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# does not remove consecutive vowels in the vowel removal function

def disemvowel(word): word_list = list(word) print (word_list) for alpha in word_list: print (alpha) if alpha in ["a","e","i","o","u","A","E","I","O","U"]: #try: word_list.remove(alpha) #except ValueError: #print("alpha {} not found".format(alpha)) print(word_list) else: print("not there") print(word_list) continue i = 0 word1="" while i<len(word_list): word1 += word1.join(word_list[i]) i=i+1 return word1

def main(): w = disemvowel("oeRGo") print (w)

main()

disemvowel.py
```def disemvowel(word):
word_list = list(word)
print (word_list)
for alpha in word_list:
print (alpha)
if alpha in ["a","e","i","o","u","A","E","I","O","U"]:
try:
word_list.remove(alpha)
except ValueError:
print("alpha {} not found".format(alpha))
print(word_list)
else:
print("not there")
i = 0
word1=""
while i<len(word_list):
word1 += word1.join(word_list[i])
i=i+1
return word1
```

OP: treehouse:~/workspace\$ python vowels.py
['o', 'e', 'R', 'G', 'o']
o
['e', 'R', 'G', 'o']
R
not there
['e', 'R', 'G', 'o']
G
not there
['e', 'R', 'G', 'o']
o
['e', 'R', 'G']
eRG

## 1 Answer

STAFF

Hi there! Yes, you're correct it will not remove consecutive vowels and there's a reason for this. You're mutating the thing you're iterating over which leads to some unexpected results. The indexing of the iterable is changing every time something is removed. The side-effect of this is that every time a `remove` happens, the subsequent letter will not be checked as it has "fallen down" into the place of the thing that just got removed. This isn't a problem when the next letter is a consonant, but obviously, is a problem if it's a vowel.

To mitigate this, you should first create a copy of the iterable. Iterate over the original iterable and use the `remove` on the copy. When you're done, rejoin the changed version into a string and return it.

Hope this helps!

yup that helped! thanks!