Loops, Conditionals
Given a word in English, return True if that word contains more vowels than non-vowels; otherwise, return False. The word will always be a single word, without any punctuation or spaces. It will contain only uppercase and lowercase letters.
If the phrase is over half vowels, it should return True:
>>> has_more_vowels("moose")
True
If it’s half vowels (or less), it’s false:
>>> has_more_vowels("mice")
False
>>> has_more_vowels("graph")
False
Don’t consider “y” as a vowel:
>>> has_more_vowels("yay")
False
Uppercase vowels are still vowels:
>>> has_more_vowels("Aal")
True
We’ve given you hasmorevowels.py, which includes the stub of a has_more_vowels function:
def has_more_vowels(word):
"""Does word contain more vowels than non-vowels?"""
Implement this function.