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 code to demonstrate working of # Unique Dictionary Value List elements# Using loop# initializing dictionarytest_dict = {"Gfg" : [6, 7, 4, 6],
"is" : [8, 9, 5],
"for" : [2, 5, 3, 7],
"Geeks" : [6, 8, 5, 2]}
# printing original dictionaryprint("The original dictionary is : " + str(test_dict))
# list to memorize elements and insert resultres = []
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))
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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 code to demonstrate working of # Unique Dictionary Value List elements# Using set() + union()# initializing dictionarytest_dict = {"Gfg" : [6, 7, 4, 6],
"is" : [8, 9, 5],
"for" : [2, 5, 3, 7],
"Geeks" : [6, 8, 5, 2]}
# printing original dictionaryprint("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))
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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 code to demonstrate working of# Unique Dictionary Value List elements# Using loop# initializing dictionarytest_dict = {"Gfg" : [6, 7, 4, 6],
"is" : [8, 9, 5],
"for" : [2, 5, 3, 7],
"Geeks" : [6, 8, 5, 2]}
# printing original dictionaryprint("The original dictionary is : " + str(test_dict))
res = []
for i in test_dict.keys():
res.extend(test_dict[i])
res=list(set(res))
# printing resultprint("Extracted items : " + str(res))
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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 code to demonstrate working of# Unique Dictionary Value List elements# Using Counter() functionfrom collections import Counter
# initializing dictionarytest_dict = {"Gfg": [6, 7, 4, 6],
"is": [8, 9, 5],
"for": [2, 5, 3, 7],
"Geeks": [6, 8, 5, 2]}
# printing original dictionaryprint("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 resultprint("Extracted items : " + str(res))
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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:
- Import the reduce() function from the functools module.
- Define the dictionary test_dict.
- Print the original dictionary test_dict.
- Use reduce() with a lambda function that performs set union (|) on each element in the dictionary’s values.
- This will create a set of all the unique values in the dictionary.
- Convert the set to a list.
- Print the list of unique values.
from functools import reduce
# initializing dictionarytest_dict = {'gfg' : [5, 6, 7, 8],
'is' : [10, 11, 7, 5],
'best' : [6, 12, 10, 8],
'for' : [1, 2, 5]}
# printing original dictionaryprint("The original dictionary is : " + str(test_dict))
# Extract Unique values dictionary valuesresult = list(reduce(lambda x, y: set(x) | set(y), test_dict.values()))
# printing resultprint("The unique values list is : " + str(result))
#This code is contributed by Rayudu . |
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.