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
def count(elements):
if elements[-1] == '.':
elements = elements[0:len(elements) - 1]
if elements in dictionary:
dictionary[elements] += 1
else:
dictionary.update({elements: 1})
Sentence = "Apple Mango Orange Mango Guava Guava Mango"
dictionary = {}
lst = Sentence.split()
for elements in lst:
count(elements)
for allKeys in dictionary:
print ("Frequency of ", allKeys, end = " ")
print (":", end = " ")
print (dictionary[allKeys], end = " ")
print()
|
Output
Frequency 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
def freq(str):
str = str.split()
str2 = []
for i in str:
if i not in str2:
str2.append(i)
for i in range(0, len(str2)):
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()
|
Output
Frequency 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
def freq(str):
str_list = str.split()
unique_words = set(str_list)
for words in unique_words :
print('Frequency of ', words , 'is :', str_list.count(words))
if __name__ == "__main__":
str ='apple mango apple orange orange apple guava mango mango'
freq(str)
|
Output
Frequency 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
def count(elements):
if elements[-1] == '.':
elements = elements[0:len(elements) - 1]
if elements in dictionary:
dictionary[elements] += 1
else:
dictionary.update({elements: 1})
Sentence = "Apple Mango Orange Mango Guava Guava Mango"
dictionary = {}
lst = Sentence.split()
for elements in lst:
count(elements)
for allKeys in dictionary:
print ("Frequency of ", allKeys, end = " ")
print (":", end = " ")
print (dictionary[allKeys], end = " ")
print()
|
Output
Frequency 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
from collections import Counter
def freq(str):
str_list = str.split()
frequency = Counter(str_list)
for word in frequency:
print('Frequency of ', word, 'is :', frequency[word])
if __name__ == "__main__":
str = 'apple mango apple orange orange apple guava mango mango'
freq(str)
|
Output
Frequency 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
def freq(str):
str_list = str.split()
frequency = {}
for word in str_list:
frequency[word] = frequency.setdefault(word, 0) + 1
for key, value in frequency.items():
print(key, ':', value)
str = 'apple mango apple orange orange apple guava mango mango'
freq(str)
|
Output
apple : 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
def freq(str):
str = str.split()
str2 = []
for i in str:
if i not in str2:
str2.append(i)
for i in range(0, len(str2)):
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()
|
Output
Frequency 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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