Cannot determine vowels from consonants

Question

With the code below, no matter what the first letter of the input is, it is always determined as a vowel:

original = raw_input(\"Please type in a word: \")
f         


        

Answer 1

First you should know how or works, it evaluates both the left and right expression and it returns the first expression that evaluates true, otherwise it returns the last expression:

From this source you can see that all of these evaluate false:

• None
• False
• zero of any numeric type, for example, 0, 0L, 0.0, 0j.
• any empty sequence, for example, '', (), [].
• any empty mapping, for example, {}.
• instances of user-defined classes, if the class defines a nonzero() or len() method, when that method returns the integer zero or bool value False.3.1

---

>>> False or True
True
>>> '' or False
False
>>> '' or 0
0

So a non-empty string will evaluate as true and it will be returned:

>>> 'abc' or False
'abc

---

 if firstLetter == "a" or "e" or "i" or "o" or "u":
     print "vowel"
 else:
     print "consonant"

For a string 'foo', 'f' is the first letter. In the first part of your code, firstLetter == "a" or "e", the left expression will evaluate as false, but 'e' is not an empty string so it evaluates as true and it will print "vowel"

You can see my other answer here which answers your question, something like this:

if c.upper() in "AEIOU"

will check if a letter is a vowel.