Python String isalpha() Method
Python String isalpha() method is used to check whether all characters in the String is an alphabet.
Python String isalpha() Method Syntax:
Syntax: string.isalpha()
Parameters: isalpha() does not take any parameters
Returns:
- True: If all characters in the string are alphabet.
- False: If the string contains 1 or more non-alphabets.
Errors and Exceptions:
- It contains no arguments, therefore an error occurs if a parameter is passed
- Both uppercase and lowercase alphabets return “True”
- Space is not considered to be the alphabet, therefore it returns “False”
Python String isalpha() Method Example:
Python3
string = "geeks"print(string.isalpha()) |
Output:
True
Example 1: Working of isalpha()
Python3
# checking for alphabetsstring = 'Ayush'print(string.isalpha()) string = 'Ayush0212'print(string.isalpha()) # checking if space is an alphabetstring = 'Ayush Saxena'print( string.isalpha()) |
Output:
True False False
Example 2: Practical Application
Given a string in Python, count the number of alphabets in the string and print the alphabets.
Input : string = 'Ayush Saxena'
Output : 11
AyushSaxena
Input : string = 'Ayush0212'
Output : 5
AyushAlgorithm:
- Initialize a new string and variable counter to 0.
- Traverse the given string character by character up to its length, check if the character is an alphabet.
- If it is an alphabet, increment the counter by 1 and add it to a new string, else traverse to the next character.
- Print the value of the counter and the new string.
Python3
# Given stringstring='Ayush Saxena'count=0# Initialising new stringsnewstring1 =""newstring2 =""# Iterating the string and checking for alphabets# Incrementing the counter if an alphabet is found# Finally printing the countfor a in string: if (a.isalpha()) == True: count+=1 newstring1+=aprint(count)print(newstring1)# Given stringstring='Ayush0212'count=0for a in string: if (a.isalpha()) == True: count+=1 newstring2+=aprint(count)print(newstring2) |
Output:
11 AyushSaxena 5 Ayush
Time Complexity: O(n)
Auxiliary Space: O(n)
Please Login to comment...