Given a list of dictionaries, perform the unlisting of records in which we just have 1 dictionary as record element.
Input : test_list = [{'best': [{'a': 6}], 'Gfg': 15}]
Output : [{'best': {'a': 6}, 'Gfg': 15}]
Explanation : The value list associated with 'best' key is changed to dictionary.
Input : test_list = [{'Gfg': [{'best' : 17}]}]
Output : [{'Gfg': {'best': 17}}]
Explanation : 'Gfg' key's value changed to single dictionary.
Method #1: Using loop + isinstance()
This is a brute-force way in which this task can be performed. In this, we test for the list type using isinstance(), and loop is used for iterations.
Python3
# Python3 code to demonstrate working of # Unlist Single Valued Dictionary List# Using loop + isinstance()# Initializing listtest_list = [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]# Printing original listprint("The original list is : " + str(test_list))# Using loop + isinstance()for dicts in test_list: for key, val in dicts.items(): # Checking for list to convert # isinstance() function if isinstance(val, list): dicts[key] = val[0]# Printing result print("The converted Dictionary list : " + str(test_list)) |
The original list is : [{'Gfg': 1, 'is': [{'b': 3, 'a': 2}]}, {'CS': 6, 'best': [{'d': 5, 'c': 4}]}]
The converted Dictionary list : [{'Gfg': 1, 'is': {'b': 3, 'a': 2}}, {'CS': 6, 'best': {'d': 5, 'c': 4}}]
Time complexity: O(N*M), where n is the length of the outer list and m is the maximum length of a list inside the dictionaries of the outer list.
Auxiliary Space: O(1), since the program does not create any additional data structures proportional to the input size. It only modifies the existing list in place.
Method #2: Using list comprehension + isinstance()
The combination of the above functions can be used to solve the problem. In this, we perform a similar task with a shorthand approach similar to the above method.
Python3
# Python3 code to demonstrate working of# Unlist Single Valued Dictionary List# Using list comprehension + isinstance()# Initializing listtest_list = [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]# Printing original listprint("The original list is : " + str(test_list))# Similar way as above, extracting first element of list# using list comprehension + isinstance() functionres = [{key: val[0] if isinstance(val, list) else val for key, val in sub.items()} for sub in test_list]# Printing the resultprint("The converted Dictionary list : " + str(res)) |
The original list is : [{'Gfg': 1, 'is': [{'b': 3, 'a': 2}]}, {'CS': 6, 'best': [{'d': 5, 'c': 4}]}]
The converted Dictionary list : [{'Gfg': 1, 'is': {'b': 3, 'a': 2}}, {'CS': 6, 'best': {'d': 5, 'c': 4}}]
Time complexity: O(N*M), where n is the length of the outer list and m is the maximum length of a list inside the dictionaries of the outer list.
Auxiliary Space: O(N), where n is the number of elements in the list “test_list”.
Method 3: Using a nested for loop and checking the type of the value.
Python3
# Python3 code to demonstrate working of# Unlist Single Valued Dictionary List# Using nested for loop# Initializing listtest_list = [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]# Printing original listprint("The original list is : " + str(test_list))# Using nested for loopres = []for sub in test_list: temp = {} for key, val in sub.items(): if isinstance(val, list) and len(val) == 1 and isinstance(val[0], dict): temp[key] = val[0] else: temp[key] = val res.append(temp)# Printing resultprint("The converted Dictionary list : " + str(res)) |
The original list is : [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]
The converted Dictionary list : [{'Gfg': 1, 'is': {'a': 2, 'b': 3}}, {'best': {'c': 4, 'd': 5}, 'CS': 6}]
Time complexity: O(NM) where N is the number of dictionaries in the list and M is the maximum number of keys in a dictionary.
Auxiliary space: O(NM) because we are creating a new list of dictionaries with the same number of keys and values as the original list.
Method 4: Using recursion to flatten and unlist the dictionary list
Approach:
- Define a function that takes a dictionary or list as input and returns the flattened and unlisted version of it.
- Check if the input is a dictionary. If it is, then iterate over its keys and values.
- If the value is a dictionary or list, call the function recursively on that value to flatten and unlist it.
- If the value is a list with only one element, then return that element.
- If the value is not a dictionary or list, return the value.
- If the input is a list, then iterate over its elements and call the function recursively on each element.
- Return the flattened and unlisted list.
Python3
# Python3 code to demonstrate working of# Unlist Single Valued Dictionary List# Using recursion# Initializing listtest_list = [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]# Printing original listprint("The original list is : " + str(test_list))# Flatten and Unlisting the dictionary list# Using recursiondef unlist_dict_list(input): if isinstance(input, dict): return {key: unlist_dict_list(val) for key, val in input.items()} elif isinstance(input, list): if len(input) == 1: return unlist_dict_list(input[0]) else: return [unlist_dict_list(elem) for elem in input] else: return inputres = unlist_dict_list(test_list)# Printing the resultant list print("The converted Dictionary list : " + str(res)) |
The original list is : [{'Gfg': 1, 'is': [{'a': 2, 'b': 3}]}, {'best': [{'c': 4, 'd': 5}], 'CS': 6}]
The converted Dictionary list : [{'Gfg': 1, 'is': {'a': 2, 'b': 3}}, {'best': {'c': 4, 'd': 5}, 'CS': 6}]
Time complexity: O(N), where n is the total number of elements in the dictionary list.
Auxiliary space: O(N), where n is the total number of elements in the dictionary list.
