Given Dictionary with value lists, the task is to write a Python program to assign each key with an index of the maximum value in the value list.
Examples:
Input : test_dict = {"gfg" : [5, 3, 6, 3], "is" : [1, 7, 5, 3], "best" : [9, 1, 3, 5]} Output : {'gfg': 2, 'is': 1, 'best': 0} Explanation : Max element in "gfg"'s value is 6 at 2nd index, hence assigned 2. Input : test_dict = {"gfg" : [9, 3, 6, 3], "is" : [1, 7, 5, 3], "best" : [9, 1, 3, 5]} Output : {'gfg': 0, 'is': 1, 'best': 0 Explanation : Max element in "gfg"'s value is 9 at 0th index, hence assigned 0.
Method #1 : Using max() + loop + index()
In this, we get the index of the maximum element from the value list using max() and index(). Loop is used for the task of iteration of keys in the dictionary.
Python3
# Python3 code to demonstrate working of # Assign keys with Maximum element index # Using max() + index() + loop # initializing dictionary test_dict = { "gfg" : [ 5 , 3 , 6 , 3 ], "is" : [ 1 , 7 , 5 , 3 ], "best" : [ 9 , 1 , 3 , 5 ]} # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) res = dict () for key in test_dict: # using index() to get required value res[key] = test_dict[key].index( max (test_dict[key])) # printing result print ( "The maximum index assigned dictionary : " + str (res)) |
Output:
The original dictionary is : {‘gfg’: [5, 3, 6, 3], ‘is’: [1, 7, 5, 3], ‘best’: [9, 1, 3, 5]}
The maximum index assigned dictionary : {‘gfg’: 2, ‘is’: 1, ‘best’: 0}
Method #2 : Using dictionary comprehension + max() + index()
In this, we perform task of getting result using dictionary comprehension shorthand variation of above method.
Python3
# Python3 code to demonstrate working of # Assign keys with Maximum element index # Using dictionary comprehension + max() + index() # initializing dictionary test_dict = { "gfg" : [ 5 , 3 , 6 , 3 ], "is" : [ 1 , 7 , 5 , 3 ], "best" : [ 9 , 1 , 3 , 5 ]} # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) # using dictionary comprehension as one liner alternative res = {key: test_dict[key].index( max (test_dict[key])) for key in test_dict} # printing result print ( "The maximum index assigned dictionary : " + str (res)) |
Output:
The original dictionary is : {‘gfg’: [5, 3, 6, 3], ‘is’: [1, 7, 5, 3], ‘best’: [9, 1, 3, 5]}
The maximum index assigned dictionary : {‘gfg’: 2, ‘is’: 1, ‘best’: 0}
Method #3: Using a loop and the enumerate() function
Step-by-step algorithm:
- Initialize a dictionary test_dict with keys as strings and values as lists of integers.
- Print the original dictionary test_dict.
- Initialize an empty dictionary res.
- For each key-value pair in the dictionary test_dict, do the following:
a. Get the maximum value of the list of integers using the max() function.
b. Get the index of the maximum value in the list of integers using a list comprehension.
c. Assign the key and the index value to the res dictionary. - Print the resulting dictionary res.
Python3
# initializing dictionary test_dict = { "gfg" : [ 5 , 3 , 6 , 3 ], "is" : [ 1 , 7 , 5 , 3 ], "best" : [ 9 , 1 , 3 , 5 ]} # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) res = dict () for key, values in test_dict.items(): max_val = max (values) max_idx = [i for i, v in enumerate (values) if v = = max_val][ 0 ] res[key] = max_idx # printing result print ( "The maximum index assigned dictionary : " + str (res)) |
The original dictionary is : {'gfg': [5, 3, 6, 3], 'is': [1, 7, 5, 3], 'best': [9, 1, 3, 5]} The maximum index assigned dictionary : {'gfg': 2, 'is': 1, 'best': 0}
Time Complexity: O(nm), where n is the number of keys in the dictionary test_dict and m is the length of the longest list of integers in the dictionary.
Auxiliary Space: O(n), where n is the number of keys in the dictionary test_dict.