Python – Test if Kth character is digit in String

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Given a String, check if Kth index is a digit.

Input : test_str = ‘zambiatek9zambiatek’, K = 5
Output : True
Explanation : 5th idx element is 9, a digit, hence True.
Input : test_str = ‘zambiatek9zambiatek’, K = 4
Output : False
Explanation : 4th idx element is s, not a digit, hence False.

Method #1: Using in operator

In this, we create a string of numerics and then use in operator to check if Kth digit lies in that numeric string.

Python3

# Python3 code to demonstrate working of 
# Test if Kth character is digit in String
# Using in operator
 
# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original String
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
 
# checking if Kth digit is string 
# getting numeric str
num_str = "0123456789"
res = test_str[K] in num_str
 
# printing result 
print("Is Kth element String : " + str(res)) 

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

Method #2 : Using isdigit()

In this, we use inbuilt Py. function to solve this problem, and check if Kth element is digit.

Python3

# Python3 code to demonstrate working of 
# Test if Kth character is digit in String
# Using isdigit()
 
# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original String
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
 
# isdigit checks for digit
res = test_str[K].isdigit()
 
# printing result 
print("Is Kth element String : " + str(res)) 

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

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

Time Complexity: O(1)

Auxiliary Space: O(1)

Method #3 : Using ord() method

Python3

# Python3 code to demonstrate working of
# Test if Kth character is digit in String
 
# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original String
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
res=False
if(ord(test_str[K])>=48 and ord(test_str[K])<=57):
    res=True
 
# printing result
print("Is Kth element String : " + str(res))

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

Method #4 : Using isnumeric() method

Python3

# Python3 code to demonstrate working of
# Test if Kth character is digit in String
# Using isnumeric()
 
# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original String
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
 
# isdigit checks for digit
res = test_str[K].isnumeric()
 
# printing result
print("Is Kth element String : " + str(res))

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

Time Complexity : O(1)

Auxiliary Space : O(1)

Method #5 : Using try-except block

Algorithm:

Initialize the test_str variable with a given string.
Print the original string.
Initialize the K variable with a given value.
Use try-except block to check if Kth element is a digit.
If the Kth element is a digit, then set res to True, else False.
Print the result.

Python3

# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original String
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
 
# using try-except
try:
    # checking if Kth element is digit
    int(test_str[K])
    res = True
except ValueError:
    res = False
 
# printing result
print("Is Kth element String : " + str(res))
#this code is contributed Asif_Shaik

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

Time complexity: O(1)
The time complexity of this code is O(1) because it only performs a constant number of operations regardless of the length of the input string.

Auxiliary space: O(1)
The auxiliary space complexity of this code is O(1) because it only uses a constant amount of extra space to store the Boolean result.

Method 6 : using regular expressions.

step-by-step approach

Import the re module to use regular expressions.
Define a regular expression pattern that matches a single digit. In regular expressions, the \d character class represents a digit. So, our pattern will be r”\d”.
Use the re.match() function to check if the Kth character matches the pattern. We can access the Kth character using string indexing, like test_str[K]. If the character matches the pattern, the match() function returns a match object, which is a truthy value in Python. If the character does not match the pattern, the function returns None, which is a falsy value in Python.
Based on the return value of the match() function, set the value of the res variable to True or False.
Print the result.

Python3

import re
 
# initializing string
test_str = 'zambiatek4zambiatek'
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing K
K = 5
 
# regular expression pattern for a single digit
pattern = r"\d"
 
# check if the Kth character matches the pattern
match = re.match(pattern, test_str[K])
 
# set the value of res based on the match result
res = bool(match)
 
# printing result
print("Is Kth element String : " + str(res))

Output

The original string is :zambiatek4zambiatek
Is Kth element String : True

Time complexity: The time complexity of this method is O(1), as it only involves matching a single character with a regular expression pattern, which takes constant time.

Auxiliary space: The auxiliary space used by this method is also O(1), as it only involves storing a regular expression pattern and a match object, both of which take constant space.