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Python – Successive Characters Frequency

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  • Last Updated : 05 Sep, 2022
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Sometimes, while working with Python strings, we can have a problem in which we need to find the frequency of next character of a particular word in string. This is quite unique problem and has the potential for application in day-day programming and web development. Let’s discuss certain ways in which this task can be performed.

Input : test_str = ‘geeks are for geeksforgeeks’, que_word = “geek” Output : {‘s’: 3} Input : test_str = ‘geek’, que_word = “geek” Output : {}

Method #1 : Using loop + count() + re.findall() The combination of above methods constitute the brute force method to perform this task. In this, we perform the task of counting using count(), and character is searched using findall(). 

Python3




# Python3 code to demonstrate working of
# Successive Characters Frequency
# Using count() + loop + re.findall()
import re
     
# initializing string
test_str = 'geeksforgeeks is best for geeks. A geek should take interest.'
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing word
que_word = "geek"
 
# Successive Characters Frequency
# Using count() + loop + re.findall()
temp = []
for sub in re.findall(que_word + '.', test_str):
    temp.append(sub[-1])
 
res = {que_word : temp.count(que_word) for que_word in temp}
 
# printing result
print("The Characters Frequency is : " + str(res))

Output : 

The original string is : geeksforgeeks is best for geeks. A geek should take interest.
The Characters Frequency is : {'s': 3, ' ': 1}

  Method #2 : Using Counter() + list comprehension + re.findall() The combination of above functions is used to perform the following task. In this, we use Counter() instead of count() to solve this problem. Works with newer versions of Python. 

Python3




# Python3 code to demonstrate working of
# Successive Characters Frequency
# Using Counter() + list comprehension + re.findall()
from collections import Counter
import re
 
# initializing string
test_str = 'geeksforgeeks is best for geeks. A geek should take interest.'
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing word
que_word = "geek"
 
# Successive Characters Frequency
# Using Counter() + list comprehension + re.findall()
res = dict(Counter(re.findall(f'{que_word}(.)', test_str,
                                    flags=re.IGNORECASE)))
 
# printing result
print("The Characters Frequency is : " + str(res))

Output : 

The original string is : geeksforgeeks is best for geeks. A geek should take interest.
The Characters Frequency is : {'s': 3, ' ': 1}

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Space Complexity: O(n)


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