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Python program to calculate the number of digits and letters in a string
  • Last Updated : 26 Nov, 2020

Prerequisites: isnumeric() method – Python

Given a string, containing digits and letters, the task is to write a Python program to calculate the number of digits and letters in a string. The following block expresses the basic idea behind it:

Input: string = "geeks2for3geeks"
Output: total digits = 2 and total letters = 13

Input: string = "python1234"
Output: total digits = 4 and total letters = 6

Input: string = "co2mpu1te10rs"
Output: total digits = 4 and total letters = 9

First Approach

The idea here is to solve this problem by iterating through all characters and check whether character is in all_digits or all_letters. 

Program:

Python3




# define all digits as string
all_digits = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
  
# define all letters
all_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l',
               'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
  
# given string
string = "geeks2for3geeks"
  
# intialized value
total_digits = 0
total_letters = 0
  
# iterate through all characters
for s in string:
  
    # if character found in all_digits then increment total_digits by one
    if s in all_digits:
        total_digits += 1
  
    # if character found in all_letters then increment total_letters by one
    elif s in all_letters:
        total_letters += 1
  
print("Total letters found :-", total_letters)
print("Total digits found :-", total_digits)

Output:



Total letters found :- 13
Total digits found :- 2

Optimized Method

Instead of checking character in all_letters, we can check:

  • if character is found in all digits, then increment total_digits value by one
  • If not it means characters is a letter, increment total_letters value by one

Program:

Python3




# define all digits as string
all_digits = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
  
# define all letters
all_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l',
               'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
  
# given string
string = "geeks2for3geeks"
  
# intialized value
total_digits = 0
total_letters = 0
  
# iterate through all characters
for s in string:
  
    # if character found in all_digits then increment total_digits by one
    if s in all_digits:
        total_digits += 1
  
    # if character not found in all_digits then increment total_letters by one
    else:
        total_letters += 1
  
print("Total letters found :-", total_letters)
print("Total digits found :-", total_digits)

Output:

Total letters found :- 13
Total digits found :- 2

Second Approach (More Optimized Method)

The idea here is to solve this problem by iterating through all characters and check whether character is letter or digits using isnumeric() function. If isnumeric is True, it means character is digit, else character is a letter.

Program:

Python3




# given string
string = "python1234"
  
# intialized value
total_digits = 0
total_letters = 0
  
# iterate through all characters
for s in string:
  
    # if character is digit (return True)
    if s.isnumeric():
        total_digits += 1
  
    # if character is letter (return False)
    else:
        total_letters += 1
  
print("Total letters found :-", total_letters)
print("Total digits found :-", total_digits)

Output:

Total letters found :- 6
Total digits found :- 4

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