initialize dict test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3} # Custom dictionary initialization in list # using {dict} + list comprehension res = [test_dict for sub in range(6)] print("The list of custom dictionaries is : " + str(res)) |
The list of custom dictionaries is : [{'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}, {'gfg': 1, 'best': 3, 'is': 2}]
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