Sometimes, while working with Python dictionaries, we can come across a problem in which we have a particular value, and we need to find frequency if it’s occurrence. Let’s discuss certain ways in which this problem can be solved.
Method #1: Using loop This problem can be solved using naive method of loop. In this we just iterate through each key in dictionary and when a match is found, the counter is increased.
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionary# Using loop# Initialize dictionarytest_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueK = 2# Using loop# Selective key values in dictionaryres = 0for key in test_dict: if test_dict[key] == K: res = res + 1 # printing resultprint("Frequency of K is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2: Using sum() + values() This can also be solved using the combination of sum() and value(). In this, sum is used to perform the summation of values filtered and values of dictionary are extracted using values()
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionary# Using sum() + values()# Initialize dictionarytest_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueK = 2# Using sum() + values()# Selective key values in dictionaryres = sum(x == K for x in test_dict.values()) # printing resultprint("Frequency of K is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #3 : Using count() and values().These methods can be used together to find number of keys with particular value.
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionary# Using count() + values()# Initialize dictionarytest_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueK = 2# Using count() + values()list1=list(test_dict.values())# Selective key values in dictionaryres = list1.count(K) # printing resultprint("Frequency of K is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(n)
Auxiliary Space : O(n)
Method #4: Using lambda functions
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionary# Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueK = 2keys = list(test_dict.keys())res = len(list(filter(lambda x: test_dict[x] == K, keys)))# printing resultprint("Frequency of K is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5: Using collections.Counter()
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionary# Using collections.Counter() # Importing collections for Counterimport collections # Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2} # printing original dictionaryprint("The original dictionary : " + str(test_dict)) # Initialize valueK = 2 # Using collections.Counter()res = collections.Counter(test_dict.values())[K] # printing resultprint("Frequency of K is : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #6: Using operator.countOf() method:
Python3
# Python3 code to demonstrate working of# Count keys with particular value in dictionaryimport operator as op# Initialize dictionarytest_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueK = 2# Using count() + values()list1=list(test_dict.values())# Selective key values in dictionaryres = op.countOf(list1,K) # printing resultprint("Frequency of K is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3
Time Complexity: O(N)
Auxiliary Space: O(N)
Method 7: Using a dictionary comprehension
Step-by-step approach:
- Create a new dictionary using a dictionary comprehension that counts the occurrence of values in the original dictionary.
- Create another dictionary using a dictionary comprehension that counts the occurrence of keys with a particular value in the original dictionary.
- Access the value of the key equal to the given value K in the second dictionary to get the frequency
Python3
# Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}# Initialize valueK = 2# Create a new dictionary with count of values as keyscount_dict = {v: list(test_dict.values()).count(v) for v in set(test_dict.values())}# Create a new dictionary with count of keys with particular value as valuesfreq_dict = {v: sum(1 for k in test_dict.keys() if test_dict[k] == v) for v in count_dict.keys()}# Access the frequency of Kres = freq_dict[K]# printing resultprint("Frequency of K is : " + str(res)) |
Frequency of K is : 3
Time complexity: O(n^2)
Auxiliary space: O(n)
