I'm pretty sure my code is correct but it doesn't seem to returning the expected output:
input anti_vowel("Hey look words") --> outputs: "Hey lk wrds".
Apparently it's not working on the 'e', can anyone explain why?
def anti_vowel(c):
newstr = ""
vowels = ('a', 'e', 'i', 'o', 'u')
for x in c.lower():
if x in vowels:
newstr = c.replace(x, "")
return newstr
The function str.replace(old, new[, max]) doesn't change the c string itself (with respect to c you call). It just returns a new string in which the occurrences of old have been replaced with new. So newstr just contains a string replaced by last vowel in c string that is the o and hence you are getting "Hey lk wrds", which is same as "Hey look words".replace('o', '').
I think you can simply write anti_vowel(c) as:
''.join([l for l in c if l not in vowels]);
What I am doing is iterating over string and if a letter is not a vowel then only include it into list(filters). After filtering I join the list back to a string.
[l for l in c if l not in vowels]?
''.join() call it is faster to use a list comprehension instead of a generator expression.
Why don't you do it with regexp? According to the documentation, something like this should work:
import re
def anti_vowel(s):
result = re.sub(r'[AEIOU]', '', s, flags=re.IGNORECASE)
return result
If you're using the function often, you could compile the regexp and use the compiled version.
Try String.translate.
>>> "Hey look words".translate(None, 'aeiouAEIOU')
'Hy lk wrds'
string.translate(s, table[, deletechars])
Delete all characters from s that are in deletechars (if present), and then translate the characters using table, which must be a 256-character string giving the translation for each character value, indexed by its ordinal. If table is None, then only the character deletion step is performed.
https://docs.python.org/2/library/string.html#string.Template.substitute
Or if you're using the newfangled Python 3:
>>> table = str.maketrans(dict.fromkeys('aeiouAEIOU'))
>>> "Hey look words".translate(table)
'Hy lk wrds'
'test'.translate(str.maketrans('aeiouAEIOU', '-'*10)) produces t-st, but can't replace with None in Python 3
Another option is to forego the vowel variable and put the characters to remove in the loop.
def anti_vowel(text):
for i in "aeiouAEIOU":
text = text.replace(i,"")
return text
print anti_vowel("HappIEAOy")
You should do this:
initialize newstr to c, and then
for x in c.lower():
if x in vowels:
newstr = newstr.replace(x, "")
That's because str.replace(old, new[, max]) returns the a copy of the string after replacing the characters:
The method replace() returns a copy of the string in which the occurrences of old have been replaced with new, optionally restricting the number of replacements to max.
So, this is the correct code:
def anti_vowel(c):
newstr = c
vowels = ('a', 'e', 'i', 'o', 'u')
for x in c.lower():
if x in vowels:
newstr = newstr.replace(x,"")
return newstr
You can also do it in a more pythonic way:
''.join([x for x in c if x not in vowels])
vowels = ('a', 'e', 'i', 'o', 'u', 'A', 'I', 'E', 'O', 'U')
for char in text:
if char in vowels:
text = text.replace(char, '')
return text
One more simpler way can be extracting the non-vowel characters from string and returning them.
def anti_vowel(text):
newstring=""
for i in text:
if i not in "aeiouAEIOU":
newstring=newstring+i
text=newstring
return text
I know there are many correct solutions on this subject but I thought to add few fun ways of solving this problem. If you come from a C++/C# or Java, you will tend to use something like compare then action using the index to remove the unwanted entry in a for loop. Python has the Remove and Del functions. Remove function uses the value and del uses the index.The pythonic solution is in the last function. Lets see how we can do that:
Here we are using the index in a for loop and del function very similar in C++:
def remove_vol(str1):
#list2 = list1 # this won't work bc list1 is the same as list2 meaning same container#
list1 = list(str1)
list2 = list(str1)
for i in range(len(list1)):
if list1[i] in volwes:
vol = list1[i]
x = list2.index(vol)
del list2[x]
print(list2)
Using the remove function:
def remove_vol(str1):
list1 = list(str1)
list2 = list(str1)
for i in list1:
if i in volwes:
list2.remove(i)
print(list2)
Building new string that does not contain the unwanted chars using their indexes:
def remove_vol(str1):
list1 = list(str1)
clean_str = ''
for i in range(len(list1)):
if list1[i] not in volwes:
clean_str += ''.join(list1[i])
print(clean_str)
Same as in the solution in above but using the value:
def remove_vol(str1):
list1 = list(str1)
clean_str = ''
for i in list1:
if i not in volwes:
clean_str += ''.join(i)
print(clean_str)
How you should do it in python? Using list comprehension! It is beautiful:
def remove_vol(list1):
clean_str = ''.join([x for x in list1 if x.lower() not in volwes])
print(clean_str)
def anti_vowel(text):
new=[]
vowels = ("aeiouAEIOU")
for i in text:
if i not in vowels:
new.append(i)
return ''.join(new)
i hope this helps..
def anti_vowel(text):
new_text = ""
for i in text:
if i == 'a' or i == 'A':
pass
elif i == 'e' or i == 'E':
pass
elif i == 'I' or i == 'i':
pass
elif i == 'o' or i == 'O':
pass
elif i == 'u' or i == 'U':
pass
else:
new_text = new_text + i
return new_text
print anti_vowel('Hey look Words!')
My implementation:
# Ask the user for input:
user_input = input("enter a string with some vowels: ")
print("input string: " + str(user_input))
vowels = ('a','e','i','o','u','A','E','I','O','U')
new_string="";
for i in range(0,len(user_input),1):
if user_input[i] in vowels:
print ('found a vowel, removing...')
else:
new_string+=user_input[i]
print("I've removed the vowels for you. You're welcome! The new string is: " + new_string)
A fairly simple approach could be;
def anti_vowel(text):
t = ''
for c in text:
if c in "aeiouAEIOU":
pass
else:
t += c
return t
def anti_vowel(text):
t=""
for c in text:
for i in "ieaouIEAOU":
if c==i:
c=""
else:
c=c
t=t+c
return t
