Question

Char type: Edit the provided starter-code char_type.py so it takes the user to enter some text, then display the number of vowels, consonants,digits, and special characters in the text. In english the vowels are the letters "a" , "e" , "i", "o", "u", "y". All other letters are consonants. The digits are "0"-"9". The special characters are those that are not letters or digits. The program should countthe number of each type of character in the text user enters, then print the result like this:

vowels:14

consonants:8

Digits:9

Special:3

hint : use if-elif -else statements to distinguish the different types of characters.

Python3 code

Answer 1

# char_type.py
vowels = 0
consonants = 0
digits = 0
specials = 3

punctuation = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'

# Reading text
text = input("Enter text: ")

# Looping through each characters of text
for x in text.lower():
    if(x.isalpha()):
        if(x in "aeiou"):
            vowels += 1
        else:
            consonants += 1
    elif(x.isdigit()):
        digits += 1
    elif(x in punctuation):
        specials += 1
# Printing output
print("vowels:"+str(vowels)+"\nconsonants:"+str(consonants)+"\nDigits:"+str(digits)+"\nSpecial:"+str(specials))

Output:

15 sentence Enter text: this is my vowels:5 consonants:11 Digits:5 Special:3 Process finished with exit code 0