On the Unix machines,there is a file /usr/dict/words that contains a list of English words,one per line.This file is used by spell-checking programs for example.
Write a python function which takes one argument,which is a word (as a string)and returns true if the word contains alternating vowels and consonants (i.e a consonant followed by a vowel followed by a consonant) Then write a short piece of code to read the above file and print out all the words like "banana" and "lever" but not words like "knock" and "print"

Very smart homework problem. Now let's see some effort on your part to solve it!

Hint ...

vowels = 'aeiouy'

word = 'polarity'

for letter in word:
    if letter in vowels:
        print( letter )

Edited 6 Years Ago by vegaseat: homework

Very smart homework problem. Now let's see some effort on your part to solve it!

Hint ...

vowels = 'aeiouy'

word = 'polarity'

for letter in word:
    if letter in vowels:
        print( letter )

I came up with this,but it does not carry out exactly what is required. Need some help to modify it. Thanks!

def check(k):
       d = ('a','e','i','o','u')
        first,second = 0,1
           if k(first) in d:
             if k(second) not in d
             first+=2
             second+=2
         return 'true'

Edited 6 Years Ago by vegaseat: added code tags

Here is another hint:

word = 'polarity'

even_letters = [word[i] for i in range(0, len(word), 2)]
odd_letters = [word[i] for i in range(1, len(word), 2)]

print(even_letters)
print(odd_letters)

Also note that on some linuxes, the dictionary is at /usr/share/dict/words .

This article has been dead for over six months. Start a new discussion instead.