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Python program to extract the Unique Dictionary Value List elements

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Given Dictionary with value lists, extract all unique values. 

Input : test_dict = {“Gfg” : [6, 7, 4, 6], “Geeks” : [6, 8, 5, 2]} 
Output : [6, 7, 4, 8, 5, 2] 
Explanation : All distincts elements extracted.


Input : test_dict = {“Gfg” : [6, 7, 6], “Geeks” : [6, 8, 5, 2]} 
Output : [6, 7, 8, 5, 2] 
Explanation : All distincts elements extracted. 

Method #1 : Using loop 

This is brute way in which this task can be performed. In this, we memoize elements when they occur and prevent them from re enter to result list.

Python3

# Python3 code to demonstrate working of
# Unique Dictionary Value List elements
# Using loop
 
# initializing dictionary
test_dict = {"Gfg" : [6, 7, 4, 6],
             "is" : [8, 9, 5],
             "for" : [2, 5, 3, 7],
             "Geeks" : [6, 8, 5, 2]}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# list to memorize elements and insert result
res = []
for sub in test_dict:
    for ele in test_dict[sub]:
         
        # testing for existence
        if ele not in res:
            res.append(ele)
 
# printing result
print("Extracted items : " + str(res))

                    

Output:

The original dictionary is : {‘Gfg’: [6, 7, 4, 6], ‘is’: [8, 9, 5], ‘for’: [2, 5, 3, 7], ‘Geeks’: [6, 8, 5, 2]}

Extracted items : [6, 7, 4, 8, 9, 5, 2, 3]

The time complexity of the given code is O(N*M), where N is the number of keys in the dictionary and M is the maximum length of the lists inside the dictionary. 

The auxiliary space complexity of the code is O(N*M), where N is the number of keys in the dictionary and M is the maximum length of the lists inside the dictionary.

Method #2 : Using set() + union()

The combination of the above functions can be used to solve this problem. In this, we convert all lists using set() and then perform a union of all elements across all keys to get the desired result.

Python3

# Python3 code to demonstrate working of
# Unique Dictionary Value List elements
# Using set() + union()
 
# initializing dictionary
test_dict = {"Gfg" : [6, 7, 4, 6],
             "is" : [8, 9, 5],
             "for" : [2, 5, 3, 7],
             "Geeks" : [6, 8, 5, 2]}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# using * operator to get union
res = list(set().union(*test_dict.values()))
 
# printing result
print("Extracted items : " + str(res))

                    

Output:

The original dictionary is : {‘Gfg’: [6, 7, 4, 6], ‘is’: [8, 9, 5], ‘for’: [2, 5, 3, 7], ‘Geeks’: [6, 8, 5, 2]}

Extracted items : [6, 7, 4, 8, 9, 5, 2, 3]

Method #3 : Using keys(),extend(),list(),set() methods

Python3

# Python3 code to demonstrate working of
# Unique Dictionary Value List elements
# Using loop
 
# initializing dictionary
test_dict = {"Gfg" : [6, 7, 4, 6],
            "is" : [8, 9, 5],
            "for" : [2, 5, 3, 7],
            "Geeks" : [6, 8, 5, 2]}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
res = []
for i in test_dict.keys():
    res.extend(test_dict[i])
res=list(set(res))
 
# printing result
print("Extracted items : " + str(res))

                    

Output
The original dictionary is : {'Gfg': [6, 7, 4, 6], 'is': [8, 9, 5], 'for': [2, 5, 3, 7], 'Geeks': [6, 8, 5, 2]}
Extracted items : [2, 3, 4, 5, 6, 7, 8, 9]

Time complexity: This method involves iterating over all the values of the dictionary, so the time complexity is O(n*m), where n is the number of keys in the dictionary and m is the maximum length of the value list.

Auxiliary space: This method requires O(nm) auxiliary space as we are storing all the values of the dictionary in a list and then creating a set out of it. However, the space required will be less than O(nm) as duplicate values will be removed while creating the set.

Method #4 : Using items(),extend(),list(),Counter() methods

Python3

# Python3 code to demonstrate working of
# Unique Dictionary Value List elements
# Using Counter() function
from collections import Counter
# initializing dictionary
test_dict = {"Gfg": [6, 7, 4, 6],
             "is": [8, 9, 5],
             "for": [2, 5, 3, 7],
             "Geeks": [6, 8, 5, 2]}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
res = []
for key, value in test_dict.items():
    res.extend(value)
freq = Counter(res)
res = list(freq.keys())
res.sort()
# printing result
print("Extracted items : " + str(res))

                    

Output
The original dictionary is : {'Gfg': [6, 7, 4, 6], 'is': [8, 9, 5], 'for': [2, 5, 3, 7], 'Geeks': [6, 8, 5, 2]}
Extracted items : [2, 3, 4, 5, 6, 7, 8, 9]

Time Complexity: O(n*m) Space Complexity: O(n*m)

Method #5 :  reduce() function from the functools module:

Algorithm:

  1. Import the reduce() function from the functools module.
  2. Define the dictionary test_dict.
  3. Print the original dictionary test_dict.
  4. Use reduce() with a lambda function that performs set union (|) on each element in the dictionary’s values.
  5. This will create a set of all the unique values in the dictionary.
  6. Convert the set to a list.
  7. Print the list of unique values.

Python3

from functools import reduce
 
 
# initializing dictionary
 
test_dict = {'gfg' : [5, 6, 7, 8],
'is' : [10, 11, 7, 5],
'best' : [6, 12, 10, 8],
'for' : [1, 2, 5]}
 
 
# printing original dictionary
 
print("The original dictionary is : " + str(test_dict))
 
 
# Extract Unique values dictionary values
 
result = list(reduce(lambda x, y: set(x) | set(y), test_dict.values()))
 
 
# printing result
 
print("The unique values list is : " + str(result))
 
#This code is contributed by Rayudu .

                    

Output
The original dictionary is : {'gfg': [5, 6, 7, 8], 'is': [10, 11, 7, 5], 'best': [6, 12, 10, 8], 'for': [1, 2, 5]}
The unique values list is : [1, 2, 5, 6, 7, 8, 10, 11, 12]

Time complexity:

The reduce() function iterates over all the values in the dictionary, and performs a set union operation on them. The time complexity of set union operation is O(n) where n is the number of elements in the sets. Therefore, the time complexity of this code is O(n*m), where n is the number of keys in the dictionary and m is the average number of elements in each list.

Auxiliary Space:

The space complexity is dominated by the space used to store the set of unique values. Since we are creating a set of all the values in the dictionary, the space complexity of this code is O(n), where n is the total number of elements in the dictionary.



Last Updated : 08 May, 2023
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