Reputation: 705
I want to replace every character that isn't "i" in the string "aeiou" with a "!"
I wrote:
def changeWord(word):
for letter in word:
if letter != "i":
word.replace(letter,"!")
return word
This just returns the original. How can I return "!!i!!"?
Upvotes: 9
Views: 40642
Reputation: 1188
Generally in python string are immutable that means it can't be changed after its creation. So we have to convert string into mutable and that can be possible using list as it is mutable.
string="Love"
editable = list(string)
editable[2]='s'
string=''.join(editable)
output:Lose
Upvotes: 0
Reputation: 113905
Nobody seems to give any love to str.translate
:
In [25]: chars = "!"*105 + 'i' + "!"*150
In [26]: 'aeiou'.translate(chars)
Out[26]: '!!i!!'
Hope this helps
Upvotes: 2
Reputation: 235984
Try this, as a one-liner:
def changeWord(word):
return ''.join(c if c == 'i' else '!' for c in word)
The answer can be concisely expressed using a generator, there's no need to use regular expressions or loops in this case.
Upvotes: 3
Reputation: 5874
Regular expressions are pretty powerful for this kind of thing. This replaces any character that isn't an "i" with "!"
import re
str = "aieou"
print re.sub('[^i]', '!', str)
returns:
!!i!!
Upvotes: 8
Reputation: 414
Strings in Python are immutable, so you cannot change them in place. Check out the documentation of str.replace:
Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.
So to make it work, do this:
def changeWord(word):
for letter in word:
if letter != "i":
word = word.replace(letter,"!")
return word
Upvotes: 13
Reputation: 250881
something like this using split()
and join()
:
In [4]: strs="aeiou"
In [5]: "i".join("!"*len(x) for x in strs.split("i"))
Out[5]: '!!i!!'
Upvotes: 3