Why does anti_vowel fail in some cases?
The most common issue is returning too soon. If you’ve indented a return to be inside of a loop, that causes problems because if it’s executed, the function exits immediately and returns whatever value is there.
Be sure to return your final result outside of any loops at the end.
2 posts were split to a new topic: Anti-vowel attempting to use remove
Hello I am having trouble with this assignement, my code looks like this :
def anti_vowel(text) :
result = list(text) # I want to work with a list
for l in result : # For every item of the list
if l in "AEIOUaeiou" : # I check if it is a vowel
result.remove(l) # If it is I remove it
return "".join(result)
print anti_vowel("ecureuil") # returns "crul"
print anti_vowel("urluberlu") # returns "rlbrl"
print anti_vowel("Un Grand bAteAU") # returns "n Grnd btA"
print anti_vowel("Hey lOOk words!") # returns "Hy lOk wrds!"
Could someone tell me what I am doing wrong? Thank you in advance!
PS: Pardon my french
can you see the split i made in this post/topic:
i named the new topic anti-vowel attempting to use remove, hey, what a coincidence, that is what you attempted, did you checkout this topic? That is the whole point of FAQ, to not having to repeat our selves.
Hello 
I think that the line {result.remove(l)} decreases the length of the list each time it gets a vowel. An alternative is to replace the line {for l in result} with {for l in text} since the length of the text does not change.
Hi everyone, I looked at the answer and don’e quite get it…
def anti_vowel(text):
t=""
for c in text:
for i in "ieaouIEAOU":
if c==i:
c="" #How does this step removing specified strings from text? thanks
else:
c=c
t=t+c
return t
Honestly, this solution is just dreadful. I know its the proposed solution, so my dreadful is not personally towards you 
i can highly recommend to run the code through a visualizer to see the steps:
http://www.pythontutor.com/visualize.html#mode=edit
lets try one together:
anti_vowel("me")
then we get to this line (the outer loop):
for c in text:
so for the first iteration, the c variable will have a value of "m".
then we loop through all the possible vowels, "m" won’t equal any vowel, so c variable will assign itself a value of "m" over and over again. Then after the inner loop, append to t variable (will contain result string)
then we get to second iteration of the outer loop, so c variable will get a value of "e"
the "e" will equal a vowel from the vowel string ("ieaouIEAOU"), so c variable will become an empty string. Then c variable will for the remaining iterations of the inner loop assign an empty string to itself. So that after the inner loop/nested loop, we can append (t=t+c) an empty string to t (variable containing the result)
the else clause is just completely redundant:
def anti_vowel(text):
t=""
for c in text:
for i in "ieaouIEAOU":
if c==i:
c=""
t=t+c
return t
print anti_vowel("hello book")
this is already better, we could do one better:
def anti_vowel(text):
t=""
for c in text:
for i in "ieaouIEAOU":
if c==i:
c=""
break
t=t+c
return t
print anti_vowel("hello book")
no need to make more iterations once determined c variable contains a vowel
but honestly, python allows to write such elegant code:
def anti_vowel(text):
result = ""
vowels = "ieaouIEAOU"
for char in text:
if char not in vowels:
result += char
return result
print anti_vowel("hello book")
its almost like reading english, if char not in vowels, you read that, you instantly understand what the line does. By using python simple not in operations + logic variable names, it makes such a difference
i would prefer to use lists for result, but you get the idea that this problem has a much more elegant way to be solved
Got it, thanks stetim94
The solution has actually been changed inside the lesson, it should now give:
def anti_vowel(text):
result = ""
vowels = "ieaouIEAOU"
for char in text:
if char not in vowels:
result += char
return result
Thank you! I made the same mistake so your comment was very helpful
i copied your code to my editor then run
and it works successfully .
We can find the index of deleted vowel and replace it for “”:
def anti_vowel(text):
vowel = list("ieaouIEAOU")
result = list(text)
#text = "Hey look Words!"
for lit in result:
if lit in vowel:
i = result.index(lit)
result.remove(lit)
result.insert(i, "")
return "".join(result)
print anti_vowel("Hey look Words!")
I want it to print the message with vowels after it returns. Does return just stop the entire function?
vowels = "aeiouAEIOU"
extrawords = 'bcd, Hy lk Wrds!'
def anti_vowel(text):
x = ""
vowels = "aeiouAEIOU"
for y in text: #iterate y over text
if y not in vowels: #compare y to vowels str, if not in
x += y #add y to x str
return x
print text
print anti_vowel("Hello World! How yah doing, how yah doing, you know I get no respect.")
It seems like it does, because I was able to do this
vowels = "aeiouAEIOU"
def anti_vowel(text):
x = ""
vowels = "aeiouAEIOU"
extrawords = "bcd Hey look Words! HEY LOOK WORDS!"
for y in text: #iterate y over text
if y not in vowels: #compare y to vowels str, if not in
x += y #add y to x str
print x
return text
print anti_vowel("Hello world")
By default, a function returns None at the end of the function. If we want to return something else at the end of the function, we can use the return keyword.
Given return is the last thing a function does, when a return keyword is reached, the function ends
Hi,
I read most of the posts and have found sound ways to do it there, but for the sake of learning, I’m still curious about why what I had on my first try didn’t work. I did:
def anti_vowel(text):
lst = []
for char in text:
lst.append(char)
for item in lst:
if item in "aeiouAEIOU":
lst.remove(item)
return str("".join(lst))
And it returns: “Hy lk Words!” I don’t really get why it removes the vowels in the first 2 words but not in the rd? Could anyone explain that? Many thanks!
the problem is actually caused by look (any word with successive vowels)
l o o k
0 1 2 3
i have written the word above, with below it the indexes.
lists can’t have empty indexes, so the moment you remove the o from index 1, everything to the right of the removed item moves one slot to the left, lets visualize:
l o k
0 1 2
so the remove action is now done, so the loop continues to the next index, which is 2. Causing the o to be skipped (or any successive vowel for that matter)
Ah that clarifies a lot thanks! Could you then enter something like a “” to put “nothing” on the removed index to continue, or would that be creating an empty index in the list and therefor this approach just doesn’t work for this problem?
There are ways to solve this exercise/challenge using remove, but then you are solving problems your code created in the first place, so you should ask yourself: isn’t there a better solution?
Good point. I’ll fix this in a more efficient way for now and get more into playing with indexes later. Thanks for the help!
First I would check all the vowels in the text and then I would remove all of them at once, I wouldn’t remove them while being inside the loop .
But having already a list of vowels there is a shorter way to do that using List Comprehensions:
def anti_vowel(text) :
new_list = [x for x in text if x not in list("AEIOUaeiou")]
return "".join(new_list)
My script was a bit different
def anti_vowel(text):
vowels = ['a','e','i','o','u','A','E','I','O','U']
for i in text:
if i in vowels:
text = text.replace(i,'')
return text