Given a list of strings, find the frequency of strings case insensitive.
Input : test_list = ["Gfg", "Best", "GFG", "is", "IS", "BEST"]
Output : {'gfg': 2, 'best': 2, 'is': 2}
Explanation : All occur twice.
Input : test_list = ["Gfg", "gfg", "GFG"]
Output : {'gfg': 3}
Explanation : Only "gfg" 3 occurrences.
Method 1: Using defaultdict() + lower()
In this, we perform lower() to all the strings, before mapping in defaultdict. This ensures case insensitivity while mapping and cumulating frequency.
Python3
# Python3 code to demonstrate working of# Strings Frequency (Case Insensitive)# Using defaultdict() + lower()from collections import defaultdict# initializing listtest_list = ["Gfg", "Best", "best", "gfg", "GFG", "is", "IS", "BEST"]# printing original listprint("The original list is : " + str(test_list))res = defaultdict(int)for ele in test_list: # lowercasing to cater for Case Insensitivity res[ele.lower()] += 1# printing resultprint("Strings Frequency : " + str(dict(res))) |
Output
The original list is : ['Gfg', 'Best', 'best', 'gfg', 'GFG', 'is', 'IS', 'BEST']
Strings Frequency : {'gfg': 3, 'best': 3, 'is': 2}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2: Using count() method
Python3
# Python3 code to demonstrate working of# Strings Frequency (Case Insensitive)# initializing listtest_list = ["Gfg", "Best", "best", "gfg", "GFG", "is", "IS", "BEST"]# printing original listprint("The original list is : " + str(test_list))x = []res = dict()for i in test_list: x.append(i.lower())a = list(set(x))for i in a: res[i] = x.count(i)# printing resultprint("Strings Frequency : " + str(dict(res))) |
Output
The original list is : ['Gfg', 'Best', 'best', 'gfg', 'GFG', 'is', 'IS', 'BEST']
Strings Frequency : {'is': 2, 'best': 3, 'gfg': 3}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: Using Counter() function
Python3
# Python3 code to demonstrate working of# Strings Frequency (Case Insensitive)from collections import Counter# initializing listtest_list = ["Gfg", "Best", "best", "gfg", "GFG", "is", "IS", "BEST"]# printing original listprint("The original list is : " + str(test_list))for i in range(len(test_list)): test_list[i] = test_list[i].lower()freq = Counter(test_list)# printing resultprint("Strings Frequency : " + str(dict(freq))) |
Output
The original list is : ['Gfg', 'Best', 'best', 'gfg', 'GFG', 'is', 'IS', 'BEST']
Strings Frequency : {'gfg': 3, 'best': 3, 'is': 2}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4: using operator.countOf() method
Python3
# Python3 code to demonstrate working of# Strings Frequency (Case Insensitive)import operator as op# initializing listtest_list = ["Gfg", "Best", "best", "gfg", "GFG", "is", "IS", "BEST"]# printing original listprint("The original list is : " + str(test_list))x = []res = dict()for i in test_list: x.append(i.lower())a = list(set(x))for i in a: res[i] = op.countOf(x,i)# printing resultprint("Strings Frequency : " + str(dict(res))) |
Output
The original list is : ['Gfg', 'Best', 'best', 'gfg', 'GFG', 'is', 'IS', 'BEST']
Strings Frequency : {'is': 2, 'gfg': 3, 'best': 3}
Time Complexity: O(N)
Auxiliary Space : O(N)
Method 5: Using a dictionary and for loop
Step-by-Step approach:
- Create an empty dictionary to store the frequency of strings.
- Loop through the strings in the list.
- Convert each string to lowercase using the str.lower() method.
- If the lowercase string is already in the dictionary, increase its value by 1.
- If the lowercase string is not in the dictionary, add it as a key with a value of 1.
- Print the dictionary.
Python3
# Strings Frequency (Case Insensitive)# initializing listtest_list = ["Gfg", "Best", "best", "gfg", "GFG", "is", "IS", "BEST"]# printing original listprint("The original list is : " + str(test_list))# create an empty dictionary to store the frequency of stringsfreq_dict = {}# loop through the strings in the listfor string in test_list: # convert each string to lowercase lowercase_string = string.lower() # if the lowercase string is already in the dictionary, increase its value by 1 if lowercase_string in freq_dict: freq_dict[lowercase_string] += 1 # if the lowercase string is not in the dictionary, add it as a key with a value of 1 else: freq_dict[lowercase_string] = 1# printing resultprint("Strings Frequency : " + str(freq_dict)) |
Output
The original list is : ['Gfg', 'Best', 'best', 'gfg', 'GFG', 'is', 'IS', 'BEST']
Strings Frequency : {'gfg': 3, 'best': 3, 'is': 2}
Time Complexity: O(n), where n is the number of strings in the list.
Auxiliary Space: O(n), where n is the number of strings in the list, to store the dictionary.
