# Example code for anti_vowel :)

#1

``````#example 1

def anti_vowel(text):
vow = "aeiouAEIOU"
word = ""
for char in text:
if (isvow(char) == False):
word = word + char
return word

def isvow(char):
vowe = "aeiouAEIOU"
for i in vowe:
if (char == i):
return True
return False

#########example 2
def anti_vowel(text):
vow = "aeiouAEIOU"
word = ""
for char in text:
if ((char in vow) == False):
word = word + char
return word``````

#2

# my example:

list1 = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
def anti_vowel(text):
list2 = []
list3 = []
for i in range(len(text)):
list2.append(text[i])
for i in range(len(text)):
for item in list1:
if item == text[i]:
list3.append(i)
count = 0
for i in range(len(list3)):
list2.pop(list3[i]-count)
count += 1
solution = "".join(list2)
return solution
print solution

#3

I thought I'd try something (which is pretty self-explanatory) from my knowledge of SQL: 'in'. I'm pretty sure we haven't touched on it before, but it worked like a charm.
Try simplifying a little without the use of multiple lists.

``````    if x in "abc"
#or
if x not in "abc"``````

SPOILER code below:

``````sentence = raw_input("Enter your thing: ")

def anti_vowel(text):
no_vowels = ""
for i in text:
if i not in "aeiouAEIOU":
no_vowels += i
return no_vowels

print anti_vowel(sentence)``````

#4

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