In this article, we will see how to get the number of keys with some given value N in a given dictionary. There are multiple methods to do this task. Let’s see them with help of examples.
Simple method –
Python3
# Python3 code to Get the number of keys# with given value N in dictionary# Initialize dictionary test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2} # printing original dictionary print("The original dictionary : " + str(test_dict)) # Initialize value N = 2# Using loop # Selective key values in dictionary res = 0for key in test_dict: if test_dict[key] == N: res = res + 1 # printing result print("Frequency of N is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of N is : 3
The time complexity of the program is O(n), where n is the number of elements in the dictionary.
The auxiliary space complexity of the program is O(1), because the program uses a constant amount of additional space (the variable res) to compute the result, regardless of the size of the input dictionary.
Method #2: Using sum() + values()
Python3
# Python3 code to Get the number of keys# with given value N in dictionary# Using sum() + values() # Initialize dictionary test_dict = {'gfg' : 1, 'is' : 2, 'best' : 3, 'for' : 2, 'CS' : 2} # printing original dictionary print("The original dictionary : " + str(test_dict)) # Initialize value N = 2# Using sum() + values() # Selective key values in dictionaryres = sum(x == N for x in test_dict.values()) # printing result print("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), where n is the length of the list test_dict
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
Method #3: Using Counter() function
Python3
# Python3 code to Get the number of keys# with given value N in dictionaryfrom collections import Counter# Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueN = 2freq = Counter(test_dict.values())res = freq[N]# printing resultprint("Frequency of N is : " + str(res)) |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of N is : 3
Time Complexity: O(N)
Auxiliary Space: O(N)
Method #4: Using list comprehension
This code uses a list comprehension to iterate through the keys in the dictionary, and for each key it checks if the value associated with that key is equal to the given value N. If it is, it adds that key to a new list. The length of this list is then calculated, which represents the number of keys in the dictionary that have a value of N.
Python3
# Method #4: Using list comprehension# Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize valueN = 2# Using list comprehension# Selective key values in dictionaryres = len([key for key in test_dict if test_dict[key] == N])# printing resultprint("Frequency of N 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 N is : 3
Time complexity: O(n) as it iterates through all the keys of the dictionary.
Auxiliary space: O(n) as it creates a new list to store the keys that have the given value N.
Method #5:Using the reduce and functools module:
Algorithm:
- Initialize the dictionary and value N
- Use reduce function to iterate through the values of the dictionary and accumulate the number of times the value N is found
- Print the count of the number of times N is found
Python3
from functools import reduce# Initialize dictionarytest_dict = {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}# printing original dictionaryprint("The original dictionary : " + str(test_dict))N = 2count = reduce(lambda acc, value: acc + (value == N), test_dict.values(), 0)# printing resultprint("Frequency of N is:", count)#This code is contributed by Jyothi Pinjala. |
The original dictionary : {'gfg': 1, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of N is: 3
Time complexity: O(n), where n is the number of values in the dictionary. This is because the reduce function needs to iterate through all the values in the dictionary to count the number of times the value N appears. The lambda function that is used to check if the current value is equal to N takes constant time.
Auxiliary Space: O(1). This is because the algorithm only uses a few constant-size variables (i.e., test_dict, N, and count) to perform the calculation, regardless of the size of the input. The reduce function doesn’t create any additional data structures, and the lambda function only uses a few constant-size variables (i.e., acc, value, and N).
