Find frequency of each word in a string in Python
Last Updated :
20 Feb, 2023
Write a python code to find the frequency of each word in a given string. Examples:
Input : str[] = "Apple Mango Orange Mango Guava Guava Mango"
Output : frequency of Apple is : 1
frequency of Mango is : 3
frequency of Orange is : 1
frequency of Guava is : 2
Input : str = "Train Bus Bus Train Taxi Aeroplane Taxi Bus"
Output : frequency of Train is : 2
frequency of Bus is : 3
frequency of Taxi is : 2
frequency of Aeroplane is : 1
Approach 1 using list():
1. Split the string into a list containing the words by using split function (i.e. string.split()) in python with delimiter space.
Note:
string_name.split(separator) method is used to split the string
by specified separator(delimiter) into the list.
If delimiter is not provided then white space is a separator.
For example:
CODE : str='This is my book'
str.split()
OUTPUT : ['This', 'is', 'my', 'book']
2. Initialize a new empty list.
3. Now append the word to the new list from previous string if that word is not present in the new list.
4. Iterate over the new list and use count function (i.e. string.
Python3
# Find frequency of each word in a string in Python
# using dictionary.
def count(elements):
# check if each word has '.' at its last. If so then ignore '.'
if elements[-1] == '.':
elements = elements[0:len(elements) - 1]
# if there exists a key as "elements" then simply
# increase its value.
if elements in dictionary:
dictionary[elements] += 1
# if the dictionary does not have the key as "elements"
# then create a key "elements" and assign its value to 1.
else:
dictionary.update({elements: 1})
# driver input to check the program.
Sentence = "Apple Mango Orange Mango Guava Guava Mango"
# Declare a dictionary
dictionary = {}
# split all the word of the string.
lst = Sentence.split()
# take each word from lst and pass it to the method count.
for elements in lst:
count(elements)
# print the keys and its corresponding values.
for allKeys in dictionary:
print ("Frequency of ", allKeys, end = " ")
print (":", end = " ")
print (dictionary[allKeys], end = " ")
print()
# This code is contributed by Ronit Shrivastava.
OutputFrequency of Apple : 1
Frequency of Mango : 3
Frequency of Orange : 1
Frequency of Guava : 2
Time complexity : O(n)
Space complexity : O(n)
(newstring[iteration])) to find the frequency of word at each iteration.
Note:
string_name.count(substring) is used to find no. of occurrence of
substring in a given string.
For example:
CODE : str='Apple Mango Apple'
str.count('Apple')
str2='Apple'
str.count(str2)
OUTPUT : 2
2
Implementation:
Python3
# Python code to find frequency of each word
def freq(str):
# break the string into list of words
str = str.split()
str2 = []
# loop till string values present in list str
for i in str:
# checking for the duplicacy
if i not in str2:
# insert value in str2
str2.append(i)
for i in range(0, len(str2)):
# count the frequency of each word(present
# in str2) in str and print
print('Frequency of', str2[i], 'is :', str.count(str2[i]))
def main():
str ='apple mango apple orange orange apple guava mango mango'
freq(str)
if __name__=="__main__":
main() # call main function
OutputFrequency of apple is : 3
Frequency of mango is : 3
Frequency of orange is : 2
Frequency of guava is : 1
Time complexity : O(n^2)
Space complexity : O(n)
Approach 2 using set():
- Split the string into a list containing the words by using split function (i.e. string.split()) in python with delimiter space.
- Use set() method to remove a duplicate and to give a set of unique words
- Iterate over the set and use count function (i.e. string.count(newstring[iteration])) to find the frequency of word at each iteration.
Implementation:
Python3
# Python3 code to find frequency of each word
# function for calculating the frequency
def freq(str):
# break the string into list of words
str_list = str.split()
# gives set of unique words
unique_words = set(str_list)
for words in unique_words :
print('Frequency of ', words , 'is :', str_list.count(words))
# driver code
if __name__ == "__main__":
str ='apple mango apple orange orange apple guava mango mango'
# calling the freq function
freq(str)
OutputFrequency of orange is : 2
Frequency of mango is : 3
Frequency of guava is : 1
Frequency of apple is : 3
Time complexity : O(n^2)
Space complexity : O(n)
Approach 3 (Using Dictionary):
Python3
# Find frequency of each word in a string in Python
# using dictionary.
def count(elements):
# check if each word has '.' at its last. If so then ignore '.'
if elements[-1] == '.':
elements = elements[0:len(elements) - 1]
# if there exists a key as "elements" then simply
# increase its value.
if elements in dictionary:
dictionary[elements] += 1
# if the dictionary does not have the key as "elements"
# then create a key "elements" and assign its value to 1.
else:
dictionary.update({elements: 1})
# driver input to check the program.
Sentence = "Apple Mango Orange Mango Guava Guava Mango"
# Declare a dictionary
dictionary = {}
# split all the word of the string.
lst = Sentence.split()
# take each word from lst and pass it to the method count.
for elements in lst:
count(elements)
# print the keys and its corresponding values.
for allKeys in dictionary:
print ("Frequency of ", allKeys, end = " ")
print (":", end = " ")
print (dictionary[allKeys], end = " ")
print()
# This code is contributed by Ronit Shrivastava.
OutputFrequency of Apple : 1
Frequency of Mango : 3
Frequency of Orange : 1
Frequency of Guava : 2
time complexity : O(m * n)
space complexity : O(k)
Approach 4: Using Counter() function:
Python3
# Python3 code to find frequency of each word
# function for calculating the frequency
from collections import Counter
def freq(str):
# break the string into list of words
str_list = str.split()
frequency = Counter(str_list)
for word in frequency:
print('Frequency of ', word, 'is :', frequency[word])
# driver code
if __name__ == "__main__":
str = 'apple mango apple orange orange apple guava mango mango'
# calling the freq function
freq(str)
OutputFrequency of apple is : 3
Frequency of mango is : 3
Frequency of orange is : 2
Frequency of guava is : 1
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach 4 (Using setdefault):
Python3
# Python3 code to find frequency of each word
def freq(str):
# break the string into list of words
str_list = str.split()
# create an empty dictionary
frequency = {}
# count frequency of each word
for word in str_list:
frequency[word] = frequency.setdefault(word, 0) + 1
# print the frequency of each word
for key, value in frequency.items():
print(key, ':', value)
str = 'apple mango apple orange orange apple guava mango mango'
# calling the function
freq(str)
#This code is contributed by Edula Vinay Kumar Reddy
Outputapple : 3
mango : 3
orange : 2
guava : 1
Time Complexity: O(n), where n is the length of the given string
Auxiliary Space: O(n)
Approach 5 :Using operator.countOf() method:
Python3
import operator as op
# Python code to find frequency of each word
def freq(str):
# break the string into list of words
str = str.split()
str2 = []
# loop till string values present in list str
for i in str:
# checking for the duplicacy
if i not in str2:
# insert value in str2
str2.append(i)
for i in range(0, len(str2)):
# count the frequency of each word(present
# in str2) in str and print
print('Frequency of', str2[i], 'is :', op.countOf(str,str2[i]))
def main():
str ='apple mango apple orange orange apple guava mango mango'
freq(str)
if __name__=="__main__":
main() # call main function
OutputFrequency of apple is : 3
Frequency of mango is : 3
Frequency of orange is : 2
Frequency of guava is : 1
Time Complexity: O(N), where n is the length of the given string
Auxiliary Space: O(N)
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