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: ")
firstLetter = original[0]
print firstLetter

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

In fact, it doesn’t matter what the boolean is in the if statement… if it is == or != , it is still return "vowel". Why?

Answer 1

Python is not the English language. If you have a bunch of expressions with or or and between them, each one must make sense on its own. Note that on its own:

if "e":
    print("something")

will always print something, even if letter doesn’t equal "e".

You need to do it like this:

if letter == "a" or letter == "e"  # (...)

Or, more concisely:

if letter in "aeiouy":