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Python – List Words Frequency in String

Last Updated : 08 May, 2023
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Given a List of Words, Map frequency of each to occurrence in String.

Input : test_str = 'geeksforgeeks is best for geeks and best for CS', count_list = ['best', 'geeksforgeeks', 'computer'] 
Output : [2, 1, 0] 
Explanation : best has 2 occ., geeksforgeeks 1 and computer is not present in string.
Input : test_str = 'geeksforgeeks is best for geeks and best for CS', count_list = ['better', 'gfg', 'computer'] 
Output : [0, 0, 0] 
Explanation : No word from list present in string.

Method #1 : Using defaultdict() + loop + list comprehension

In this, we compute words frequency using loop + defaultdict() and then use list comprehension to get all the counts corresponding to list of words.

Python3




# Python3 code to demonstrate working of
# Divide String into Equal K chunks
# Using list comprehension
from collections import defaultdict
 
# Initializing strings
test_str = 'geeksforgeeks is best for geeks and best for CS'
 
# Printing original string
print("The original string is : " + str(test_str))
 
# Initializing count_list
count_list = ['best', 'geeksforgeeks', 'computer', 'better', 'for', 'and']
 
# Computing frequency
res = defaultdict(int)
 
for sub in test_str.split():
    res[sub] += 1
 
# Assigning to list words
res = [res[sub] for sub in count_list]
 
# Printing result
print("The list words frequency : " + str(res))


Output

The original string is : geeksforgeeks is best for geeks and best for CS
The list words frequency : [2, 1, 0, 0, 2, 1]

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the length of the list.

Method #2 : Using Counter() + list comprehension

In this, Counter() is used to perform the task of computing frequency, post that, list comprehension is used to assign a frequency to list words.

Python3




# Python3 code to demonstrate working of
# Divide String into Equal K chunks
# Using list comprehension
from collections import Counter
 
# initializing strings
test_str = 'geeksforgeeks is best for geeks and best for CS'
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing count_list
count_list = ['best', 'geeksforgeeks', 'computer', 'better', 'for', 'and']
 
# computing frequency using Counter()
res = Counter(test_str.split())
     
# assigning to list words
res = [res[sub] for sub in count_list]
 
# printing result
print("The list words frequency : " + str(res))


Output

The original string is : geeksforgeeks is best for geeks and best for CS
The list words frequency : [2, 1, 0, 0, 2, 1]

Time complexity: O(N) since using a loop
Auxiliary Space: O(1)

Method #3 : Using count() method

Approach

  1. Split the string test_str which results in a list(x)
  2. Initiate a for loop to traverse the list of strings.
  3. Now append the occurrence of each string in x to the output list.
  4. Display output list.

Python3




# Python3 code to demonstrate working of
# Divide String into Equal K chunks
 
# Initializing strings
test_str = 'geeksforgeeks is best for geeks and best for CS'
 
# Printing original string
print("The original string is : " + str(test_str))
x=test_str.split()
 
# Iitializing count_list
count_list = ['best', 'geeksforgeeks', 'computer', 'better', 'for', 'and']
 
# Cmputing frequency
res=[]
 
for i in count_list:
    res.append(x.count(i))
 
# Pinting result
print("The list words frequency : " + str(res))


Output

The original string is : geeksforgeeks is best for geeks and best for CS
The list words frequency : [2, 1, 0, 0, 2, 1]

Time Complexity : O(M*N) M – length of x N – length of count_list
Auxiliary Space : O(N) N – length of output list

Method #4: Using dictionary comprehension

In this method, we can use a dictionary comprehension to count the frequency of each word in the given string. The keys of the dictionary will be the words from the count_list, and the values will be the frequency of each word in the given string. 

Python3




# Python3 code to demonstrate working of
# Divide String into Equal K chunks
 
 
# initializing strings
test_str = 'geeksforgeeks is best for geeks and best for CS'
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing count_list
count_list = ['best', 'geeksforgeeks', 'computer', 'better', 'for', 'and']
 
# computing frequency using dictionary comprehension
res = {i: test_str.split().count(i) for i in count_list}
 
# printing result
print("The list words frequency : " + str([res[i] for i in count_list]))


Output

The original string is : geeksforgeeks is best for geeks and best for CS
The list words frequency : [2, 1, 0, 0, 2, 1]

Time complexity: O(N), where n is the length of the given string.
Auxiliary space: O(K), where k is the number of words in the count_list.



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