Given a list of dictionaries, write a Python program to get the maximum of each key.
Examples:
Input : test_list = [{“Gfg” : 8, “is” : 1, “Best” : 9}, {“Gfg” : 2, “is” : 9, “Best” : 1}, {“Gfg” : 5, “is” : 10, “Best” : 7}]
Output : {‘Gfg’: 8, ‘is’: 10, ‘Best’: 9}
Explanation : Maximum of Gfg key is 8 among possible 8, 2 and 5.Input : test_list = [{“Gfg” : 8, “is” : 1, “Best” : 9}, {“Gfg” : 5, “is” : 10, “Best” : 7}]
Output : {‘Gfg’: 8, ‘is’: 10, ‘Best’: 9}
Explanation : Maximum of Best key is 7 among possible 9, 7.
Method #1 : Using items() + loop + max()
In this, we iterate for each dictionary and keys in them using loop and keep updating maximum values for each key using max().
Python3
# Python3 code to demonstrate working of# All Keys Maximum in Dictionary List# Using items() + loop + max()# initializing Matrixtest_list = [{"Gfg": 8, "is": 1, "Best": 9}, {"Gfg": 2, "is": 9, "Best": 1}, {"Gfg": 5, "is": 10, "Best": 7}]# printing original listprint("The original list is : " + str(test_list))res = {}for dic in test_list: for key, val in dic.items(): # checking for key presence and updating max if key in res: res[key] = max(res[key], val) else: res[key] = val# printing resultprint("All keys maximum : " + str(res)) |
Output:
The original list is : [{‘Gfg’: 8, ‘is’: 1, ‘Best’: 9}, {‘Gfg’: 2, ‘is’: 9, ‘Best’: 1}, {‘Gfg’: 5, ‘is’: 10, ‘Best’: 7}] All keys maximum : {‘Gfg’: 8, ‘is’: 10, ‘Best’: 9}The original list is : [{‘Gfg’: 8, ‘is’: 1, ‘Best’: 9}, {‘Gfg’: 2, ‘is’: 9, ‘Best’: 1}, {‘Gfg’: 5, ‘is’: 10, ‘Best’: 7}] All keys maximum : {‘Gfg’: 8, ‘is’: 10, ‘Best’: 9}
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Method #2 : Using defaultdict()
In this, we omit step of conditional check for key existence by using defaultdict(). Rest all the functionalities are similar to above method.
Python3
# Python3 code to demonstrate working of# All Keys Maximum in Dictionary List# Using defaultdict()from collections import defaultdict# initializing Matrixtest_list = [{"Gfg": 8, "is": 1, "Best": 9}, {"Gfg": 2, "is": 9, "Best": 1}, {"Gfg": 5, "is": 10, "Best": 7}]# printing original listprint("The original list is : " + str(test_list))res = defaultdict(int)for dic in test_list: for key, val in dic.items(): # defaultdict helps to avoid conditional check here res[key] = max(res[key], val)# printing resultprint("All keys maximum : " + str(dict(res))) |
Output:
The original list is : [{‘Gfg’: 8, ‘is’: 1, ‘Best’: 9}, {‘Gfg’: 2, ‘is’: 9, ‘Best’: 1}, {‘Gfg’: 5, ‘is’: 10, ‘Best’: 7}] All keys maximum : {‘Gfg’: 8, ‘is’: 10, ‘Best’: 9}
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.
Method 3: using dictionary comprehension
Approach:
- Initialize the list of dictionaries.
- Use dictionary comprehension to create a new dictionary with the maximum value for each key.
- The key is obtained by iterating over the keys of the first dictionary in the list using the keys() method.
- The value is obtained by using the max() function to find the maximum value for that key in all the dictionaries in the list. This is done using a generator expression that iterates over the list of dictionaries and extracts the value for the current key from each dictionary.
- Print the resulting dictionary.
Python3
test_list = [{"Gfg": 8, "is": 1, "Best": 9}, { "Gfg": 2, "is": 9, "Best": 1}, {"Gfg": 5, "is": 10, "Best": 7}]# Using dictionary comprehensionmax_dict = {k: max(d[k] for d in test_list) for k in test_list[0].keys()}print("All keys maximum: ", max_dict) |
All keys maximum: {'Gfg': 8, 'is': 10, 'Best': 9}
Time complexity: O(NK), where N is the number of dictionaries in the list and K is the number of keys in each dictionary.
Auxiliary space: O(K), where K is the number of keys in each dictionary, to store the resulting dictionary.
Method #4: Using set() and max() with list comprehension
Step-by-step approach:
- Convert the list of dictionaries into a set of all keys using set() function.
- Loop through the set of keys and use list comprehension to get a list of values for each key from all dictionaries.
- Use the max() function to find the maximum value from the list of values.
- Create a dictionary using dictionary comprehension to store the maximum value for each key.
- Print the final dictionary
Below is the implementation of the above approach:
Python3
# Python3 code to demonstrate working of# All Keys Maximum in Dictionary List# Using set() and max() with list comprehension# initializing Matrixtest_list = [{"Gfg": 8, "is": 1, "Best": 9}, {"Gfg": 2, "is": 9, "Best": 1}, {"Gfg": 5, "is": 10, "Best": 7}]# printing original listprint("The original list is : " + str(test_list))# create a set of all keysall_keys = set().union(*test_list)# create a dictionary to store the maximum value for each keymax_values = {key: max(dic.get(key, float('-inf')) for dic in test_list) for key in all_keys}# printing resultprint("All keys maximum : " + str(max_values)) |
The original list is : [{'Gfg': 8, 'is': 1, 'Best': 9}, {'Gfg': 2, 'is': 9, 'Best': 1}, {'Gfg': 5, 'is': 10, 'Best': 7}]
All keys maximum : {'Gfg': 8, 'is': 10, 'Best': 9}
Time complexity: O(nklogk), where n is the number of dictionaries in the list and k is the maximum number of keys in a dictionary.
Auxiliary space: O(k), where k is the maximum number of keys in a dictionary. We need to store the set of all keys and the dictionary to store the maximum value for each key.
Method #5: Using reduce() and lambda function from functools module
Steps:
- Import the “reduce” and “functools” modules.
- Initialize the “test_list” with dictionaries as elements.
- Print the original list using the print() function.
- Define a lambda function that takes two dictionaries and returns a new dictionary with the maximum value for each key.
- Use the “reduce” function to apply the lambda function to all dictionaries in the “test_list” and obtain a single dictionary with the maximum value for each key.
- Print the maximum values dictionary using the print() function.
Python3
# Python3 code to demonstrate working of# All Keys Maximum in Dictionary List# Using reduce() and lambda function# from functools module# import reduce and functoolsfrom functools import reduce# initializing Matrixtest_list = [{"Gfg": 8, "is": 1, "Best": 9}, {"Gfg": 2, "is": 9, "Best": 1}, {"Gfg": 5, "is": 10, "Best": 7}]# printing original listprint("The original list is : " + str(test_list))# define lambda function to get maximum# values for each keydef max_dict(dict1, dict2): return {key: max(dict1.get(key, float( '-inf')), dict2.get(key, float('-inf'))) for key in set(dict1.keys()) | set(dict2.keys())}# use reduce function to get maximum values# for all dictionaries in the listmax_values = reduce(max_dict, test_list)# printing resultprint("All keys maximum : " + str(max_values)) |
The original list is : [{'Gfg': 8, 'is': 1, 'Best': 9}, {'Gfg': 2, 'is': 9, 'Best': 1}, {'Gfg': 5, 'is': 10, 'Best': 7}]
All keys maximum : {'Best': 9, 'Gfg': 8, 'is': 10}
Time Complexity: O(N*K*log(K)), where n is the length of the “test_list” and k is the number of keys in each dictionary.
Auxiliary Space: O(k), where k is the number of keys in each dictionary.
