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We are given a dictionary with lists as values and the task is to transform each element in these lists into individual dictionary records. Specifically, each list element should become a key in a new dictionary with an empty list as its value.
For example, given {'gfg': [4, 5], 'best': [8, 10, 7, 9]}, the output should be {'gfg': [{4: []}, {5: []}], 'best': [{8: []}, {10: []}, {7: []}, {9: []}]}.
This brute-force method iterates through all values (lists) in the dictionary and for each element in the list it creates a new dictionary with the element as the key and an empty list as its value. The enumerate() function is used to track the index while iterating allowing us to replace each element directly.
{'gfg': [{4: []}, {5: []}], 'is': [{8: []}], 'best': [{10: []}]}
Explanation:
{element: []} using its index (i).dictionary comprehensionIn this method we utilize dictionary comprehension to create a new dictionary where the items()function is used to iterate over key-value pairs of the original dictionary. For each list in the values, a nested list comprehension converts its elements into dictionaries {element: []}.
{'gfg': [{4: []}, {5: []}], 'is': [{8: []}], 'best': [{10: []}]}
Explanation:
k, li) using items().v) in the list (li) into a dictionary {v: []}.We can solve the given problem with recursion by traversing the dictionary and checking if the value is a list, then convert each element into a dictionary with the element as the key and an empty list as its value but if the value is a dictionary then recursively process it.
{'gfg': [{4: []}, {5: []}], 'is': [{8: []}], 'best': [{10: []}]}
Explanation:
convert_to_list iterates through the dictionary, If a value is a list then it converts each element into a dictionary with an empty list and if the value is a dictionary then it calls itself to process that dictionary.{v: []} if it's not already a dictionary.