Python – Successive Characters Frequency

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 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 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}