Python – Test for Even values dictionary values lists

Given a dictionary with lists as values, map Boolean values depending upon all values in List are Even or not.

Input : {“Gfg” : [6, 8, 10], “is” : [8, 10, 12, 16], “Best” : [10, 16, 14, 6]} 
Output : {‘Gfg’: True, ‘is’: True, ‘Best’: True} 
Explanation : All lists have even numbers. 

Input : {“Gfg” : [6, 5, 10], “is” : [8, 10, 11, 16], “Best” : [10, 16, 14, 6]} 
Output : {‘Gfg’: False, ‘is’: False, ‘Best’: True} 
Explanation : Only “Best” has even numbers.

Method #1: Using loop

This is brute way in which this task can be performed. In this, we iterate for all the values and check if all list values are Even if yes, we assign key as True else False.

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : {'Gfg': False, 'is': True, 'Best': True}

Time Complexity: O(N)
Auxiliary Space: O(N)

Method #2:  Using all() + dictionary comprehension

This is yet another way in which this task can be performed. In this, we check for all the elements using all() and dictionary comprehension is used to remake the result.

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : {'Gfg': False, 'is': True, 'Best': True}

Time Complexity: O(N)
Auxiliary Space: O(N)

Method #3:  Using items()+filter()+ lambda functions

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : ['is', 'Best']

Time Complexity: O(N)
Auxiliary Space: O(N)

Method #4: Using NumPy library

Output:

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : {'Gfg': False, 'is': True, 'Best': True}

Time Complexity: O(N)
Auxiliary Space: O(N)

Method #5: Using reduce() and map() functions:

Step-by-step approach:

• Import the “reduce” function from the “functools” library.
• Initialize an empty list called “res” to store the keys of the dictionary whose values are all even.
• Initialize the input dictionary called “test_dict”.
• Iterate through the items of the dictionary using the “items()” method.
• Use the “reduce()” and “map()” functions to check if all the values in the current key’s value list are even.
• If all the values are even, append the key to the “res” list.
• Print the final result.

Below is the implementation of the above approach:

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : ['is', 'Best']

Time Complexity:

• The time complexity of the “reduce()” function is O(n), where n is the number of items in the input list.\The time complexity of the “map()” function is also O(n), where n is the number of items in the input list.
• Therefore, the time complexity of the algorithm is O(m*n), where m is the number of key-value pairs in the dictionary and n is the length of the longest value list in the dictionary.

Auxiliary Space:

• The space complexity of the algorithm is O(m), where m is the number of key-value pairs in the dictionary.
• This is because we are storing the keys of the dictionary whose values are all even in the “res” list, which has a maximum size of m.

Method #6: Using list comprehension and all() function

• Start by initializing the dictionary test_dict with some key-value pairs.
• Use a dictionary comprehension to create a new dictionary res by iterating over the key-value pairs in test_dict.
• For each key-value pair, we use a list comprehension to check if all the elements in the list value are even. We do this by checking if the remainder of each element with 2 is 0, using the modulo operator %.
• Use the all() function to check if all the elements in the list comprehension are True. If they are, we set the value of the corresponding key in res to True, otherwise we set it to False.
• Finally, print the original dictionary and the computed dictionary.

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : {'Gfg': False, 'is': True, 'Best': True}

Time complexity: O(n*m), where n is the number of keys in the dictionary and m is the maximum length of any list value in the dictionary.
Auxiliary space: O(n), where n is the number of keys in the dictionary, to store the result in the dictionary res.

Method #7: Using dictionary comprehension and any() function:

Step-by-step approach:

• Initialize the dictionary
• Create a dictionary comprehension that iterates over the key-value pairs of the original dictionary
• For each key-value pair, use the any() function with a generator expression to check if any of the elements in the value list are odd
• The result of the any() function will be used as the value for the new dictionary
• Return the new dictionary

Below is the implementation of the above approach:

The original dictionary is : {'Gfg': [6, 7, 3], 'is': [8, 10, 12, 16], 'Best': [10, 16, 14, 6]}
The computed dictionary : {'Gfg': False, 'is': True, 'Best': True}

Time complexity: O(n), where n is the total number of elements in the dictionary
Auxiliary space: O(n), where n is the total number of elements in the dictionary