I'm trying to write a function that gets rid of the vowels in a given string, but it seems to not behave the way it should do...
def anti_vowel(text):
for c in text:
if c in "aeiouAEIOU":
no_vowel = text.replace(c, '')
return no_vowel
print(anti_vowel('Hello World')
so instead of printing
Hll Wrld
It prints
Hell Wrld
Thanks (in advance) for helping
The problem is that no_vowel only has the value of the last time that text.replace(c, '') was executed. Another issue is that no_vowel only gets a value when there is actually a vowel to remove; the code would fail on anti_vowel('vwllss'). Furthermore, you don't have to check whether a character is contained in the text before calling str.replace().
This should work:
def anti_vowel(text):
for vowel in "aeiouAEIOU":
text = text.replace(vowel, '')
return text
print(anti_vowel('Hello World'))
As others indicated, another approach would be to write code in a different way:
def anti_vowel(text):
''.join(c for c in text if c not in 'aeiouAEIOU')
Please do use a generator expression in ''.join() and not a list comprehension; such a list comprehension would allocate memory unnecessarily.
str.join. The argument for join will always be converted to a sequence so a generator expression is slower and as memory efficient as a list comprehension.
str.join: github.com/python/cpython/blob/master/Objects/…, see also my other commentYou can use string.translate() for this. For example:
def anti_vowel(text):
return text.translate(None, "aeiouAEIOU")
print(anti_vowel("hello world"))
With Python 3 the delete argument is gone, but you can still do it by mapping a character to None.
def anti_vowel_py3(text):
return text.translate({ord(i): None for i in "aeiouAEIOU"})
print(anti_vowel_py3("hello world"))
Your code doesnt work because every iteration you assign no_vowel with the text all over again and you iterate the text's letters what you shouldnt because replace already does it. You should write it like that:
def anti_vowel(text):
no_vowel = text
for c in 'aeiouAEIOU':
no_vowel = no_vowel.replace(c, '')
return no_vowel
Or, you could use a list comprehension. More Pythonic and faster to run:
def anti_vowel(text):
return ''.join([c for c in text if c not in 'aeiouAEIOU])
"['H', 'l', 'l', ' ', 'W', 'r', 'l', 'd']". This does: ''.join(c for c in text if c not in 'aeiouAEIOU')
str.join needs a sequence (not a generator) so using a generator only adds needless overhead when it's converted to a list.list(generator) which is definetly slower than doing a list comprehension.In every iteration of the loop, text is "Hello World", and the last vowel of text is "o", so at the end of the loop, no_vowel is "Hell Wrld".
In python2.7, use method translate instead. Here is the official document:
translate(...)
S.translate(table [,deletechars]) -> string Return a copy of the string S, where all characters occurring in the optional argument deletechars are removed, and the remaining characters have been mapped through the given translation table, which must be a string of length 256 or None. If the table argument is None, no translation is applied and the operation simply removes the characters in deletechars.
"Hello World".translate(None, "aeiouAEIOU") gives the correct result "Hll Wrld"
Also, re.sub('[aeiouAEIOU]', "", "Hello World") works for both python2.7 and python3
TypeError: translate() takes exactly one argument (2 given).
deletechars of method translate is gone.re.sub('[aeiouAEIOU]', "", "Hello World") will work for both python2.7 and python3
no_voweldefined?