Sometimes while working with Python strings, we can have a problem in which we need to extract the frequency of all the words in a string. This problem has been solved earlier. This discusses the shorthands to solve this problem as this has applications in many domains ranging from web development and competitive programming. Let’s discuss certain ways in which this problem can be solved.
Input : test_str = 'Gfg is best'
Output : {'Gfg': 1, 'is': 1, 'best': 1}
Input : test_str = 'Gfg Gfg Gfg'
Output : {'Gfg': 3}
Method #1: Using dictionary comprehension + count() + split()
The combination of the above functions can be used to solve this problem. In this, we first split all the words and then perform a count of them using count() method.
Python3
# Python3 code to demonstrate working of# Words Frequency in String Shorthands# Using dictionary comprehension + count() + split()# Initializing stringtest_str = 'Gfg is best . Geeks are good and Geeks like Gfg'# Printing original stringprint("The original string is : " + str(test_str))# Words Frequency in String Shorthands# Using dictionary comprehension + count() + split()res = {key: test_str.count(key) for key in test_str.split()}# Printing resultprint("The words frequency : " + str(res)) |
The original string is : Gfg is best . Geeks are good and Geeks like Gfg The words frequency : {‘Gfg’: 2, ‘is’: 1, ‘best’: 1, ‘.’: 1, ‘Geeks’: 2, ‘are’: 1, ‘good’: 1, ‘and’: 1, ‘like’: 1}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using Counter() + split()
The combination of the above methods can also be used to solve this problem. In this, we perform the task of counting using Counter() and separation of words using split().
Python3
# Python3 code to demonstrate working of# Words Frequency in String Shorthands# Using Counter() + split()from collections import Counter# initializing stringtest_str = 'Gfg is best . Geeks are good and Geeks like Gfg'# printing original stringprint("The original string is : " + str(test_str))# Words Frequency in String Shorthands# using Counter() + split()res = Counter(test_str.split())# Printing resultprint("The words frequency : " + str(dict(res))) |
The original string is : Gfg is best . Geeks are good and Geeks like Gfg The words frequency : {‘Gfg’: 2, ‘is’: 1, ‘best’: 1, ‘.’: 1, ‘Geeks’: 2, ‘are’: 1, ‘good’: 1, ‘and’: 1, ‘like’: 1}
Method #3 : Using dictionary comprehension + operator.countOf() + split() :
The combination of the above functions can be used to solve this problem. In this, we first split all the words and then perform count of them using operator.countOf() method.
Python3
# Python3 code to demonstrate working of# Words Frequency in String Shorthands# Using dictionary comprehension + operator.countOf() + split()import operator as op# Initializing stringtest_str = 'Gfg is best . Geeks are good and Geeks like Gfg'# Printing original stringprint("The original string is : " + str(test_str))listString = test_str.split()# Words Frequency in String Shorthands# Using dictionary comprehension + operator.countOf()res = {key: op.countOf(listString, key) for key in listString}# Printing the resultprint("The words frequency : " + str(res)) |
The original string is : Gfg is best . Geeks are good and Geeks like Gfg
The words frequency : {'Gfg': 2, 'is': 1, 'best': 1, '.': 1, 'Geeks': 2, 'are': 1, 'good': 1, 'and': 1, 'like': 1}
Time Complexity: O(N)
Auxiliary Space : O(N)
Method #4: Using defaultdict
Python3
from collections import defaultdict# Initializing stringtest_str = 'Gfg is best . Geeks are good and Geeks like Gfg'# Printing original stringprint("The original string is : " + str(test_str))# Split the string into a list of wordslistString = test_str.split()# Creating a defaultdict with default value 0freq = defaultdict(int)# Iterating through the list of words and# count the frequency of each wordfor word in listString: freq[word] += 1# Converting the defaultdict to a regular dictionaryres = dict(freq)# Printing resultprint("The words frequency : " + str(res)) |
The original string is : Gfg is best . Geeks are good and Geeks like Gfg
The words frequency : {'Gfg': 2, 'is': 1, 'best': 1, '.': 1, 'Geeks': 2, 'are': 1, 'good': 1, 'and': 1, 'like': 1}
Time Complexity: O(n), where n is the number of words in the string.
Auxiliary Space: O(n), where n is the number of unique words in the string.
Method #5: Using set() and list comprehension
Step-by-step approach:
- Split the string into a list of words.
- Use set() to get a unique set of words.
- Use a list comprehension to count the frequency of each word in the original list.
- Store the results in a dictionary using dictionary comprehension.
- Print the final result.
Python3
# Initializing stringtest_str = 'Gfg is best . Geeks are good and Geeks like Gfg'# Printing original stringprint("The original string is : " + str(test_str))# Split the string into a list of wordslistString = test_str.split()# Using set() and list comprehension to count the frequency of each wordfreq = {word: listString.count(word) for word in set(listString)}# Printing resultprint("The words frequency : " + str(freq)) |
The original string is : Gfg is best . Geeks are good and Geeks like Gfg
The words frequency : {'are': 1, 'good': 1, 'and': 1, 'like': 1, 'best': 1, 'Gfg': 2, 'is': 1, 'Geeks': 2, '.': 1}
Time complexity: O(n^2) where n is the length of the list of words.
Auxiliary space: O(n) where n is the length of the list of words.
