Python anti-vowel function - some vowels cannot be removed

Question

I just started to learn Python in codacademy. I was trying to do anti-vowel function, but found the following problem with 'u'.

def anti_vowel(text):
    a = []
    for i in text:
        a.append(i)
    for item in a:
        if item in "aeiouAEIOU":
            a.remove(item)
    print ''.join(a)

print anti_vowel("Hey You!")
print anti_vowel("Hey look Words!")
print anti_vowel("aeiouAEIOU")

It printed

"Hy Yu!"
"Hy lk Words!"
"eoAIU"

Instead of

"Hy Y!"
"Hy lk Wrds!"
""

Somehow, some vowels was not removed es expected.

I found many alternatives for the function. However, please help me identify the mistake of the current code.

Answer 1

There's no need to use remove, and there's no need to iterate twice. Rather, as you iterate, check whether the item is a vowel and only append if it is not.

def anti_vowel(text):
    a = []
    for i in text:
        if i not in "aeiouAEIOU":
            a.append(i)
    print ''.join(a)

Answer 2

When you look very closely at the remaining vowels, you can see that all those vowels remain that immediately follow one another. In your last example, a (removed) e (stays) i (removed) o (stays) and so on.

This is because you are iterating over a list and at the same time you are modifying it.

To solve the problem, you should make a copy of the list. Then you can iterate over the original one while modifying the copy.

Answer 3

Removing items while iterating is not a good idea.

Do it in one line with a generator comprehension passed to str.join

def anti_vowel(text):
    return ''.join(item for item in text if item not in "aeiouAEIOU")

or maybe more performant using a set for faster letter lookup (not sure converting to lowercase would speed this up since it creates a new string for that)

s=set("aeiouAEIOU")
def anti_vowel(text):
    return ''.join(item for item in text if item not in s)

Answer 4

Like others have already stated, you are modifying your list while iterating through it. I did want to suggest a python built-in option for this task, though for 3.x > Python > 2.6:

print "Hey You!".translate(None, "aeiouAEIOU")

In Python 3.x you'll need to take account for the standard Unicode string and translate it first:

translation = dict.fromkeys(map(ord, "aeiouAEIOU"), None)
print("Hey You!".translate(translation))

Answer 5

def anti_vowel(text):
    a = []

    for i in text:
        a.append(i)

    b = a[:]
    for item in a:
        if item in "aeiouAEIOU":
            b.remove(item)
    print (''.join(b))

print (anti_vowel("Hey You!"))
print (anti_vowel("Hey look Words!"))
print (anti_vowel("aeiouAEIOU"))

Answer 6

Why not try recursively like this?

def anti_vowel(s):
    if not s:
        return s
    elif s[0] in "aeiouAEIOU":
        return anti_vowel(s[1:])
    return s[0] + anti_vowel(s[1:])

print (anti_vowel("Hey You!"))
print (anti_vowel("Hey look Words!"))
print (anti_vowel("aeiouAEIOU"))

Output:

Hy Y!
Hy lk Wrds!

Answer 7

you are removing item from list while traversing so thats creating a problem. Moreover there is no need to traverse the so many times.

you could use:

def anti_vowel(text):
    a = []
    for i in text:
        if i not in "aeiouAEIOU":
            a.append(i)
    print ''.join(a)

or I would say instead of using list use string like this :

def anti_vowel(text):
    a = ""
    for item in text:
        if item not in "aeiouAEIOU":
            a+=item
    print a

Edited:

using ''.join(list) is faster than using string cancatenation as pointed out by @jean in the comment section.

Answer 8

You could also join the result of a list comprehension, like:

def anti_vowel(text):
    return "".join([char for char in text if char not in "aeiouAEIOU"])

Be sure to let your functions return their results if you want the function to do more than just printing to the terminal (for example if you want to use the returned value somehow). There's no need to print the results of a function that prints its result and returns None (which is the case if there are no return statements inside).

The [char for char in text ...] part of the list comprehension is simply a for loop over the characters in the string named text. The [... if char not in "aeiouAEIOU"] excludes the characters who are present in the "aeiouAEIOU" string. Finally, "".join([ ... ]) merges the non-excluded characters together to form a string which by the return statement is returned.