Counting vowels and consonants in a file (Python)

Question

I need to make a program that will read a text file and print how many vowels and consonants there are. I make a text file to test and the only thing in it is "This is a test". However the output it always:

Enter the file to check: test.txt

The number of Vowels is: 1

The number of consonants is: 0

fileName = input("Enter the file to check: ").strip()

infile = open(fileName, "r")


vowels = set("A E I O U a e i o u")
cons = set("b c d f g h j k l m n p q r s t v w x y z B C D F G H J K L M N P Q R S T V W X Y Z")

text = infile.read().split()


countV = 0
for V in text:
    if V in vowels:
        countV += 1

countC = 0
for C in text:
    if C in cons:
        countC += 1

print("The number of Vowels is: ",countV,"\nThe number of consonants is: ",countC)

If there is a better way to enter the values for vowels and cons I would also like to know, as I get an error when I try to user .lower() to convert everything in the file to lower case.....

Solution

1. set("A E I O U a e i o u") will result in {' ', 'A', 'E', 'I', 'O', 'U', 'a', 'e', 'i', 'o', 'u'}. If you'll notice, the space is also considered. You'll need to remove the spaces between the letters.

2. infile.read().split() will split based on whitespace so you get a list of words. You then proceed to iterate over the words, and try a membership comparison between the words and the letters. This will not work out for you.

3. You don't need to iterate twice. Once is enough.

---

Here's a cleaned up version of your code.

vowels = set("AEIOUaeiou")
cons = set("bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ")

countV = 0
countC = 0
for c in infile.read():
    if c in vowels:
        countV += 1
    elif c in cons:
        countC += 1

---

As an improvement, consider the use of collections.Counter. It does the counting for you, and you just sum up the counts.

import collections
c = collections.Counter(infile.read())

countV = sum(c[k] for k in c if k in vowels)
countC = sum(c[k] for k in c if k in cons)

Comments

[Willem Van Onsem]

Because you count spaces as well??

[PyPunk]

Thank you @COLDSPEED this helped. I did it slightly different but getting rid of the .split() and removing the spaces was a big help. I thought that sets don't repeat themselves though? So doesn't that mean that it would count the spaces too rather than showing 0 as a result?

[cs95]

@EvanH It counts the spaces once, but they are still counted. You will have spaces erroneously counted as both vowels and characters unless you removed them.

[PyPunk]

@COLDSPEED Got it. Thank you very much for the help and quick response!